Q: The least number more than 5000 which is divisible by 73 is

Concept & Strategy: Finding a number "more than" a base value that is perfectly divisible by a divisor is mathematically identical to finding the least number to *add* to that base value. We find the remainder of the base value, then calculate the difference needed to step up to the next perfect multiple. Required = Base + (Divisor - Remainder) Step-by-Step Solution: • Step 1: Set up long division for 5000 \div 73. • Step 2: Execute the division steps. 500 \div 73 \approx 6. (73 × 6 = 438) Remainder = 500 - 438 = 62. Bring down the 0 to make 620. 620 \div 73 \approx 8. (73 × 8 = 584) Remainder = 620 - 584 = 36. • Step 3: Calculate the amount required to reach the next multiple. Amount to add = Divisor - Remainder Amount to add = 73 - 36 = 37. • Step 4: Add this amount to the original base. 5000 + 37 = 5037. Exam Strategy & Shortcut: Approximation via Multiplication: Rather than performing long division, use base-10 estimations. We need a multiple of 73 near 5000. 73 × 100 = 7300 (Too high) 73 × 50 = 3650 (Too low) Let's try 73 × 70 = 5110. 5110 is very close to 5000. Now, step backward by subtracting 73 to find the first multiple just above 5000. 5110 - 73 = 5037. This matches option (b) directly. This method is often faster and less prone to subtraction errors than standard long division. Common Pitfall: Students often subtract the remainder directly from the base number (5000 - 36 = 4964). While 4964 is perfectly divisible by 73, the question explicitly asks for a number *more than* 5000. Subtracting drops you into the range below the target base. Final Answer: Therefore, the correct answer is 5037.

Q: The nearest integer to 58701 which is exactly divisible by 567 is

Concept & Logic: "Nearest integer" problems require evaluating two candidate multiples: the one immediately below the base number, and the one immediately above it. After dividing the base by the divisor to find the remainder R, we compare the distances: • Distance to lower multiple = R • Distance to upper multiple = Divisor - R The required answer corresponds to whichever distance is numerically smaller. Step-by-Step Solution: • Step 1: Divide the base number by the divisor to find the remainder. 58701 \div 567 • Step 2: Execute long division. 587 \div 567 = 1. Remainder is 20. Bring down 0 to make 200. 200 \div 567 = 0. Remainder is 200. Bring down 1 to make 2001. 2001 \div 567 = 3. (567 × 3 = 1701). Remainder = 2001 - 1701 = 300. • Step 3: Calculate the two distances. Distance down = 300 Distance up = 567 - 300 = 267 • Step 4: Compare the distances to find the nearest integer. Since 267 < 300, the upper multiple is strictly closer to the base number. We find it by adding the upward distance: 58701 + 267 = 58968. Exam Strategy & Shortcut: Block Approximation: Look at the first three digits of the dividend and divisor. We know 567 × 100 = 56700. Let's subtract this large chunk from our target. 58701 - 56700 = 2001. Now we only need to find the nearest multiple of 567 to 2001. 567 × 3 = 1701 (Gap of 300) 567 × 4 = 2268 (Gap of 267) The multiple 2268 is closer to 2001. So, 56700 + 2268 = 58968 is the nearest integer. This avoids formal long division entirely. Common Pitfall: A major trap is always subtracting the remainder by default, which yields the nearest *lower* integer (58701 - 300 = 58401). The phrase "nearest integer" is bidirectional. You must always check whether rounding up or rounding down provides a shorter absolute distance. Final Answer: Therefore, the correct answer is 58968.

Q: The smallest 6-digit number exactly divisible by 111 is

Concept & Formula: To find the smallest n-digit number perfectly divisible by a divisor D, start with the smallest possible n-digit base number. Find the remainder upon division, and add the difference between the divisor and the remainder to bridge the gap to the next multiple. Required Number = Base + (Divisor - Remainder) Step-by-Step Solution: • Step 1: Identify the absolute smallest 6-digit number. The base number is 100000. • Step 2: Execute long division to find the remainder. 100000 \div 111 We know 111 × 9 = 999. So, 111 × 900 = 99900. Subtracting this from the base: 100000 - 99900 = 100. The remainder is 100. • Step 3: Calculate the amount required to reach the next perfect multiple. Number to add = Divisor - Remainder Number to add = 111 - 100 = 11. • Step 4: Add this value to the original base number. 100000 + 11 = 100011. Exam Strategy & Shortcut: Option Elimination: Look at the options and apply the divisibility rule for 3 (since 111 is divisible by 3, the answer must also be). (a) 111111 \rightarrow Sum is 6 (Divisible, but is it the smallest?) (b) 110011 \rightarrow Sum is 4 (Not divisible) (c) 100011 \rightarrow Sum is 3 (Divisible) (d) 110101 \rightarrow Sum is 4 (Not divisible) Comparing the valid candidates, 100011 is significantly smaller than 111111. A quick mathematical check (100011 = 99900 + 111) confirms it is built from exact multiples. Common Pitfall: Students often incorrectly assume a number like 111111 is the smallest simply because it repeats the digits of the divisor and feels conceptually "clean". Never assume visual patterns bypass mathematical rules; always start your logic from the fundamental numerical base 100000. Final Answer: Therefore, the correct answer is 100011.

Q: The sum of all 2-digit numbers divisible by 5 is

Concept & Formula: A sequence of numbers with a constant difference forms an Arithmetic Progression (A.P.). To find the sum of an A.P., we use the formula: S = (n/2)(a + l) Where n is the total number of terms, a is the first term, and l is the last term. Step-by-Step Solution: • Step 1: Identify the A.P. series. The 2-digit numbers perfectly divisible by 5 are: 10, 15, 20, \dots, 95. Here, the first term a = 10, the last term l = 95, and the common difference d = 5. • Step 2: Calculate the total number of terms (n). Use the general n-th term formula: l = a + (n-1)d 95 = 10 + (n-1)5 85 = 5(n-1) 17 = n - 1 \Rightarrow n = 18. There are 18 valid terms in total. • Step 3: Calculate the sum of the series. S = (18/2) × (10 + 95) S = 9 × 105 S = 945. Exam Strategy & Shortcut: Average Rule: The sum of any evenly spaced sequence is simply the average of the first and last terms multiplied by the total number of terms. Average = (10 + 95/2) = 52.5. The number of terms can be found mentally: (95 - 10) \div 5 + 1 = 18. Sum = 52.5 × 18 = 945. This avoids writing out formal A.P. notation entirely. Common Pitfall: The most common mistake is miscounting the number of terms in the set. Many students instinctively divide 100 by 5 to get 20 terms, forgetting that 5 (a 1-digit number) and 100 (a 3-digit number) must be explicitly excluded from the 2-digit constraint requested by the prompt. Final Answer: Therefore, the correct answer is 945.

Question 1

If $p$ is a prime number greater than 3, then $(p^2 - 1)$ is always divisible by

Accepted Answer

Concept & Logic: Any prime number greater than 3 can be expressed in the form 6k \pm 1. Alternatively, you can analyze the algebraic expansion of the difference of squares: a^2 - b^2 = (a-b)(a+b). Step-by-Step Solution: • Expand the given expression: p^2 - 1 = (p - 1)(p + 1) • Since p is a prime number greater than 3, p must be an odd number. • Therefore, both (p - 1) and (p + 1) are consecutive even numbers. • In any two consecutive even numbers, one is a multiple of 2 and the other is a multiple of 4. Their product is therefore divisible by 2 × 4 = 8. • Additionally, consider the three consecutive integers: (p - 1), p, and (p + 1). Exactly one of them must be a multiple of 3. • Since p is a prime > 3, p cannot be the multiple of 3. Thus, either (p - 1) or (p + 1) must be divisible by 3. • Combining these factors, the product (p - 1)(p + 1) is perfectly divisible by both 8 and 3, meaning it is divisible by 8 × 3 = 24. Exam Strategy & Shortcut: Use value substitution for instant results. Take the first prime number greater than 3, which is p = 5. Calculate 5^2 - 1 = 25 - 1 = 24. Check the next prime, p = 7. Calculate 7^2 - 1 = 49 - 1 = 48. Both results are divisible by 24, firmly pointing to option (c). Common Pitfall: Students often pick p = 2 or p = 3 without reading the constraint "> 3". If p = 3, 3^2 - 1 = 8, which is not divisible by 24, leading test-takers to incorrectly choose "None of these". Always honor the constraints in the prompt. Final Answer: Therefore, the correct answer is 24.

Question 2

$(x^n - a^n)$ is divisible by $(x - a)$

Accepted Answer

Concept & Formula: According to the Remainder Theorem (and Factor Theorem), a polynomial P(x) is divisible by (x - a) if and only if P(a) = 0. Step-by-Step Solution: • Let our polynomial be defined as P(x) = x^n - a^n. • We want to test divisibility by (x - a). To do this, we find the root of the divisor by setting it to zero: x - a = 0 \Rightarrow x = a. • Substitute x = a into our polynomial P(x): P(a) = (a)^n - a^n P(a) = a^n - a^n = 0 • Because P(a) evaluates to exactly 0, there is no remainder. • Notice that this subtraction equals 0 regardless of whether n is odd, even, or prime. As long as n is a natural number, the terms cancel out perfectly. Exam Strategy & Shortcut: Substitute small integer values for n to quickly verify the rule. • If n = 1: (x^1 - a^1) = (x - a), which is obviously divisible by (x - a). • If n = 2: (x^2 - a^2) = (x - a)(x + a), which contains (x - a) as a factor. • If n = 3: (x^3 - a^3) = (x - a)(x^2 + ax + a^2), which contains (x - a) as a factor. Since it works for both odd (1, 3) and even (2) numbers, it must be true for all values of n. Common Pitfall: A frequent error is confusing the rules for (x^n - a^n) with (x^n + a^n). Remember that (x^n + a^n) is only divisible by (x + a) when n is *odd*. However, with a negative sign in the middle, (x^n - a^n) is unconditionally divisible by (x - a). Final Answer: Therefore, the correct answer is for all values of n.

Question 3

If the square of an odd natural number is divided by 8, then the remainder will be

Accepted Answer

Concept & Logic: Any odd natural number can be algebraically represented as (2k + 1). When squared, its expanded form contains the product of two consecutive integers, forcing a specific divisibility property. Step-by-Step Solution: • Let the odd natural number be x = 2k + 1, where k is an integer (k \ge 0). • Square the number: (2k + 1)^2 = 4k^2 + 4k + 1 • Factor out 4k from the first two terms: 4k(k + 1) + 1 • Observe the term k(k + 1). This is the product of two consecutive integers. In any pair of consecutive integers, one must be even. Therefore, k(k + 1) is always a multiple of 2 (i.e., 2m). • Substitute 2m into the expression: 4(2m) + 1 = 8m + 1 • This proves that dividing the square of an odd number by 8 will always exactly leave a remainder of 1. Exam Strategy & Shortcut: Do not waste time on algebra if you can use pure arithmetic testing. Take the first few odd natural numbers, square them, and divide by 8: • 3^2 = 9 \Rightarrow 9 \div 8 leaves remainder 1. • 5^2 = 25 \Rightarrow 25 \div 8 leaves remainder 1. • 7^2 = 49 \Rightarrow 49 \div 8 leaves remainder 1. The pattern is instantly clear and verifiable. Common Pitfall: Students often test only the number 1 (1^2 = 1). While 1 \div 8 mathematically leaves a remainder of 1, it can be confusing in the pressure of an exam. Testing 3 and 5 provides unambiguous confirmation. Final Answer: Therefore, the correct answer is 1.

Question 4

The largest number that exactly divides each number of the sequence $1^5 - 1, 2^5 - 2, 3^5 - 3, \dots, n^5 - n, \dots$ is

Accepted Answer

Concept & Formula: To find the largest number that divides every term in a sequence generated by a formula like n^5 - n, you must find the Greatest Common Divisor (GCD) of the first few valid non-zero terms. Step-by-Step Solution: • The sequence is defined by the term T_n = n^5 - n. Let us calculate the first few terms by substituting n = 1, 2, 3, \dots • For n = 1: 1^5 - 1 = 0 *(0 is divisible by everything, so move to the next term).* • For n = 2: 2^5 - 2 = 32 - 2 = 30 • For n = 3: 3^5 - 3 = 243 - 3 = 240 • For n = 4: 4^5 - 4 = 1024 - 4 = 1020 • We now need the largest number that divides 30, 240, and 1020. • Notice that the first non-zero term is 30. The greatest divisor of a sequence cannot be larger than its smallest non-zero term. • Let us verify if 30 divides the next terms: 240 \div 30 = 8 and 1020 \div 30 = 34. It works perfectly. Exam Strategy & Shortcut: The largest common divisor for a sequence of this type is universally dictated by its first non-zero result. Simply calculate n = 2 \Rightarrow 2^5 - 2 = 30. The answer cannot exceed 30. Look at the options: 1, 15, 30, 120. The largest option that is \le 30 and divides 30 is 30 itself. Common Pitfall: A frequent mistake is calculating the GCD algebraically by factoring n^5 - n into n(n-1)(n+1)(n^2+1), which proves divisibility by 6 (since n(n-1)(n+1) is divisible by 3! = 6), but missing the hidden factor of 5 that makes the full divisor 30. Value substitution avoids this blind spot entirely. Final Answer: Therefore, the correct answer is 30.

Question 5

The smallest 4-digit number exactly divisible by 7 is

Accepted Answer

Concept & Strategy: To find the smallest n-digit number exactly divisible by a given divisor d, we start with the absolute smallest n-digit number base. We then find the remainder when this base is divided by d. The number we need to *add* to the base to reach the next perfect multiple is (d - remainder). Formula: Required Number = Base + (Divisor - Remainder) Step-by-Step Solution: • Step 1: Identify the smallest 4-digit number. The smallest 4-digit number is 1000. • Step 2: Divide this base number by the given divisor (7) to find the remainder. 1000 \div 7 7 × 142 = 994 Remainder = 1000 - 994 = 6 • Step 3: Calculate how much needs to be added to make it perfectly divisible. Number to add = Divisor - Remainder Number to add = 7 - 6 = 1 • Step 4: Add this value to the original base number. 1000 + 1 = 1001 Exam Strategy & Shortcut: Option Elimination via Direct Division: Since the options are close to 1000, start dividing the smallest options directly by 7. Option (a) is 1001. Let's apply the divisibility rule for 7, or simply divide: 100 \div 7 = 14 remainder 2. Bring down 1 to make 21. 21 \div 7 = 3. It divides perfectly. Since it's the smallest option, it must be the answer. No complex formulas required! Common Pitfall: A very common mistake is simply subtracting the remainder (6) from the base number (1000). While 1000 - 6 = 994 is indeed divisible by 7, it is a 3-digit number. The question specifically asks for a 4-digit number, meaning we must *add* to reach the next multiple. Final Answer: Therefore, the correct answer is 1001.

Question 6

Find the least 6-digit number which is exactly divisible by 349.

Accepted Answer

Concept & Logic: Similar to finding a 4-digit number, to find the least 6-digit number divisible by a divisor D, we start with the smallest theoretical 6-digit base number, which is 100000. We divide this base by the divisor to find the remainder. Because subtracting the remainder would drop us down to a 5-digit number, we must *add* the difference between the divisor and the remainder to our base. Required = 100000 + (Divisor - Remainder) Step-by-Step Solution: • Step 1: Set up the long division of the smallest 6-digit number by 349. 100000 \div 349 • Step 2: Execute the division. 1000 \div 349 \approx 2. (349 × 2 = 698) Remainder = 1000 - 698 = 302 Bring down a 0 to make 3020. 3020 \div 349 \approx 8. (349 × 8 = 2792) Remainder = 3020 - 2792 = 228 Bring down the final 0 to make 2280. 2280 \div 349 \approx 6. (349 × 6 = 2094) Remainder = 2280 - 2094 = 186 • Step 3: Calculate the amount required to reach the next perfect multiple. Number to add = Divisor - Remainder Number to add = 349 - 186 = 163 • Step 4: Add this to the base number. 100000 + 163 = 100163 Exam Strategy & Shortcut: Unit Digit Check: Look closely at the options. We are looking for a multiple of 349 that is just above 100000. Our quotient was roughly 286 (giving a remainder). The next multiple must be 349 × 287. Look at the unit digits: 9 × 7 = 63. Therefore, the correct answer MUST end in a 3. Options (a), (b), and (c) all end in 63. Let's look at the magnitude. 349 × 1000 \approx 349000. 160000 (Option c) is way too large. 101000 (Option b) is also likely too large since 349 is a relatively small jump from 100000. 100163 (Option a) represents a jump of only 163, which is less than the divisor 349, perfectly matching the logic of finding the *least* number. Common Pitfall: Long division with 3-digit divisors is prone to arithmetic errors. A simple mistake in subtraction (e.g., 3020 - 2792) cascades and guarantees an incorrect remainder, subsequently ruining the final addition. Take an extra 10 seconds to verify your subtraction steps during the long division. Final Answer: Therefore, the correct answer is 100163.

Question 7

The least number, which must be added to the greatest 6-digit number so that the sum may be exactly divisible by 327 is

Accepted Answer

Concept & Logic: When a question asks for the least number to be *added* to a base number to make it exactly divisible, we must determine how far the current remainder is from reaching the next full multiple of the divisor. We must bridge the gap between the remainder and the divisor itself. Number to add = Divisor - Remainder Step-by-Step Solution: • Step 1: Identify the greatest 6-digit number, which is 999999. • Step 2: Set up and execute long division by 327. 999 \div 327 = 3 (327 × 3 = 981). Remainder is 18. Bring down 9 to make 189. Since 189 < 327, the quotient is 0. Bring down the next 9 to make 1899. 1899 \div 327 = 5 (327 × 5 = 1635). Remainder is 264. Bring down the final 9 to make 2649. 2649 \div 327 = 8 (327 × 8 = 2616). Remainder is 33. • Step 3: Note the divisor (327) and the final remainder (33). • Step 4: Calculate the required addition. Number to add = 327 - 33 = 294. Exam Strategy & Shortcut: Divisibility Rule of 3: The divisor 327 is a multiple of 3 (since 3+2+7 = 12). The base number 999999 is obviously a multiple of 3. If we add an unknown number x to 999999, the sum (999999 + x) must be divisible by 327, and therefore must be divisible by 3. Since 999999 is already a multiple of 3, the number x we add MUST also be a multiple of 3. Check the options: • (a) 194 \rightarrow 1+9+4 = 14 (No) • (b) 264 \rightarrow 2+6+4 = 12 (Yes) • (c) 292 \rightarrow 2+9+2 = 13 (No) • (d) 294 \rightarrow 2+9+4 = 15 (Yes) This instantly narrows the field down to two choices, drastically reducing the time spent if you need to guess or do partial division. Common Pitfall: The most common mistake is stopping at the remainder (33) and selecting it as the answer, or selecting a distractor option that represents a partial calculation step (like 264, which appeared as an intermediate remainder during step 2). Always apply the formula Divisor - Remainder for "must be added" questions. Final Answer: Therefore, the correct answer is 294.

Question 8

The least number more than 5000 which is divisible by 73 is

Accepted Answer

Concept & Strategy: Finding a number "more than" a base value that is perfectly divisible by a divisor is mathematically identical to finding the least number to *add* to that base value. We find the remainder of the base value, then calculate the difference needed to step up to the next perfect multiple. Required = Base + (Divisor - Remainder) Step-by-Step Solution: • Step 1: Set up long division for 5000 \div 73. • Step 2: Execute the division steps. 500 \div 73 \approx 6. (73 × 6 = 438) Remainder = 500 - 438 = 62. Bring down the 0 to make 620. 620 \div 73 \approx 8. (73 × 8 = 584) Remainder = 620 - 584 = 36. • Step 3: Calculate the amount required to reach the next multiple. Amount to add = Divisor - Remainder Amount to add = 73 - 36 = 37. • Step 4: Add this amount to the original base. 5000 + 37 = 5037. Exam Strategy & Shortcut: Approximation via Multiplication: Rather than performing long division, use base-10 estimations. We need a multiple of 73 near 5000. 73 × 100 = 7300 (Too high) 73 × 50 = 3650 (Too low) Let's try 73 × 70 = 5110. 5110 is very close to 5000. Now, step backward by subtracting 73 to find the first multiple just above 5000. 5110 - 73 = 5037. This matches option (b) directly. This method is often faster and less prone to subtraction errors than standard long division. Common Pitfall: Students often subtract the remainder directly from the base number (5000 - 36 = 4964). While 4964 is perfectly divisible by 73, the question explicitly asks for a number *more than* 5000. Subtracting drops you into the range below the target base. Final Answer: Therefore, the correct answer is 5037.

Question 9

The nearest integer to 58701 which is exactly divisible by 567 is

Accepted Answer

Concept & Logic: "Nearest integer" problems require evaluating two candidate multiples: the one immediately below the base number, and the one immediately above it. After dividing the base by the divisor to find the remainder R, we compare the distances: • Distance to lower multiple = R • Distance to upper multiple = Divisor - R The required answer corresponds to whichever distance is numerically smaller. Step-by-Step Solution: • Step 1: Divide the base number by the divisor to find the remainder. 58701 \div 567 • Step 2: Execute long division. 587 \div 567 = 1. Remainder is 20. Bring down 0 to make 200. 200 \div 567 = 0. Remainder is 200. Bring down 1 to make 2001. 2001 \div 567 = 3. (567 × 3 = 1701). Remainder = 2001 - 1701 = 300. • Step 3: Calculate the two distances. Distance down = 300 Distance up = 567 - 300 = 267 • Step 4: Compare the distances to find the nearest integer. Since 267 < 300, the upper multiple is strictly closer to the base number. We find it by adding the upward distance: 58701 + 267 = 58968. Exam Strategy & Shortcut: Block Approximation: Look at the first three digits of the dividend and divisor. We know 567 × 100 = 56700. Let's subtract this large chunk from our target. 58701 - 56700 = 2001. Now we only need to find the nearest multiple of 567 to 2001. 567 × 3 = 1701 (Gap of 300) 567 × 4 = 2268 (Gap of 267) The multiple 2268 is closer to 2001. So, 56700 + 2268 = 58968 is the nearest integer. This avoids formal long division entirely. Common Pitfall: A major trap is always subtracting the remainder by default, which yields the nearest *lower* integer (58701 - 300 = 58401). The phrase "nearest integer" is bidirectional. You must always check whether rounding up or rounding down provides a shorter absolute distance. Final Answer: Therefore, the correct answer is 58968.

Question 10

The smallest 6-digit number exactly divisible by 111 is

Accepted Answer

Concept & Formula: To find the smallest n-digit number perfectly divisible by a divisor D, start with the smallest possible n-digit base number. Find the remainder upon division, and add the difference between the divisor and the remainder to bridge the gap to the next multiple. Required Number = Base + (Divisor - Remainder) Step-by-Step Solution: • Step 1: Identify the absolute smallest 6-digit number. The base number is 100000. • Step 2: Execute long division to find the remainder. 100000 \div 111 We know 111 × 9 = 999. So, 111 × 900 = 99900. Subtracting this from the base: 100000 - 99900 = 100. The remainder is 100. • Step 3: Calculate the amount required to reach the next perfect multiple. Number to add = Divisor - Remainder Number to add = 111 - 100 = 11. • Step 4: Add this value to the original base number. 100000 + 11 = 100011. Exam Strategy & Shortcut: Option Elimination: Look at the options and apply the divisibility rule for 3 (since 111 is divisible by 3, the answer must also be). (a) 111111 \rightarrow Sum is 6 (Divisible, but is it the smallest?) (b) 110011 \rightarrow Sum is 4 (Not divisible) (c) 100011 \rightarrow Sum is 3 (Divisible) (d) 110101 \rightarrow Sum is 4 (Not divisible) Comparing the valid candidates, 100011 is significantly smaller than 111111. A quick mathematical check (100011 = 99900 + 111) confirms it is built from exact multiples. Common Pitfall: Students often incorrectly assume a number like 111111 is the smallest simply because it repeats the digits of the divisor and feels conceptually "clean". Never assume visual patterns bypass mathematical rules; always start your logic from the fundamental numerical base 100000. Final Answer: Therefore, the correct answer is 100011.

Question 11

The sum of all 2-digit numbers divisible by 5 is

Accepted Answer

Concept & Formula: A sequence of numbers with a constant difference forms an Arithmetic Progression (A.P.). To find the sum of an A.P., we use the formula: S = (n/2)(a + l) Where n is the total number of terms, a is the first term, and l is the last term. Step-by-Step Solution: • Step 1: Identify the A.P. series. The 2-digit numbers perfectly divisible by 5 are: 10, 15, 20, \dots, 95. Here, the first term a = 10, the last term l = 95, and the common difference d = 5. • Step 2: Calculate the total number of terms (n). Use the general n-th term formula: l = a + (n-1)d 95 = 10 + (n-1)5 85 = 5(n-1) 17 = n - 1 \Rightarrow n = 18. There are 18 valid terms in total. • Step 3: Calculate the sum of the series. S = (18/2) × (10 + 95) S = 9 × 105 S = 945. Exam Strategy & Shortcut: Average Rule: The sum of any evenly spaced sequence is simply the average of the first and last terms multiplied by the total number of terms. Average = (10 + 95/2) = 52.5. The number of terms can be found mentally: (95 - 10) \div 5 + 1 = 18. Sum = 52.5 × 18 = 945. This avoids writing out formal A.P. notation entirely. Common Pitfall: The most common mistake is miscounting the number of terms in the set. Many students instinctively divide 100 by 5 to get 20 terms, forgetting that 5 (a 1-digit number) and 100 (a 3-digit number) must be explicitly excluded from the 2-digit constraint requested by the prompt. Final Answer: Therefore, the correct answer is 945.

Question 12

How many 3-digit numbers are completely divisible by 6?

Accepted Answer

Concept & Formula: To find the count of numbers in an Arithmetic Progression (A.P.) that are divisible by a certain number, we first identify the smallest and largest numbers in that range divisible by the target number. The number of terms n in an A.P. is calculated using the formula: n = (L - A/D) + 1 Where: • L = Last term • A = First term • D = Common difference Step-by-Step Solution: • Given: The range is all 3-digit numbers, which spans from 100 to 999. The divisor is 6. • Step 1: Find the first 3-digit number divisible by 6. Divide 100 by 6. The remainder is 4. To make it divisible by 6, we must add (6 - 4) = 2. So, the first term A = 100 + 2 = 102. • Step 2: Find the last 3-digit number divisible by 6. Divide 999 by 6. The remainder is 3. Subtract the remainder from 999 to get the largest multiple. So, the last term L = 999 - 3 = 996. • Step 3: Calculate the total number of terms. Using the A.P. formula with D = 6: n = (996 - 102/6) + 1 n = (894/6) + 1 n = 149 + 1 = 150 Exam Strategy & Shortcut: For any range starting from 1 to N, the number of multiples of X is simply the integer part of (N/X). To find multiples specifically for 3-digit numbers, subtract the multiples up to 99 from the multiples up to 999: Multiples up to 999 = \lfloor (999/6) \rfloor = 166 Multiples up to 99 = \lfloor (99/6) \rfloor = 16 Total = 166 - 16 = 150. This is much faster than finding the exact boundary terms! Common Pitfall: Students often forget to add 1 at the end of the A.P. formula, calculating (L - A/D) and stopping there, which leads to selecting 149 instead of 150. Always remember that inclusive boundary counting requires the +1. Final Answer: Therefore, the correct answer is 150.

Question 13

The number of terms between 11 and 200 which are divisible by 7 but not by 3 are

Accepted Answer

Concept & Logic: This problem applies the principle of set theory and divisibility. To find numbers divisible by X but NOT by Y, you must first find the total count of numbers divisible by X, and then subtract the count of numbers that are divisible by both X and Y. Numbers divisible by both 7 and 3 are exactly the numbers divisible by their Least Common Multiple (LCM). LCM(7, 3) = 21 Step-by-Step Solution: • Given: Range is between 11 and 200. Target is divisible by 7 but not 3. • Step 1: Find total numbers divisible by 7. The first multiple of 7 after 11 is 14. The last multiple of 7 before 200 is 196 (since 200 \div 7 = 28 remainder 4, 200 - 4 = 196). Count = (196 - 14/7) + 1 = (182/7) + 1 = 26 + 1 = 27. • Step 2: Find numbers divisible by both 7 and 3 (i.e., divisible by 21). The first multiple of 21 after 11 is 21. The last multiple of 21 before 200 is 189 (since 200 \div 21 = 9 remainder 11, 200 - 11 = 189). Count = (189 - 21/21) + 1 = (168/21) + 1 = 8 + 1 = 9. • Step 3: Subtract the common multiples. Terms divisible by 7 but not 3 = (Total divisible by 7) - (Total divisible by 21) Count = 27 - 9 = 18 Exam Strategy & Shortcut: Use the quotient method for ranges: Multiples of 7 up to 200 = \lfloor (200/7) \rfloor = 28. Multiples of 7 up to 11 = \lfloor (11/7) \rfloor = 1. Total multiples of 7 = 28 - 1 = 27. Multiples of 21 up to 200 = \lfloor (200/21) \rfloor = 9. Multiples of 21 up to 11 = \lfloor (11/21) \rfloor = 0. Total multiples of 21 = 9 - 0 = 9. Required answer = 27 - 9 = 18. This prevents having to deduce the exact starting and ending multiples. Common Pitfall: A common error is trying to exclude numbers that are just multiples of 3 without recognizing that we only care about the overlap (multiples of 21). Subtracting all multiples of 3 from the range instead of from the multiples of 7 will yield completely incorrect negative values. Final Answer: Therefore, the correct answer is 18.

Question 14

Out of the numbers divisible by 3 between 14 and 95 if the numbers with 3 at unit's place are removed, then how many numbers will remain?

Accepted Answer

Concept & Strategy: This problem tests your ability to filter sequences based on specific digit constraints. First, determine the total number of multiples of a given divisor within the specified range. Then, manually identify the specific terms that violate the secondary condition (in this case, having a unit digit of 3) and subtract their count from the total. Step-by-Step Solution: • Given: The range is between 14 and 95. We need numbers divisible by 3, but must exclude any number ending in 3. • Step 1: Total numbers divisible by 3 in the range. The first multiple of 3 after 14 is 15. The last multiple of 3 before 95 is 93. Using the A.P. formula: n = (Last - First/3) + 1 n = (93 - 15/3) + 1 n = (78/3) + 1 = 26 + 1 = 27 So, there are 27 multiples of 3 in this range. • Step 2: Identify and remove numbers with 3 at the unit's place. List the numbers in the range ending in 3: 23, 33, 43, 53, 63, 73, 83, 93. Now, check which of these are actually divisible by 3 (sum of digits must be a multiple of 3): • 33 (3+3=6, Yes) • 63 (6+3=9, Yes) • 93 (9+3=12, Yes) There are exactly 3 numbers that fit both criteria. • Step 3: Calculate the final remaining numbers. Remaining = (Total multiples of 3) - (Multiples of 3 ending in 3) Remaining = 27 - 3 = 24 Exam Strategy & Shortcut: An arithmetic progression with a common difference of 3 repeats its unit digits in a cycle of 10 (e.g., 3 × 1 = 3, 3 × 11 = 33, 3 × 21 = 63). Within every 30 numbers, there will be exactly one multiple of 3 ending in 3. From 14 to 95 spans roughly 80 numbers, meaning there should be 2 or 3 such occurrences. Spot-checking 33, 63, and 93 is immediate and removes the need to write out non-multiples like 23 or 43. Common Pitfall: Students often subtract *all* numbers ending in 3 (which is 8 numbers) from the total 27, resulting in 19. You must remember that you are only removing numbers that were in your original list of multiples of 3. Final Answer: Therefore, the correct answer is 24.

Question 15

How many numbers less than 1000 are multiples of both 10 and 13?

Accepted Answer

Concept & Formula: When a question asks for a number to be a multiple of *both* X and Y, it implies the number must be a multiple of their Least Common Multiple (LCM). LCM(10, 13) Since 10 and 13 share no common prime factors (they are co-prime), their LCM is simply their product: 10 × 13 = 130. Step-by-Step Solution: • Given: We need to find the count of numbers less than 1000 that are multiples of both 10 and 13. • Step 1: Determine the base multiple. Any number divisible by both 10 and 13 must be divisible by 130. • Step 2: Find the maximum multiple under 1000. We want to find how many times 130 goes into numbers less than 1000. Divide 999 by 130: (999/130) \approx 7.68 • Step 3: Interpret the integer quotient. The integer part of the quotient represents the exact number of multiples starting from 1 up to 999. These numbers are: 130, 260, 390, 520, 650, 780, 910. There are exactly 7 numbers. Exam Strategy & Shortcut: Simply evaluate \lfloor (1000/130) \rfloor. Since 13 × 7 = 91, 130 × 7 = 910, which is under 1000. The next multiple, 130 × 8 = 1040, exceeds 1000. Thus, 7 is immediately the answer without listing them. Common Pitfall: Some might try to find the numbers divisible by 10, then those divisible by 13, and use a complex Venn diagram overlap method. This takes too long. Recognizing that "multiple of both" strictly means "multiple of the LCM" bypasses the unnecessary steps entirely. Final Answer: Therefore, the correct answer is 7.

Question 16

How many integers between 100 and 150, both inclusive, can be evenly divided by neither 3 nor 5?

Accepted Answer

Concept & Logic: To find the count of numbers divisible by "neither A nor B", we use Set Theory and the Principle of Inclusion-Exclusion. First, determine the total numbers in the range. Then find the count of numbers divisible by A, divisible by B, and divisible by both (LCM of A and B). The total numbers divisible by A OR B is: n(A \cup B) = n(A) + n(B) - n(A \cap B) Finally, subtract this union from the total pool to get the "neither" count. Step-by-Step Solution: • Given: Range is integers between 100 and 150, inclusive. Divisors are 3 and 5. • Step 1: Find total integers in the range. Total = 150 - 100 + 1 = 51 numbers. • Step 2: Calculate multiples of 3 (n(3)). First multiple \ge 100 is 102. Last \le 150 is 150. Count = (150 - 102/3) + 1 = (48/3) + 1 = 16 + 1 = 17. • Step 3: Calculate multiples of 5 (n(5)). First multiple \ge 100 is 100. Last \le 150 is 150. Count = (150 - 100/5) + 1 = (50/5) + 1 = 10 + 1 = 11. • Step 4: Calculate multiples of both 3 and 5 (n(15)). First multiple of 15 \ge 100 is 105. Last \le 150 is 150. Count = (150 - 105/15) + 1 = (45/15) + 1 = 3 + 1 = 4. • Step 5: Apply Inclusion-Exclusion. Numbers divisible by 3 or 5: n(3 \cup 5) = 17 + 11 - 4 = 24. • Step 6: Find the "neither" count. Neither = Total - n(3 \cup 5) = 51 - 24 = 27. Exam Strategy & Shortcut: Use the quotient method for fast range counts: Let F(N, X) = \lfloor (N/X) \rfloor Multiples in range 100 to 150 for divisor X = F(150, X) - F(99, X). • n(3) = 50 - 33 = 17 • n(5) = 30 - 19 = 11 • n(15) = 10 - 6 = 4 Divisible by 3 or 5 = 17 + 11 - 4 = 24. Total numbers = 150 - 99 = 51. Neither = 51 - 24 = 27. This relies entirely on fast division rather than setting up A.P. formulas. Common Pitfall: The word "inclusive" means both boundaries are part of the count. Many students simply subtract 150 - 100 = 50 total numbers, omitting the +1. This throws off the final arithmetic and leads to choosing option 26 instead of 27. Final Answer: Therefore, the correct answer is 27.

Question 17

If all the numbers from 501 to 700 are written, what is the total number of times the digit 6 appears?

Accepted Answer

Concept & Strategy: When asked to find how many times a specific digit appears in a given range of numbers, it is most efficient to count the occurrences of that digit in each positional place value independently (hundreds, tens, and units). Because we are counting the literal number of times the digit is written, a number like 666 will correctly be counted three times using this positional method. Step-by-Step Solution: • Given: The numerical range is from 501 to 700. We need to count the frequency of the digit 6. • Step 1: Count occurrences in the Hundreds place. The digit 6 appears in the hundreds place for all numbers from 600 to 699. Total occurrences = 100 times. • Step 2: Count occurrences in the Tens place. The digit 6 appears in the tens place for the 560s (560 to 569) and the 660s (660 to 669). There are 10 occurrences in the 500s and 10 occurrences in the 600s. Total occurrences = 20 times. • Step 3: Count occurrences in the Units place. The digit 6 appears in the units place whenever a number ends in 6 (e.g., 506, 516, 526, \dots, 596). In the 500s block, this happens 10 times. In the 600s block, this also happens 10 times (606, 616, \dots, 696). Total occurrences = 20 times. • Step 4: Sum the occurrences. Total = 100 (Hundreds) + 20 (Tens) + 20 (Units) = 140. Exam Strategy & Shortcut: Memorize the standard distribution of digits: In any full 100-number block (like 500 to 599), any given digit from 0-9 will appear exactly 20 times in the tens and units places combined. For 501 to 700, we have exactly two 100-number blocks (500s and 600s). Tens/Units count = 20 × 2 = 40. Then, just add the 100 occurrences where 6 is the leading digit in the 600s. 100 + 40 = 140. Solved in 5 seconds. Common Pitfall: A very common mistake is trying to list out numbers with 6 and mistakenly treating a number like 666 as a single occurrence, or conversely, subtracting overlaps. The question asks how many times the *digit* is written, not how many *numbers* contain the digit. Positional counting avoids this trap entirely. Final Answer: Therefore, the correct answer is 140.

Question 18

The total number of integers between 200 and 400, each of which either begins with 3 or ends with 3 or both is

Accepted Answer

Concept & Strategy: This problem relies on set theory, specifically the Principle of Inclusion-Exclusion, or simple logical division of a range to avoid double counting. We must count numbers that start with 3, numbers that end with 3, and ensure the overlap (numbers that start AND end with 3) is only counted once. Step-by-Step Solution: • Given: The target range is between 200 and 400. We are looking for integers that either begin with 3, end with 3, or both. • Step 1: Count numbers that begin with 3. Within the 200 to 400 range, all numbers in the 300s begin with 3. These span from 300 to 399. Total numbers = 100. *(Note: This block already includes all numbers that begin AND end with 3, like 303, 313, etc.)* • Step 2: Count numbers that end with 3, but do NOT begin with 3. Since we already completely covered the 300s, we only need to look at the remaining part of the range: the 200s (from 200 to 299). Numbers in this block ending in 3 are: 203, 213, 223, 233, 243, 253, 263, 273, 283, 293. Total numbers = 10. • Step 3: Combine the counts. Total valid integers = (Numbers starting with 3) + (Numbers ending with 3 in the remaining range) Total = 100 + 10 = 110. Exam Strategy & Shortcut: Break the problem into mutually exclusive sets to bypass inclusion-exclusion formulas entirely. Set A: The entire 300 series (100 numbers). Set B: The 200 series ending in 3. Since it's exactly one 100-number block, any specific unit digit appears exactly 10 times. 100 + 10 = 110. This mental math takes less than 10 seconds. Common Pitfall: Using the union formula n(A \cup B) = n(A) + n(B) - n(A \cap B) but miscalculating the sets. Students might incorrectly say there are 20 numbers ending in 3 between 200 and 400, add that to 100, and then subtract the 10 overlaps. While mathematically sound (100 + 20 - 10 = 110), it introduces unnecessary steps and room for arithmetic error. Final Answer: Therefore, the correct answer is 110.

Question 19

A number divided by 68 gives the quotient 260 and remainder zero. If the same number is divided by 65, the remainder is

Accepted Answer

Concept & Logic: This problem applies the Division Algorithm and modular arithmetic properties. The Division Algorithm states: Dividend = Divisor × Quotient + Remainder. When evaluating the remainder of a product, if any individual factor of that product is perfectly divisible by the new divisor, the entire product is perfectly divisible (leaving a remainder of 0). Step-by-Step Solution: • Given: A number N when divided by 68 gives a quotient of 260 and remainder 0. • Step 1: Express the number as an equation. N = 68 × 260 + 0 N = 68 × 260 • Step 2: Test divisibility by the new divisor (65). We need to find the remainder of (68 × 260) \div 65. Check the factors 68 and 260 to see if they relate to 65. 68 \div 65 leaves a remainder of 3. 260 \div 65 leaves a remainder of 0 (since 65 × 4 = 260). • Step 3: Calculate the final remainder. Since 260 is a multiple of 65, the entire product (68 × 260) is a multiple of 65. Therefore, the remainder is 0. Exam Strategy & Shortcut: Never multiply the divisor and quotient out (e.g., doing 68 × 260). Immediately check if the new divisor (65) is a factor of either the old divisor (68) or the quotient (260). Noticing that 260 is exactly 4 times 65 makes the answer instantly 0. Common Pitfall: A major time-waster is multiplying 68 × 260 = 17,680 and then performing long division by 65. This introduces massive room for error on a question designed to take seconds. Final Answer: Therefore, the correct answer is 0.

Question 20

Which of the following prime numbers while dividing 2176 leaves 9 as remainder?

Accepted Answer

Concept & Strategy: By the division algorithm, if a number N divided by a divisor D leaves a remainder R, then the number (N - R) must be perfectly divisible by D. Once we find (N - R), we can test the options to see which prime perfectly divides it. Step-by-Step Solution: • Given: The original number is 2176 and the target remainder is 9. • Step 1: Subtract the remainder from the number. Perfectly divisible number = 2176 - 9 = 2167. The target prime number must be a factor of 2167. • Step 2: Check for prime factors using basic divisibility rules. Instead of testing large options immediately, quickly scan 2167 for small prime factors. Test for divisibility by 11: (Sum of digits in odd places) - (Sum of digits in even places). (2 + 6) - (1 + 7) = 8 - 8 = 0. Since the result is 0, 2167 is perfectly divisible by 11. • Step 3: Divide by the found factor to find the paired factor. 2167 \div 11 = 197. Thus, 2167 = 11 × 197. Both 11 and 197 are prime numbers that would leave a remainder of 9. • Step 4: Check options. 197 is present in the options. Exam Strategy & Shortcut: Subtract the remainder first to get 2167. Then, look at the unit digits of the options. • 17 × (something ending in 1) \rightarrow 7 (e.g. 17 × 100 = 1700, too small. 17 × 127?) • 29 × (something ending in 3) \rightarrow 7 A faster route is running the divisibility rule for 11 on 2167. The moment you spot it is divisible by 11, division yields 197 directly, matching the last option. Common Pitfall: Attempting to perform long division of 2176 by 167 and 197 directly without subtracting the remainder first. This makes the arithmetic significantly clunkier. Final Answer: Therefore, the correct answer is 197.

Number System Questions

Questions

Practice smarter