Question 1

How many integers from 300 to 700 (inclusive) are divisible by 2, 3, and 7?

Accepted Answer

Introduction / Context: This question is about counting numbers in a given interval that satisfy multiple divisibility conditions. Here, each required number must be divisible by 2, 3, and 7 simultaneously. This is best solved using the least common multiple (LCM) and counting multiples of that LCM within the range from 300 to 700. Given Data / Assumptions: - We consider all integers from 300 to 700 inclusive. - We want numbers divisible by 2, 3, and 7. - Divisibility by all three means the number must be a multiple of LCM(2, 3, 7). Concept / Approach: First, compute LCM(2, 3, 7). Any number divisible by 2, 3, and 7 must be a multiple of this LCM. Then, find the smallest and largest multiples of this LCM within the interval [300, 700]. Finally, count how many such multiples exist by using a simple arithmetic progression count. Step-by-Step Solution: Step 1: Find LCM of 2, 3, and 7. 2 = 2, 3 = 3, 7 = 7. LCM(2, 3, 7) = 2 * 3 * 7 = 42. Step 2: Find the smallest multiple of 42 that is at least 300. 42 * 7 = 294, which is less than 300. 42 * 8 = 336, which lies within the range. So the first valid multiple is 336. Step 3: Find the largest multiple of 42 that is at most 700. 42 * 16 = 672, which is less than or equal to 700. 42 * 17 = 714, which exceeds 700. So the last valid multiple is 672. Step 4: Count the multiples from 336 to 672 inclusive. These are 42 * 8, 42 * 9, ..., 42 * 16. The indices run from 8 to 16 inclusive, giving 16 - 8 + 1 = 9 numbers. Verification / Alternative check: List them explicitly to verify: 336, 378, 420, 462, 504, 546, 588, 630, 672. All are divisible by 42, hence by 2, 3, and 7, and all lie between 300 and 700 inclusive. Counting them confirms there are 9 such numbers. Why Other Options Are Wrong: - 7 or 8: These underestimate the count; we clearly find 9 valid multiples. - 10: This would require another multiple of 42 in the range, but 42 * 17 = 714 is outside the interval. Common Pitfalls: A common error is to count numbers divisible by 2, by 3, and by 7 separately and try to combine the counts incorrectly. Another mistake is forgetting that the range includes both endpoints (300 and 700), or miscomputing the first or last multiple inside the range. Using LCM with index counting is the most systematic approach. Final Answer: There are 9 integers between 300 and 700 that are divisible by 2, 3, and 7.

Question 2

The product of three consecutive odd integers is 1287. What is the largest of these three odd integers?

Accepted Answer

Introduction / Context: This question involves consecutive odd integers and their product. We are told that the product of three consecutive odd numbers equals 1287 and asked to determine the largest of the three. It checks pattern recognition, number sense, and sometimes the ability to guess intelligently around cube roots. Given Data / Assumptions: - There are three consecutive odd integers. - Let them be of the form n - 2, n, and n + 2 or n, n + 2, and n + 4. - Their product is 1287. - We must find the largest of these three odd integers. Concept / Approach: Consecutive odd integers are close together, so their product will be near the cube of the middle number. We can use approximate cube roots of 1287 to locate likely candidates and then test small sets of three consecutive odd numbers around that estimate. Direct trial is quite efficient here because the numbers are modest. Step-by-Step Solution: Step 1: Estimate the middle number. The cube root of 1000 is 10 and of 1331 is 11, so the middle number is likely around 11. Step 2: Try 11 as the middle odd integer. Then the three consecutive odd integers would be 9, 11, and 13. Step 3: Compute the product 9 * 11 * 13. 9 * 11 = 99. 99 * 13 = 99 * 10 + 99 * 3 = 990 + 297 = 1287. So 9, 11, and 13 indeed multiply to 1287. Step 4: Identify the largest of these integers. The three integers are 9, 11, and 13, and the largest is 13. Verification / Alternative check: To be sure, we can check other nearby sets. For example, 11, 13, 15 have product 11 * 13 * 15 = 2145, which is too large. Similarly, 7, 9, 11 have product 693, too small. So 9, 11, 13 is the unique suitable triple that produces 1287, confirming that 13 is the largest. Why Other Options Are Wrong: - 9: This is actually the smallest of the three integers. - 11: This is the middle integer, not the largest. - 17: 15, 17, 19 or 13, 15, 17 would produce much larger products than 1287. Common Pitfalls: A typical mistake is assuming the numbers are consecutive integers (like 10, 11, 12) instead of consecutive odd integers. Another error is choosing the wrong triple around the estimate or miscomputing the product. Careful multiplication and a logical search around the estimated cube root avoid these problems. Final Answer: The largest of the three consecutive odd integers is 13.

Question 3

When the number 3521 is divided by 8, what is the remainder?

Accepted Answer

Introduction / Context: This question is about finding the remainder when a large number is divided by 8. It reinforces the idea that to test divisibility by 8 or to find the remainder, you only need to look at the last three digits of the number, because 1000 is divisible by 8. Given Data / Assumptions: - The number given is 3521. - We are dividing this number by 8. - We must find the remainder after division. Concept / Approach: A number is divisible by 8 if the integer formed by its last three digits is divisible by 8. Similarly, the remainder when the whole number is divided by 8 is the same as the remainder when the last three digits are divided by 8. This is because 1000, 2000, etc. are multiples of 8. Step-by-Step Solution: Step 1: Extract the last three digits of 3521. The last three digits are 521. Step 2: Divide 521 by 8 and find the remainder. 8 × 65 = 520. 521 - 520 = 1. So the remainder when 521 is divided by 8 is 1. Step 3: Conclude for the whole number 3521. Since the remainder for the last three digits is 1, the remainder when 3521 is divided by 8 is also 1. Verification / Alternative check: Alternatively, perform the full division roughly: 8 × 400 = 3200. 3521 - 3200 = 321. 8 × 40 = 320. 321 - 320 = 1. We again obtain a remainder of 1, confirming our earlier result based on the last three digits. Why Other Options Are Wrong: - 3, 7, 9: These are simply incorrect remainders; detailed division shows the remainder must be 1. None of them is compatible with the exact calculations above. Common Pitfalls: Some students try to divide the entire number step by step without using the three digit shortcut, which increases the chance of arithmetic mistakes. Others confuse the divisibility rule for 4 (using last two digits) with that for 8 (using last three digits). Remember that powers of 10 and their divisibility by 8 explain why only the last three digits matter. Final Answer: The remainder when 3521 is divided by 8 is 1.

Question 4

For prime numbers greater than 5, how many different digits are possible in the units place?

Accepted Answer

Introduction / Context: This conceptual question asks about the units digits that prime numbers greater than 5 can have. It aims to test understanding of divisibility rules and the basic structure of prime numbers in the decimal system. Given Data / Assumptions: - We consider only prime numbers greater than 5. - We are interested in the digit in the units place (the last digit). - We must count how many different digits can appear as this last digit. Concept / Approach: If a number ends with an even digit (0, 2, 4, 6, 8), it is divisible by 2 and thus cannot be prime unless it is 2 itself. If a number ends with 5 or 0, it is divisible by 5 and cannot be prime unless it is 5 itself. Since we are considering primes greater than 5, their units digits must avoid these divisibility traps. Step-by-Step Solution: Step 1: List all possible units digits from 0 to 9. These digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Step 2: Exclude digits that force divisibility by 2 for numbers greater than 2. Digits 0, 2, 4, 6, and 8 at the end of a number make it even. Any such number greater than 2 is composite, not prime. Step 3: Exclude digits that force divisibility by 5 for numbers greater than 5. Digits 0 and 5 at the end of a number make it divisible by 5. Any such number greater than 5 is composite, not prime. Step 4: For prime numbers greater than 5, the only possible last digits are 1, 3, 7, and 9. So there are 4 possible units digits. Verification / Alternative check: Look at some examples: Primes greater than 5 include 7, 11, 13, 17, 19, 23, 29, 31, 37, and so on. Their units digits are 7, 1, 3, 7, 9, 3, 9, 1, 7, and so on, always from the set {1, 3, 7, 9}. This confirms that no other units digit occurs for such primes. Why Other Options Are Wrong: - 3: Too small; we have identified four possible digits (1, 3, 7, 9). - 5: Too large; there are not five distinct digits that can appear at the end of primes greater than 5. - 6: Also too large; it includes digits that would make the number composite. Common Pitfalls: Some learners mistakenly include 2 and 5 as possible units digits for primes in general without noting the restriction “greater than 5”. Others forget that ending in 0 implies divisibility by 10. Understanding divisibility rules and the exceptions (2 and 5 themselves) is the key to this question. Final Answer: The number of possible digits that can appear in the units place of a prime number greater than 5 is 4.

Question 5

The polynomial x² + a x + b leaves the same remainder when divided by (x − 1) and by (x + 1). What are the values of a and b respectively?

Accepted Answer

Introduction / Context: This algebra question involves remainders of a polynomial when divided by linear factors. It uses the Remainder Theorem, which states that the remainder when a polynomial P(x) is divided by (x − c) is simply P(c). Here, we know that the remainders for two different divisors, (x − 1) and (x + 1), are equal, and we must deduce conditions on the coefficients a and b. Given Data / Assumptions: - The polynomial is P(x) = x² + a x + b. - When P(x) is divided by (x − 1), the remainder is P(1). - When P(x) is divided by (x + 1), the remainder is P(−1). - These two remainders are equal. - We must find a and b that satisfy this condition. Concept / Approach: By the Remainder Theorem, set P(1) equal to P(−1). This gives an equation in a and b. Solve this equation to find the relationship between a and b. If only a is fixed and b remains unrestricted, then b can be any integer. We then match this with the most appropriate option. Step-by-Step Solution: Step 1: Compute P(1). P(1) = 1² + a * 1 + b = 1 + a + b. Step 2: Compute P(−1). P(−1) = (−1)² + a * (−1) + b = 1 − a + b. Step 3: Use the condition that the remainders are the same. P(1) = P(−1). So 1 + a + b = 1 − a + b. Step 4: Simplify this equation. 1 + a + b = 1 − a + b implies a = −a. So 2a = 0 and hence a = 0. Step 5: Observe that b cancels out and is not restricted by this equation. Therefore, b can be any integer. Verification / Alternative check: Take any specific integer value for b, such as b = 3, and with a = 0, form P(x) = x² + 3. Then: P(1) = 1 + 3 = 4. P(−1) = 1 + 3 = 4. The remainders are equal. If we choose a different b, the equality still holds, showing that only a must be zero and b remains arbitrary. Why Other Options Are Wrong: - 4 and 0: This fixes a = 4, but we have shown a must be 0. - 0 and 3: This is just one special case; the problem allows any integer b, not only 3. - 3 and 0: Again this forces a = 3, contradicting a = 0. Common Pitfalls: A common error is to misapply the Remainder Theorem, or to forget that both remainders must be equal, leading to incorrect equations. Some students mistakenly assume that both a and b must be uniquely determined, not recognising that one coefficient can remain free. Always follow the algebra carefully to see whether each parameter is fully determined. Final Answer: The correct condition is that a = 0 and b can be any integer, corresponding to 0 and any integer.

Question 6

Consider all two digit numbers that remain unchanged when their digits are reversed (for example, 11, 22, ..., 99). What is the average of all such two digit numbers?

Accepted Answer

Introduction / Context: This question deals with a special set of two digit numbers: those whose digits are the same and hence remain unchanged when the digits are reversed. Examples include 11, 22, and so on up to 99. We are asked to find the average (arithmetic mean) of all such numbers. Given Data / Assumptions: - We consider all two digit numbers with identical tens and units digits. - These numbers are 11, 22, 33, 44, 55, 66, 77, 88, and 99. - We must compute the average of this set. Concept / Approach: The average of a set of numbers is the sum of the numbers divided by how many numbers there are. For a set of evenly spaced numbers forming an arithmetic progression, the average is also equal to the mean of the first and last numbers. These repeated digit numbers form such an arithmetic progression with common difference 11. Step-by-Step Solution: Step 1: List the numbers. The numbers are 11, 22, 33, 44, 55, 66, 77, 88, and 99. Step 2: Recognise the pattern. This is an arithmetic progression with first term 11 and last term 99, and common difference 11. Step 3: Use the formula for the average of an arithmetic progression. The average of an arithmetic progression = (first term + last term) / 2. So average = (11 + 99) / 2. Compute: 11 + 99 = 110. 110 / 2 = 55. Therefore, the average is 55. Verification / Alternative check: We can also use the sum and count directly: There are 9 numbers: 11, 22, 33, 44, 55, 66, 77, 88, 99. Sum = 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99. Pairing 11 with 99, 22 with 88, 33 with 77, and 44 with 66 gives four pairs of sum 110 each, plus 55 in the middle. Total sum = 4 * 110 + 55 = 440 + 55 = 495. Average = 495 / 9 = 55. This confirms our earlier result. Why Other Options Are Wrong: - 33, 44, 66: These are individual members of the set but not the average of the entire group. The symmetry of the set about 55 ensures that the average must be 55. Common Pitfalls: Some students may miscount the numbers or incorrectly believe the average is the middle term without checking that the sequence is symmetric. Others might attempt to compute the full sum and make arithmetic errors. Using the arithmetic progression average formula is a quick and reliable method here. Final Answer: The average of all such two digit numbers is 55.

Question 7

What is the sum of all prime numbers between 60 and 80?

Accepted Answer

Introduction / Context: This question focuses on identifying prime numbers in a small interval and then summing them. It tests both knowledge of basic primality checks and careful arithmetic when adding the identified primes together. Given Data / Assumptions: - We consider integers strictly between 60 and 80. - We must identify which of these integers are prime numbers. - Then we must add all these prime numbers to find their total sum. Concept / Approach: A prime number is a number greater than 1 with exactly two positive divisors: 1 and itself. To test whether a number in this range is prime, it is sufficient to check divisibility by smaller primes such as 2, 3, 5, 7, and 11. Once we list all primes from 61 to 79, we can add them together. Step-by-Step Solution: Step 1: List integers from 61 to 79. The numbers are 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79. Step 2: Identify primes in this list. - 61: Not divisible by 2, 3, 5, or 7 → prime. - 67: Not divisible by 2, 3, 5, or 7 → prime. - 71: Not divisible by 2, 3, 5, or 7 → prime. - 73: Not divisible by 2, 3, 5, or 7 → prime. - 79: Not divisible by 2, 3, 5, or 7 → prime. Other numbers in the range are composite (for example, 63 is divisible by 3, 65 by 5, 69 by 3, 77 by 7, etc.). Step 3: Sum these primes. Sum = 61 + 67 + 71 + 73 + 79. First, 61 + 67 = 128. 128 + 71 = 199. 199 + 73 = 272. 272 + 79 = 351. So the sum is 351. Verification / Alternative check: Check quickly that no prime is missed: The only candidates in this range that are not obviously even or multiples of 5 are 61, 67, 71, 73, 79. All have been tested for divisibility by small primes, confirming their primality. Hence, the list is complete and the sum 351 is correct. Why Other Options Are Wrong: - 272, 284, 414: These numbers do not equal the sum of the prime numbers between 60 and 80. They likely arise from partial sums or misidentified primes. Common Pitfalls: Some learners may accidentally include composite numbers like 69 or 77, or forget one of the primes such as 79. Others miscalculate the sum. A systematic scan for divisibility and careful addition step by step prevents these errors. Final Answer: The sum of all prime numbers between 60 and 80 is 351.

Question 8

What is the digit in the units place of 7 raised to the power 73, that is 7^73?

Accepted Answer

Introduction / Context: This question asks for the units digit of a very large power, 7^73. Direct multiplication is impossible, so the goal is to use patterns in the units digits of powers of 7. This is a classic modular arithmetic or cyclic pattern problem in number systems. Given Data / Assumptions: - We consider 7^73. - We only need the units digit of this large number. - No full expansion is required or practical. Concept / Approach: The units digits of powers of 7 repeat in a fixed cycle. By computing a few small powers of 7, we can see this repeating pattern and then use the exponent modulo the cycle length to determine the units digit of 7^73. This is a standard cyclicity approach. Step-by-Step Solution: Step 1: Compute the first few powers of 7 and record only their units digits. 7^1 = 7 → units digit = 7. 7^2 = 49 → units digit = 9. 7^3 = 343 → units digit = 3. 7^4 = 2401 → units digit = 1. Step 2: Observe the pattern. The units digits go 7, 9, 3, 1. If we continued, 7^5 = 7^4 * 7 would end with 1 * 7 = 7, so the cycle repeats every 4 steps. Step 3: Use the cycle length to find the position of 7^73 in the pattern. The cycle length is 4, so we find 73 modulo 4. 73 / 4 = 18 remainder 1 (because 4 * 18 = 72 and 73 − 72 = 1). So 73 ≡ 1 (mod 4). Step 4: Match the remainder with the pattern. Remainder 1 corresponds to the first element of the cycle, which has units digit 7. Therefore, 7^73 has units digit 7. Verification / Alternative check: We know that 7^(4k + 1) always has the same units digit as 7^1, which is 7. Since 73 = 4 * 18 + 1, 7^73 must have the same units digit as 7^1, confirming our earlier reasoning. Why Other Options Are Wrong: - 1: This is the units digit for exponents that are multiples of 4 (such as 7^4, 7^8, etc.), not 73. - 3: Appears as the units digit for exponents congruent to 3 modulo 4 (for example, 7^3, 7^7, ...). - 9: Appears for exponents congruent to 2 modulo 4 (such as 7^2, 7^6, ...). Common Pitfalls: A frequent mistake is to miscompute the cycle length or to incorrectly compute 73 modulo 4. Some students stop after finding only a partial pattern or wrongly assume a different cycle length. Carefully listing the first four powers and using remainder arithmetic avoids these errors. Final Answer: The units digit in the expansion of 7^73 is 7.

Question 9

Consider two integers p and q, both greater than 1, which are relatively prime (that is, gcd(p, q) = 1). Examine the following three statements: (1) Both p and q may be prime numbers. (2) Both p and q may be composite numbers. (3) One of p and q may…

Accepted Answer

Introduction / Context: This question explores the concept of relatively prime integers (also called coprime integers). Two integers p and q are relatively prime if their greatest common divisor is 1. The problem asks which combinations of “prime” and “composite” are possible for p and q when they are coprime and both are greater than 1. Given Data / Assumptions: p and q are integers, both greater than 1.p and q are relatively prime, meaning gcd(p, q) = 1.Statement (1): both p and q may be prime numbers.Statement (2): both p and q may be composite numbers.Statement (3): one of p and q may be prime and the other composite. Concept / Approach: To check whether each statement is possible, we do not need a general proof for all p and q; it is enough to exhibit one valid example for each statement that satisfies gcd(p, q) = 1. If we can find such examples, then the statements are correct. We will construct simple concrete pairs for each of the three cases. Step-by-Step Solution: Step 1: For statement (1), consider p = 2 and q = 3. Both are prime and gcd(2, 3) = 1, so statement (1) is possible.Step 2: For statement (2), consider p = 4 and q = 9. Both are composite (4 = 2 * 2 and 9 = 3 * 3), and gcd(4, 9) = 1, so both can be composite and still relatively prime.Step 3: For statement (3), consider p = 2 and q = 9. Here 2 is prime and 9 is composite, and gcd(2, 9) = 1, so one prime and one composite is also possible.Step 4: Because we have valid examples for each of the three statements, all three statements are correct. Verification / Alternative check: The only requirement for being relatively prime is that p and q share no common prime factor. This condition does not depend on whether p or q themselves are prime or composite. As long as their prime factorizations have no overlap, they are coprime. This general reasoning confirms that all three combinations are possible. Why Other Options Are Wrong: Any option that omits one of the statements (such as “1 and 2 only” or “1 and 3 only”) is incorrect because we have explicit examples showing that statements (1), (2) and (3) all occur. “2 and 3 only” wrongly excludes the very common case where both numbers are prime and coprime, such as 2 and 3. Common Pitfalls: A typical misconception is that if two numbers are relatively prime, both must be prime. This is false; many pairs of composite numbers, like 8 and 9 or 25 and 16, are relatively prime. Another misconception is to think that a prime number cannot be relatively prime with a composite number; again, 2 and 9 is a simple counterexample. Final Answer: All three situations are possible, so the correct choice is 1, 2 and 3.

Question 10

Which one of the following is an irrational number (that is, a real number that cannot be expressed as a ratio of two integers)?

Accepted Answer

Introduction / Context: This question checks your understanding of rational and irrational numbers. A rational number can be written as a fraction of two integers, while an irrational number cannot be expressed exactly as such a fraction and has a non terminating, non repeating decimal expansion. Given Data / Assumptions: Option (a): √59059.Option (b): 231/593, a fraction of two integers.Option (c): 0.454545...., a repeating decimal.Option (d): 0.121122111222, a terminating decimal with finitely many digits shown. Concept / Approach: All terminating decimals and repeating decimals represent rational numbers, because they can be converted into fractions. In contrast, the square root of a positive integer that is not a perfect square is known to be irrational. So we must check which option fits that pattern and which are clearly rational based on their form. Step-by-Step Solution: Step 1: Consider option (b), 231/593. It is explicitly written as a ratio of two integers, so it is a rational number by definition.Step 2: Consider option (c), 0.454545.... The notation with repeating 45 shows a recurring decimal. Any recurring decimal can be expressed as a fraction (for example, 0.4545.... can be converted to 45/99), so this is also rational.Step 3: Consider option (d), 0.121122111222. As written, there are a finite number of digits after the decimal point and no ellipsis. Any terminating decimal can be written as an integer divided by a power of 10, so it is rational.Step 4: Consider option (a), √59059. The integer 59059 is not a perfect square (there is no integer whose square equals 59059). The square root of a positive integer that is not a perfect square is always irrational. Verification / Alternative check: You can quickly check whether 59059 is a perfect square by estimating nearby squares. For example, 240^2 = 57600 and 250^2 = 62500, so 59059 lies between these, and no integer between 240 and 250 has a square equal to 59059. Therefore 59059 is not a perfect square, and its square root is indeed irrational. Why Other Options Are Wrong: Option (b) is clearly rational because it is already in fractional form. Option (c) is a classic repeating decimal, which always represents a rational number. Option (d) is a finite decimal, which can always be written as a rational number. None of these can be irrational. Common Pitfalls: Some learners mistakenly assume that any decimal with a complex looking pattern is irrational, even if it terminates. Others may think that any decimal written with dots must be irrational, without checking whether the pattern repeats. Remember: repeating or terminating decimals are always rational. Final Answer: The only irrational number among the options is √59059.

Question 11

How many ordered pairs of digits (A, B) are possible in the number 479865AB so that the entire number is divisible by 9 and, in addition, the last digit B is odd?

Accepted Answer

Introduction / Context: This problem combines the divisibility rule for 9 with conditions on the last digit of a number. We are given a seven digit number 479865AB, where A and B are digits from 0 to 9. The number must be divisible by 9, and the last digit B must be odd. We are asked to count how many ordered pairs (A, B) satisfy these conditions. Given Data / Assumptions: The number has the form 4 7 9 8 6 5 A B.Digits A and B can be any digits from 0 to 9.The number must be divisible by 9.The last digit B must be an odd digit. Concept / Approach: A number is divisible by 9 if and only if the sum of its digits is a multiple of 9. We first compute the sum of the known digits, then include A and B in that sum. We also restrict B to odd digits (1, 3, 5, 7, 9). Working modulo 9, we find all pairs (A, B) that make the total digit sum a multiple of 9, and then count how many such pairs are possible. Step-by-Step Solution: Step 1: Compute the sum of the fixed digits: 4 + 7 + 9 + 8 + 6 + 5 = 39.Step 2: The total sum of digits is 39 + A + B.Step 3: For divisibility by 9, we require 39 + A + B to be a multiple of 9.Step 4: Since 39 is already a multiple of 9 (39 = 9 × 4 + 3, so 39 ≡ 3 (mod 9)), we need A + B ≡ 6 (mod 9) so that 39 + A + B ≡ 3 + (A + B) ≡ 0 (mod 9).Step 5: The digits A and B must satisfy A + B = 6, 15 or 24, because these are the possible two digit sums in 0–9 range that are congruent to 6 modulo 9.Step 6: List all digit pairs (A, B) with A, B between 0 and 9 and B odd, satisfying these sums.Step 7: For A + B = 6 with B odd: possible pairs are (1, 5), (3, 3) and (5, 1).Step 8: For A + B = 15 with B odd: possible pairs are (6, 9), (8, 7). (Pairs like (7, 8) or (9, 6) are invalid because B must be odd and 8 or 6 are even.)Step 9: The sum 24 cannot be obtained with B odd and digits 0–9 simultaneously, because the only combinations involve at least one two digit or invalid digit.Step 10: Collecting valid pairs, we have (1, 5), (3, 3), (5, 1), (6, 9) and (8, 7), which is 5 pairs in total. Verification / Alternative check: We can verify by quickly checking the digit sum for each pair. For example, for (A, B) = (1, 5), the sum is 39 + 1 + 5 = 45, divisible by 9. For (6, 9), the sum is 39 + 6 + 9 = 54, also divisible by 9. All five pairs yield digit sums that are multiples of 9 and have B odd. Why Other Options Are Wrong: An answer of 6 or 9 overcounts and suggests the inclusion of invalid pairs where B is even or outside the 0–9 range. An answer of 11 grossly overcounts possible combinations and does not respect the digit constraints. Only 5 pairs meet all conditions. Common Pitfalls: Common mistakes include forgetting that B must be odd, miscomputing the sum of the fixed digits (39), or mis-applying modulo 9 arithmetic. Another frequent error is not considering all sums congruent to 6 modulo 9, or accidentally including pairs where A or B are not single digits. Final Answer: The number of ordered pairs (A, B) that satisfy all the conditions is 5.

Question 12

Consider the multiplication 999 × abc = def132 in decimal notation, where a, b, c, d, e and f are all single digits and abc and def represent three digit numbers. What are the values of a, b, c, d, e and f respectively?

Accepted Answer

Introduction / Context: This question involves a neat property of multiplying a three digit number by 999. The product 999 × abc is written as def132, where both abc and def are three digit numbers and all letters represent individual digits. We must deduce the digits a, b, c, d, e and f using an algebraic identity and place value reasoning. Given Data / Assumptions: We have 999 × abc = def132.abc and def are three digit numbers made of digits a, b, c and d, e, f respectively.The last three digits of the product are 1, 3 and 2 in that order.All letters represent decimal digits 0 to 9, with a and d non zero. Concept / Approach: Using the identity 999 = 1000 − 1, we can write 999 × abc as 1000 × abc − abc. This means that 999 × abc is equal to the number obtained by writing abc000 and then subtracting abc. For any three digit number abc, this subtraction has a special pattern: the product 999 × abc can be written as the concatenation of (abc − 1) followed by (1000 − abc), each part being three digits. Step-by-Step Solution: Step 1: Let N be the three digit number abc.Step 2: Compute 999 × N as (1000 − 1) × N = 1000N − N.Step 3: In decimal notation, 1000N is N followed by three zeros, and subtracting N gives a six digit number that can be viewed as (N − 1)(1000 − N), where both parts are written with three digits (including leading zeros if necessary).Step 4: In the given product 999 × abc = def132, the last three digits are 132, so 1000 − N must be 132.Step 5: Solve for N: 1000 − N = 132 implies N = 1000 − 132 = 868.Step 6: So abc = 868, meaning a = 8, b = 6 and c = 8.Step 7: The first three digits def must equal N − 1, so def = 868 − 1 = 867.Step 8: Therefore d = 8, e = 6 and f = 7. Verification / Alternative check: Verify directly by multiplication: 999 × 868 = (1000 − 1) × 868 = 868000 − 868. Subtracting, we get 867132. This matches the pattern def132 with def = 867 exactly. The digits are therefore (a, b, c, d, e, f) = (8, 6, 8, 8, 6, 7). Why Other Options Are Wrong: Each wrong option assigns a different order to the digits 6, 7 and 8. Substituting these into the pattern would not satisfy the relation 1000 − abc = 132 or would give a product different from def132. Only the combination 8, 6, 8, 8, 6, 7 simultaneously satisfies the identity and the final three digits requirement. Common Pitfalls: Some learners try to brute force the multiplication with each option instead of using the 1000 − 1 identity, which is more time consuming. Others may misapply the concatenation idea or miscalculate 1000 − N. Carefully applying the identity 999N = (N − 1)(1000 − N) provides a fast and reliable path to the answer. Final Answer: The correct digits are 8, 6, 8, 8, 6, 7 for (a, b, c, d, e, f) respectively.

Question 13

Which of the following statements about prime numbers is not true?

Accepted Answer

Introduction / Context: This question tests your understanding of basic facts and theorems about prime numbers. Several statements are given, and you must identify which one is not always true. The key is to recall known properties of primes and to look for a counterexample to any suspicious generalization. Given Data / Assumptions: Statement (a): The difference of two primes greater than 2 is divisible by 2.Statement (b): If a prime p divides m × n, then p divides m or p divides n.Statement (c): Every number of the form 6n − 1 (n ∈ N) is a prime.Statement (d): There is only one set of three primes with gaps of 2 between consecutive primes. Concept / Approach: We analyze each statement. Statements (a) and (b) are standard facts: (a) follows because all primes greater than 2 are odd, and (b) is a fundamental theorem about primes dividing products. Statement (d) refers to prime triplets. Statement (c) is a strong claim that all numbers of a certain form are prime, which is suspicious because most such simple forms include composite numbers as well. Step-by-Step Solution: Step 1: Check statement (a). Any prime greater than 2 is odd. The difference of two odd numbers is even, and every even number is divisible by 2. So statement (a) is always true.Step 2: Check statement (b). This is a well known property: if p is prime and p divides m × n, then p must divide at least one of m or n. This is true for all integers m and n.Step 3: Check statement (d). The only triple of primes of the form p, p + 2 and p + 4 is 3, 5 and 7. For larger primes, at least one of the three consecutive odd numbers is divisible by 3, so they cannot all be prime. Thus there is indeed only one such prime triplet, and statement (d) is true.Step 4: Check statement (c). It claims that all numbers of the form 6n − 1 are prime. Take n = 6; 6n − 1 = 36 − 1 = 35, which equals 5 × 7 and is composite. Therefore statement (c) is not always true. Verification / Alternative check: Other counterexamples for statement (c) also exist, such as n = 10 giving 6 × 10 − 1 = 59 (prime) but n = 12 giving 71 (prime) and n = 15 giving 89 (prime), while n = 11 gives 65 = 5 × 13 (composite). The form 6n − 1 includes both primes and composites, so the claim “it is a prime number” for all n is false. Why Other Options Are Wrong: Options (a), (b) and (d) describe correct facts about primes and prime gaps. They cannot be the “not true” statement. Only option (c) makes an over-generalized claim that fails when tested with specific values of n. Common Pitfalls: Some learners forget that many primes greater than 3 can be written as 6n ± 1 and incorrectly conclude that every number of this form is prime. This is not correct; the form includes many composite numbers. Always look for counterexamples when a statement claims that “every number of this form is prime.” Final Answer: The statement that is not always true is “If a number is of the form 6n − 1, then it is a prime number.”

Question 14

Twice a certain fraction plus five times its reciprocal adds up to 7. What is the value of this fraction?

Accepted Answer

Introduction / Context: This question is a classic algebraic equation involving a fraction and its reciprocal. You are told that twice the fraction plus five times its reciprocal sums to 7, and you must find the fraction. The problem tests your skill in setting up equations and solving quadratic equations. Given Data / Assumptions: Let the fraction be x, where x is a nonzero rational number.Twice the fraction is 2x.The reciprocal of the fraction is 1/x.Five times its reciprocal is 5/x.The equation given is 2x + 5/x = 7. Concept / Approach: Because the equation involves both x and 1/x, multiplying through by x will clear the denominator and convert the equation into a quadratic equation in x. Quadratic equations of the form ax^2 + bx + c = 0 can be solved using factorization or the quadratic formula. We then check which roots match the given options. Step-by-Step Solution: Step 1: Start from the equation 2x + 5/x = 7.Step 2: Multiply both sides by x (x ≠ 0) to eliminate the denominator: 2x^2 + 5 = 7x.Step 3: Rearrange the equation into standard quadratic form: 2x^2 − 7x + 5 = 0.Step 4: Compute the discriminant: D = b^2 − 4ac = (−7)^2 − 4 × 2 × 5 = 49 − 40 = 9.Step 5: Take the square root of the discriminant: √D = √9 = 3.Step 6: Solve for x using the quadratic formula x = [7 ± 3] / (2 × 2) = (7 ± 3) / 4.Step 7: The two solutions are x = (7 + 3) / 4 = 10 / 4 = 5/2 and x = (7 − 3) / 4 = 4 / 4 = 1.Step 8: Among the given options, only 5/2 appears; 1 is not listed as an option. Verification / Alternative check: Substitute x = 5/2 back into the original equation. Twice the fraction is 2 × 5/2 = 5. The reciprocal is 2/5, so five times the reciprocal is 5 × 2/5 = 2. Adding these gives 5 + 2 = 7, which satisfies the equation. This confirms that 5/2 is indeed a correct solution. Since 1 is not provided in the options, 5/2 is the only valid choice. Why Other Options Are Wrong: For x = 2/5, we get 2 × 2/5 + 5 × 5/2 = 4/5 + 25/2, which is far greater than 7. For x = 5/4 or 4/5, similar substitutions do not yield 7 as the total. Hence, none of these options satisfy the equation. Common Pitfalls: It is easy to make algebraic mistakes when clearing the denominator, such as forgetting to multiply every term by x. Another common error is mishandling the quadratic formula or miscomputing the discriminant. Carefully following each algebraic step and checking solutions in the original equation helps avoid these mistakes. Final Answer: The fraction that satisfies the condition is 5/2.

Question 15

In an arithmetic progression (A.P.), the 3rd term is 19 and the 6th term is 37. What is the value of the 13th term of this A.P.?

Accepted Answer

Introduction / Context: This problem involves an arithmetic progression (A.P.), where each term after the first is obtained by adding a constant common difference. We are given two non consecutive terms and asked to find another later term. This tests your ability to set up equations for terms of an A.P. and solve for the first term and common difference. Given Data / Assumptions: Let the first term of the A.P. be a.Let the common difference be d.The 3rd term (T3) is 19.The 6th term (T6) is 37.We must find the 13th term (T13). Concept / Approach: In an A.P., the n-th term is given by Tn = a + (n − 1)d. Using this formula, we express T3 and T6 in terms of a and d, creating a system of two linear equations. Solving these equations yields the values of a and d, which then allow us to compute T13 directly. Step-by-Step Solution: Step 1: Use the formula Tn = a + (n − 1)d.Step 2: For the 3rd term, T3 = a + 2d = 19.Step 3: For the 6th term, T6 = a + 5d = 37.Step 4: Subtract the equation in Step 2 from the equation in Step 3: (a + 5d) − (a + 2d) = 37 − 19.Step 5: This simplifies to 3d = 18, so d = 6.Step 6: Substitute d = 6 into a + 2d = 19: a + 2 × 6 = 19, so a + 12 = 19 and a = 7.Step 7: Now find the 13th term: T13 = a + 12d = 7 + 12 × 6.Step 8: Compute 12 × 6 = 72, so T13 = 7 + 72 = 79. Verification / Alternative check: We can list some terms to check the pattern. The sequence begins as 7, 13, 19, 25, 31, 37, ... The 3rd term is 19 and the 6th term is 37, which matches the given data. Continuing with difference 6, the 7th term is 43, 8th is 49, and so on, up to the 13th term, which will indeed be 79. Why Other Options Are Wrong: Values like 43, 45 or 49 occur earlier in the sequence (43 as T7, 49 as T8). They are not the 13th term. Any answer smaller than 79 ignores the common difference or misapplies the term formula. Common Pitfalls: Some learners mistakenly plug n directly into Tn = a + nd instead of a + (n − 1)d. Others miscalculate the difference when subtracting equations, leading to an incorrect value for d. Writing the general term formula correctly and checking at least one intermediate term helps confirm the pattern. Final Answer: The 13th term of the arithmetic progression is 79.

Question 16

Convert the binary number 10110 (base 2) into its equivalent decimal (base 10) number.

Accepted Answer

Introduction / Context: This question is about number system conversion from binary to decimal. Binary numbers use base 2 with digits 0 and 1. Each position represents a power of 2. We must convert the binary number 10110 into its base 10 (decimal) equivalent by expanding it in powers of 2. Given Data / Assumptions: The given binary number is 10110₂.We must express it as a decimal number.Binary positions from right to left represent 2^0, 2^1, 2^2, 2^3 and 2^4 respectively. Concept / Approach: To convert a binary number to decimal, multiply each binary digit by 2 raised to the power of its position index (starting from 0 at the rightmost digit) and then add all these products. Only positions with a 1 contribute to the sum; positions with a 0 contribute nothing. Step-by-Step Solution: Step 1: Write 10110 as digits with place values: 1 0 1 1 0.Step 2: Label the positions from right to left: the rightmost 0 is at 2^0, the next 1 is 2^1, then 1 at 2^2, 0 at 2^3 and 1 at 2^4.Step 3: Compute the contribution of each digit: 1 × 2^4 = 1 × 16 = 16; 0 × 2^3 = 0; 1 × 2^2 = 1 × 4 = 4; 1 × 2^1 = 1 × 2 = 2; 0 × 2^0 = 0.Step 4: Add these contributions: 16 + 0 + 4 + 2 + 0 = 22.Step 5: Therefore, 10110₂ equals 22₁₀ in decimal. Verification / Alternative check: We can also do the conversion using repeated doubling: starting from the left, process digits one by one. Begin with 1. Multiply by 2 and add the next digit: 1 × 2 + 0 = 2; then 2 × 2 + 1 = 5; 5 × 2 + 1 = 11; 11 × 2 + 0 = 22. This verifies that the decimal equivalent is 22. Why Other Options Are Wrong: The value 20 would correspond to a different binary number (e.g., 10100₂). The values 18 and 16 likewise correspond to other binary patterns. Only 22 matches the positional value calculation from 10110₂. Common Pitfalls: Common mistakes include assigning the wrong powers of 2 to each digit, especially mixing up the order, or forgetting to include one of the 1-digits in the sum. Carefully labeling each position with its correct power of 2 avoids these errors. Final Answer: The binary number 10110 corresponds to the decimal number 22.

Question 17

What is the reflection of the point (6, −3) in the horizontal line y = 2 in the Cartesian plane?

Accepted Answer

Introduction / Context: This coordinate geometry question asks for the reflection of a point across a horizontal line. Reflecting a point across a line involves finding a point on the opposite side of the line at the same perpendicular distance. Here, we reflect the point (6, −3) across the line y = 2. Given Data / Assumptions: Original point P has coordinates (6, −3).Line of reflection is the horizontal line y = 2.We seek the coordinates of the reflected point P'. Concept / Approach: For reflection across a horizontal line y = k, the x coordinate remains unchanged and only the y coordinate is adjusted. The distance from the original point to the line is measured vertically, and the reflected point lies the same distance on the other side of the line. If the original y coordinate is y₁, the reflected y coordinate y₂ satisfies y₂ − k = −(y₁ − k). Step-by-Step Solution: Step 1: Identify the distance from point P to the line y = 2.Step 2: The original y coordinate is −3. The line y = 2 has y = 2.Step 3: Compute the vertical distance: 2 − (−3) = 2 + 3 = 5.Step 4: The reflected point must lie 5 units above the line if the original point is 5 units below it.Step 5: Therefore, the new y coordinate is 2 + 5 = 7.Step 6: Reflection across a horizontal line does not change the x coordinate, so x remains 6.Step 7: The reflected point has coordinates (6, 7). Verification / Alternative check: We can also use the formula for reflection across y = k: if the original y coordinate is y₁, then the reflected y coordinate is y₂ = 2k − y₁. Here k = 2 and y₁ = −3, so y₂ = 2 × 2 − (−3) = 4 + 3 = 7. This matches the coordinate found earlier. The x coordinate remains 6, so the reflected point is (6, 7). Why Other Options Are Wrong: Options with x = −2 or x = −6 incorrectly change the x coordinate, which should remain constant when reflecting across a horizontal line. The point (−2, 3) is not the correct symmetric point about y = 2 and also changes x. Only (6, 7) maintains the correct horizontal position and symmetric vertical distance. Common Pitfalls: A common mistake is to reflect about the x axis instead of y = 2, or to change both coordinates instead of only the y coordinate. Another error is to subtract the distance twice, leading to an incorrect y value. Always calculate the vertical distance from the point to the line and then move the same distance on the opposite side. Final Answer: The reflection of (6, −3) in the line y = 2 is the point (6, 7).

Question 18

What is the smallest number that must be subtracted from 518 so that the resulting number is exactly divisible by 13?

Accepted Answer

Introduction / Context: This question is about divisibility and remainders. We are asked to find the least number that must be subtracted from 518 so that the result is divisible by 13. This tests understanding of how remainders behave when adjusting a number by subtraction. Given Data / Assumptions: Original number is 518.We seek an integer k such that 518 − k is divisible by 13.Among the options, we must choose the smallest such k. Concept / Approach: When a number N is divided by a divisor d, we can write N = dq + r, where r is the remainder. To get the nearest smaller multiple of d, we must subtract the remainder r from N. Therefore, the smallest non negative number that must be subtracted is exactly the remainder when N is divided by d. Step-by-Step Solution: Step 1: Divide 518 by 13 to find the remainder.Step 2: Compute 13 × 39 = 507.Step 3: Subtract 507 from 518: 518 − 507 = 11.Step 4: This shows that 518 = 13 × 39 + 11, so the remainder when 518 is divided by 13 is 11.Step 5: To get the largest multiple of 13 that is less than or equal to 518, subtract the remainder from 518.Step 6: So the required multiple is 518 − 11 = 507, which is already known to be divisible by 13.Step 7: Therefore, the least number that must be subtracted is 11. Verification / Alternative check: Check divisibility: 507 ÷ 13 = 39 exactly, with no remainder. If we were to subtract a smaller number, say 10, we would get 508, and 508 ÷ 13 is not an integer. Similarly, subtracting 9 gives 509, which is also not divisible by 13. Therefore, 11 is the smallest subtraction that makes the number divisible by 13. Why Other Options Are Wrong: Subtracting 10 yields 508, and 13 × 39 = 507 and 13 × 40 = 520, so 508 is not a multiple of 13. Subtracting 9 or 12 gives results with remainders 2 or 10 when divided by 13. These are not exact multiples. Only subtracting 11 eliminates the remainder completely. Common Pitfalls: Some learners mistakenly divide in the wrong direction or try random subtractions from the options. Others may miscalculate the product 13 × 39 or forget that the remainder itself is the quantity that must be removed. Always compute the remainder first, then subtract exactly that amount to reach the nearest smaller multiple. Final Answer: The smallest number that must be subtracted is 11.

Question 19

In an arithmetic progression, the sum of the first 13 terms is required. The 5th term of the A.P. is 1 and the 8th term is −17. What is the sum of the first 13 terms?

Accepted Answer

Introduction / Context: This problem deals with an arithmetic progression (A.P.) where two specific terms are known and we must find the sum of the first 13 terms. It combines the formula for the n-th term of an A.P. with the formula for the sum of the first n terms, requiring you to solve for both the first term and common difference. Given Data / Assumptions: Let the first term be a and common difference be d.The 5th term T5 is 1.The 8th term T8 is −17.We must find the sum S13 of the first 13 terms. Concept / Approach: The n-th term of an A.P. is Tn = a + (n − 1)d. The sum Sn of the first n terms is Sn = n * (first term + last term) / 2, or Sn = n/2 * [2a + (n − 1)d]. First, we use T5 and T8 to solve for a and d. Then we find the 13th term T13 and finally use the sum formula. Step-by-Step Solution: Step 1: Write T5 and T8 using Tn = a + (n − 1)d.Step 2: T5 = a + 4d = 1.Step 3: T8 = a + 7d = −17.Step 4: Subtract the equation for T5 from the equation for T8: (a + 7d) − (a + 4d) = −17 − 1.Step 5: This gives 3d = −18, so d = −6.Step 6: Substitute d = −6 into a + 4d = 1: a + 4(−6) = 1, so a − 24 = 1 and a = 25.Step 7: Now find the 13th term T13 = a + 12d = 25 + 12(−6) = 25 − 72 = −47.Step 8: Use the sum formula S13 = 13 * (first term + last term) / 2.Step 9: S13 = 13 * (25 + (−47)) / 2 = 13 * (−22) / 2 = 13 * (−11) = −143. Verification / Alternative check: We can check some terms to ensure the pattern is correct. With a = 25 and d = −6, the sequence begins: 25, 19, 13, 7, 1, −5, −11, −17, ... T5 is 1 and T8 is −17, matching the given values. Continuing this pattern down to T13 and summing the first 13 terms using the formula confirms S13 = −143. Why Other Options Are Wrong: Options like −140 or 61 result from miscalculations, such as using the wrong number of terms or incorrect values of a or d. A positive sum like 166 contradicts the fact that many later terms in the progression are negative and of larger magnitude, so the overall sum must be negative. Common Pitfalls: Students often forget that Tn = a + (n − 1)d, not a + nd. Another frequent error is to mis-handle the negative common difference or to use the wrong formula for the sum. Careful algebra and explicit substitution into the formulas help avoid these issues. Final Answer: The sum of the first 13 terms of the A.P. is −143.

Question 20

Which of the following numbers is not a prime number?

Accepted Answer

Introduction / Context: This question asks you to identify which of the given numbers is not prime. A prime number has exactly two distinct positive divisors: 1 and the number itself. A composite number has additional divisors. We must test each option and find the one that has at least one non trivial factor. Given Data / Assumptions: The numbers to test are 731, 227, 347 and 461.We need to decide which is composite (not prime).We can use divisibility tests and trial division up to the square root of each number. Concept / Approach: To test whether a number N is prime, it is enough to check divisibility by all prime numbers less than or equal to √N. If none divide N, then N is prime. We will test each candidate, focusing on obvious factors first. If we find a factorization, the number is composite. Step-by-Step Solution: Step 1: Test 731. Check divisibility by small primes.Step 2: 731 is not even, and the sum of its digits is 7 + 3 + 1 = 11, not divisible by 3, so 2 and 3 do not divide it. It does not end with 5, so 5 does not divide it.Step 3: Check divisibility by 7: 7 × 100 = 700; 7 × 104 = 728, remainder 3, so 7 is not a factor.Step 4: Check 11: 11 × 60 = 660; 11 × 66 = 726, remainder 5, so 11 is not a factor.Step 5: Check 17: 17 × 40 = 680 and 17 × 3 = 51; 680 + 51 = 731, so 17 × 43 = 731. Thus 731 = 17 × 43 and is composite.Step 6: Now test 227. Try dividing by primes up to about 15 (since √227 is a little less than 16). It is not divisible by 2, 3 (digit sum 2 + 2 + 7 = 11), 5, 7, 11 or 13. Hence 227 is prime.Step 7: Test 347. Its square root is less than 19. It is not divisible by 2, 3 (digit sum 3 + 4 + 7 = 14), 5, 7, 11, 13 or 17. So 347 is prime.Step 8: Test 461. Its square root is a little more than 21. It is not divisible by 2, 3 (digit sum 4 + 6 + 1 = 11), 5, 7, 11, 13, 17 or 19. Thus 461 is prime. Verification / Alternative check: Once we find that 731 = 17 × 43, we have a clear composite factorization. For the other numbers, systematic checking up to the square root shows no factors. This confirms that 227, 347 and 461 each have no non trivial divisors and are prime. Why Other Options Are Wrong: Options (b), (c) and (d) are primes, not composites. They do not have any divisors other than 1 and themselves. Therefore they are not correct answers to the question asking for a non prime number. Common Pitfalls: Students sometimes stop testing too early or skip some primes, accidentally labelling a prime as composite or vice versa. Another mistake is to rely solely on divisibility by 2, 3 or 5 without checking higher primes. Using a systematic trial division up to the square root ensures accuracy. Final Answer: The number that is not prime is 731, since 731 = 17 × 43.

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