Q: Consecutive Even Integers — Sum to 130 The sum of five consecutive even integers A, B, C, D, and E equals 130. Determine the product A * E (smallest times largest).

Introduction / Context: This question assesses comfort with sequences of consecutive even integers and the technique of using the middle term to simplify calculations. For five consecutive even numbers, the central value is the average of the five and the endpoints are symmetric around it. Given Data / Assumptions: Five consecutive even integers: A, B, C, D, E. Sum is 130. We must compute the product of the smallest (A) and largest (E). Concept / Approach: Let the middle even integer be C = x. Then the five numbers are x-4, x-2, x, x+2, x+4. Their sum is 5x. So x = (sum) / 5. With x known, endpoints are found directly and the product is straightforward. Step-by-Step Solution: 1) Compute x: x = 130 / 5 = 26.2) Determine the five numbers: A = 26 - 4 = 22; B = 24; C = 26; D = 28; E = 30.3) Compute product: A * E = 22 * 30 = 660.4) Check the sum: 22 + 24 + 26 + 28 + 30 = 130 (verified). Verification / Alternative check: Because the set is symmetric, A + E = 52 and B + D = 52; hence total = (A + E) + (B + D) + C = 52 + 52 + 26 = 130, confirming correctness. Why Other Options Are Wrong: 720, 616, 672, and 640 correspond to products of incorrect endpoint pairs that arise from miscomputing the center or using odd steps. Common Pitfalls: Using increments of 1 instead of 2 (even numbers); dividing by 2 or 4 instead of 5 to get the middle term; arithmetic slips when multiplying endpoints. Final Answer: 660

Q: Consecutive Integers — Sum to 170 The sum of five consecutive integers is 170. Find the product of the smallest and the largest integer in this sequence.

Introduction / Context: This is a classic consecutive-integers problem that rewards recognizing that the average of equally spaced numbers equals the middle term. With five consecutive numbers, the central integer is simply total/5. Once the sequence is identified, the endpoints multiply easily. Given Data / Assumptions: Total of 5 consecutive integers: 170. Let the sequence be n-2, n-1, n, n+1, n+2. We seek (n-2) * (n+2). Concept / Approach: The sum equals 5n, so n = 170 / 5 = 34. The integers are 32, 33, 34, 35, 36. The smallest times the largest is 32 * 36. Alternatively, use the identity (n-2)*(n+2) = n^2 - 4 to compute quickly once n is known. Step-by-Step Solution: 1) Compute middle n: n = 170 / 5 = 34.2) List the sequence: 32, 33, 34, 35, 36.3) Multiply extremes: 32 * 36 = (32 * 30) + (32 * 6) = 960 + 192 = 1152.4) Optional identity: n^2 - 4 = 34^2 - 4 = 1156 - 4 = 1152. Verification / Alternative check: Confirm the sum: 32 + 33 + 34 + 35 + 36 = 170, ensuring the sequence is correct, thus validating the product. Why Other Options Are Wrong: 1512, 1102, 1210, and 1120 are products from incorrect endpoint choices or arithmetic mistakes (e.g., mixing numbers or using 31/37 by error). Common Pitfalls: Forgetting that consecutive means step 1; miscomputing the average; arithmetic slips in multiplication under time pressure. Final Answer: 1152

Q: Consecutive Odd Squares — Difference Property Which number always divides the difference between the squares of two consecutive odd integers?

Introduction / Context: This question probes number properties and the algebra of squares. For consecutive odd integers, their separation is 2. Squaring and subtracting reveals a fixed factorization that exposes guaranteed divisibility by a specific integer regardless of which odd numbers are chosen. Given Data / Assumptions: Let the two consecutive odd integers be n and n+2. n is odd, so n+1 is even. We analyze (n+2)^2 - n^2. Concept / Approach: Use the identity a^2 - b^2 = (a - b)(a + b). Here, a - b = 2 and a + b = 2n + 2 = 2(n + 1). Hence the difference equals 2 * 2(n + 1) = 4(n + 1). Since n is odd, n + 1 is even, so write n + 1 = 2k. Then 4(n + 1) = 4 * 2k = 8k, which is divisible by 8 for all choices of n. Step-by-Step Solution: 1) Compute difference: (n+2)^2 - n^2 = 4n + 4.2) Factor: 4n + 4 = 4(n + 1).3) Since n is odd, let n + 1 = 2k, giving 4 * 2k = 8k.4) Therefore the difference is always a multiple of 8. Verification / Alternative check: Try sample pairs: 3 and 5 → 25 - 9 = 16 (divisible by 8). 7 and 9 → 81 - 49 = 32 (divisible by 8). The pattern holds universally. Why Other Options Are Wrong: 7 and 5 do not always divide the result; 3 and 6 divide some cases but not all. Only 8 is guaranteed for every pair of consecutive odd integers. Common Pitfalls: Forgetting that “consecutive odd” means a gap of 2; applying the result to consecutive integers (gap of 1) which yields a different divisibility pattern. Final Answer: 8

Q: Arithmetic Simplification — Common Factor Method Evaluate: 9856 × 156 + 9856 × 844. Use factoring to simplify the computation.

Introduction / Context: This expression is designed to reward recognition of the distributive property. Rather than multiply two large products separately, factor out the common term to turn the problem into a single, easy multiplication. Given Data / Assumptions: Expression: 9856 × 156 + 9856 × 844. All numbers are integers. Standard arithmetic rules apply. Concept / Approach: Use the distributive property: a*b + a*c = a*(b + c). Here a = 9856, b = 156, c = 844. Adding the small numbers first is simpler and reduces computational effort dramatically. Step-by-Step Solution: 1) Factor: 9856 × 156 + 9856 × 844 = 9856 × (156 + 844).2) Add inside the parentheses: 156 + 844 = 1000.3) Multiply: 9856 × 1000 = 9,856,000.4) Therefore, the value is 9,856,000. Verification / Alternative check: If you computed each product separately and added, you would reach the same total, but with more work. The factorization method is quicker and less error-prone. Why Other Options Are Wrong: Other totals result from arithmetic slips, mis-adding 156 + 844, or misplacing zeros when multiplying by 1000. Common Pitfalls: Ignoring the common factor; performing two large multiplications and making place-value mistakes; forgetting to append three zeros when multiplying by 1000. Final Answer: 9856000

Q: Multiplication by 625 — Power-of-10 Trick Compute 6674 × 625 quickly by using 625 = 10^4 / 16.

Introduction / Context: Mental math with numbers like 25, 125, and 625 benefits from rewriting them as fractions of powers of 10. Since 625 = 5^4 = 10000 / 16, you can multiply by 10000 (add four zeros) and then divide by 16, which is efficient and accurate. Given Data / Assumptions: We need 6674 × 625. Use 625 = 10000 / 16. All operations are in base-10 decimal arithmetic. Concept / Approach: Compute 6674 × 10000 first (easy place-value shift), then perform an exact division by 16. Division by 16 can be done by successive halving: divide by 2 four times. Step-by-Step Solution: 1) Multiply: 6674 × 10000 = 66,740,000.2) Divide by 2: 33,370,000.3) Divide by 2 again (÷4 total): 16,685,000.4) Divide by 2 again (÷8 total): 8,342,500.5) Divide by 2 again (÷16 total): 4,171,250. Verification / Alternative check: Long multiplication or a calculator yields the same product 4,171,250, confirming the power-of-10 trick is valid. Why Other Options Are Wrong: Other choices stem from halving errors or misplacing zeros during the 10000 multiplication step. Common Pitfalls: Forgetting that four halvings are required to divide by 16; dropping a zero when scaling by 10000; arithmetic slips during repeated division. Final Answer: 4171250

Q: Near-1000 Multiplication — Complement Method Evaluate 9998 × 999 efficiently by using (1000 − 2) × (1000 − 1).

Introduction / Context: Numbers close to a power of 10 are perfect for the complement method. Computing 9998 × 999 is much easier if we rewrite both factors relative to 1000 and expand carefully, avoiding long multiplication. Given Data / Assumptions: 9998 = 1000 − 2. 999 = 1000 − 1. We want (1000 − 2) × (1000 − 1). Concept / Approach: Use distributive expansion: (a − b)(a − c) = a^2 − a(b + c) + bc. Let a = 1000, b = 2, c = 1. This keeps arithmetic simple and exact while leveraging place-value structure. Step-by-Step Solution: 1) Compute a^2 = 1000^2 = 1,000,000.2) Compute a(b + c) = 1000 × (2 + 1) = 3000.3) Compute bc = 2 × 1 = 2.4) Combine: 1,000,000 − 3000 + 2 = 997,002? Wait—this is for (1000 − 2)(1000 − 3). Re-evaluate directly: (1000 − 2)(1000 − 1) = 1,000,000 − 1000 − 2000 + 2 = 997,002? That still seems off—let’s expand directly:(1000 − 2) × (1000 − 1) = 1000×1000 − 1000 − 2000 + 2 = 1,000,000 − 3000 + 2 = 997,002. But our original factors are 9998 and 999 (not 998 and 997). Recognize a slip: 9998 = 10000 − 2, not 1000 − 2. Correct expansion follows.5) Correct form: 9998 × 999 = (10000 − 2) × (1000 − 1) = 10000×1000 − 10000 − 2000 + 2 = 10,000,000 − 12,000 + 2 = 9,988,002. Verification / Alternative check: Use 9998 × 999 = 9998 × (1000 − 1) = 9,998,000 − 9,998 = 9,988,002, confirming the corrected computation. Why Other Options Are Wrong: 9997001 and 9987012 reflect misapplied complements; 9898012 is far off due to place-value errors; 9990002 ignores the subtraction step. Common Pitfalls: Confusing 9998 with 1000 − 2 instead of 10000 − 2; dropping thousands when subtracting; mixing terms in the distributive expansion. Final Answer: 9988002

Question 1

From the integers 1 to 100 inclusive, how many numbers are divisible by 5 and also contain the digit 5 at least once?

Accepted Answer

Introduction / Context: This counting problem combines divisibility with a digit-presence condition. Numbers divisible by 5 end in 0 or 5; among them, we must count those that also have the digit 5 somewhere in their decimal representation. Given Data / Assumptions: Range: 1 to 100 inclusive. Divisible by 5 ⇒ last digit is 0 or 5. Digit 5 must appear in at least one position. Concept / Approach: Partition divisible-by-5 numbers into those ending with 5 and those ending with 0. The first group automatically contains a 5. The second group contains a 5 only if the tens digit is 5 (i.e., the number 50 within 1–100). Step-by-Step Solution: List multiples of 5 ending with 5: 5, 15, 25, 35, 45, 55, 65, 75, 85, 95 → 10 numbers.Check multiples of 5 ending with 0 that include a digit 5: only 50 qualifies (tens digit = 5).Total = 10 + 1 = 11. Verification / Alternative check: Scan all multiples of 5 from 5 to 100; the only candidate ending with 0 that contains a 5 is 50. Count remains 11. Why Other Options Are Wrong: 10 ignores 50; 12 overcounts by adding a non-qualifying multiple; 20 counts all multiples of 5 regardless of the digit condition; 9 misses more than one qualifying number. Common Pitfalls: Forgetting to include 50 or accidentally including 100 (which has no digit 5). Final Answer: 11

Question 2

A two-digit number has a digit sum of 14, and the absolute difference between its digits is 2. What is the product of the two digits of this two-digit number?

Accepted Answer

Introduction / Context: This is a classic two-digit digit-algebra problem. You are given the sum and the absolute difference of the digits and asked to determine their product. Such problems measure skill in setting up and solving small linear systems for digits constrained between 0 and 9. Given Data / Assumptions: Let the tens digit be a and the units digit be b. a + b = 14. |a - b| = 2. Digits are integers with 0 ≤ a, b ≤ 9 and a ≠ 0 (two-digit number). Concept / Approach: Assume without loss of generality a ≥ b, so a - b = 2. Solve the system with two equations in two unknowns, then compute the product a * b. Step-by-Step Solution: Add equations: (a + b) + (a - b) = 14 + 2 ⇒ 2a = 16.Solve for a: a = 8.Find b: b = 14 - a = 14 - 8 = 6.Compute product: a * b = 8 * 6 = 48. Verification / Alternative check: Check the difference: |8 - 6| = 2 and sum: 8 + 6 = 14; both conditions satisfied. Why Other Options Are Wrong: 56 corresponds to 7 * 8 (sum 15). 45 corresponds to 9 * 5 (difference 4). 36 corresponds to 9 * 4 (sum 13). “None of the above” is unnecessary because 48 fits. Common Pitfalls: Using the wrong sign for the difference or swapping digits without checking the absolute difference condition. Final Answer: 48

Question 3

Number System — Divisibility Adjustment by 33 What is the smallest whole number that must be added to 231228 so that the result is exactly divisible by 33?

Accepted Answer

Introduction / Context: This problem checks quick divisibility using remainders (modular arithmetic). We are asked to find the least non-negative integer that, when added to 231228, produces a multiple of 33. Such questions are common in aptitude tests because they reward number sense and speed without heavy computation. Given Data / Assumptions: Dividend candidate: 231228. Divisor: 33. We want the smallest added value k such that 231228 + k is divisible by 33. k must be a whole number (0, 1, 2, ...). Concept / Approach: Use the remainder idea: if N % 33 = r, then N + (33 - r) is the next multiple of 33, unless r = 0 (in which case k = 0). The goal is to compute the remainder efficiently and adjust to the next multiple. Step-by-Step Solution: 1) Compute r = 231228 % 33.2) Divide 231228 by 33 to find the remainder. The remainder is 30.3) The next multiple is obtained by adding 33 - r = 33 - 30 = 3.4) Therefore k = 3 is the least number to add. Verification / Alternative check: Compute (231228 + 3) / 33 = 231231 / 33 = 701 (an integer), confirming exact divisibility. Any smaller k (0, 1, 2) would keep a non-zero remainder. Why Other Options Are Wrong: 1 and 2 leave remainders 31 and 32 respectively; 4 overshoots the next multiple; 5 also overshoots and is not minimal. Common Pitfalls: Adding until you find a multiple by trial rather than computing the remainder; confusing subtract with add when asked for the next multiple (upward adjustment is required here). Final Answer: 3

Question 4

Number System — Density of Rational Numbers Between the real numbers 1 and 1000 (exclusive), how many rational numbers exist?

Accepted Answer

Introduction / Context: This question tests understanding of the density of rational numbers on the real number line. A rational number is any number that can be expressed as p/q where p and q are integers and q != 0. The concept of density means that between any two distinct real numbers, there are infinitely many rational numbers. Given Data / Assumptions: Interval considered: (1, 1000), open interval (endpoints excluded). Rational numbers are of the form p/q with integers p, q and q != 0. No restriction on numerator/denominator size beyond being integers. Concept / Approach: Rational numbers are dense in the reals: given any two reals a < b, there exists a rational r with a < r < b. Moreover, you can construct infinitely many rationals in any interval by, for example, taking midpoints repeatedly or considering sequences p/n for suitable integers p and large n that land between a and b. Step-by-Step Solution: 1) Pick any two numbers between 1 and 1000, say 2 and 3; there are infinitely many rationals in (2, 3).2) Because (1, 1000) contains (2, 3), it likewise contains infinitely many rationals.3) More generally, for any n, the set {k/n : k = n+1, ..., 1000n-1} yields many rationals in (1, 1000).4) As n grows, the count grows without bound, proving infinitude. Verification / Alternative check: Repeated bisection: start with 1 and 1000; the midpoint 500.5 is rational. Bisect again to get more rationals. This process never ends, establishing an infinite set. Why Other Options Are Wrong: 998, 999, and 1000 suggest a finite count, which contradicts density. “None” is clearly false because many rationals (e.g., 3/2, 5/3) lie in the interval. Common Pitfalls: Confusing “integers” with “rationals” and counting only whole numbers; misreading the question as asking for integers between 1 and 1000. Final Answer: infinite

Question 5

Consecutive Even Integers — Sum to 130 The sum of five consecutive even integers A, B, C, D, and E equals 130. Determine the product A * E (smallest times largest).

Accepted Answer

Introduction / Context: This question assesses comfort with sequences of consecutive even integers and the technique of using the middle term to simplify calculations. For five consecutive even numbers, the central value is the average of the five and the endpoints are symmetric around it. Given Data / Assumptions: Five consecutive even integers: A, B, C, D, E. Sum is 130. We must compute the product of the smallest (A) and largest (E). Concept / Approach: Let the middle even integer be C = x. Then the five numbers are x-4, x-2, x, x+2, x+4. Their sum is 5x. So x = (sum) / 5. With x known, endpoints are found directly and the product is straightforward. Step-by-Step Solution: 1) Compute x: x = 130 / 5 = 26.2) Determine the five numbers: A = 26 - 4 = 22; B = 24; C = 26; D = 28; E = 30.3) Compute product: A * E = 22 * 30 = 660.4) Check the sum: 22 + 24 + 26 + 28 + 30 = 130 (verified). Verification / Alternative check: Because the set is symmetric, A + E = 52 and B + D = 52; hence total = (A + E) + (B + D) + C = 52 + 52 + 26 = 130, confirming correctness. Why Other Options Are Wrong: 720, 616, 672, and 640 correspond to products of incorrect endpoint pairs that arise from miscomputing the center or using odd steps. Common Pitfalls: Using increments of 1 instead of 2 (even numbers); dividing by 2 or 4 instead of 5 to get the middle term; arithmetic slips when multiplying endpoints. Final Answer: 660

Question 6

Consecutive Integers — Sum to 170 The sum of five consecutive integers is 170. Find the product of the smallest and the largest integer in this sequence.

Accepted Answer

Introduction / Context: This is a classic consecutive-integers problem that rewards recognizing that the average of equally spaced numbers equals the middle term. With five consecutive numbers, the central integer is simply total/5. Once the sequence is identified, the endpoints multiply easily. Given Data / Assumptions: Total of 5 consecutive integers: 170. Let the sequence be n-2, n-1, n, n+1, n+2. We seek (n-2) * (n+2). Concept / Approach: The sum equals 5n, so n = 170 / 5 = 34. The integers are 32, 33, 34, 35, 36. The smallest times the largest is 32 * 36. Alternatively, use the identity (n-2)*(n+2) = n^2 - 4 to compute quickly once n is known. Step-by-Step Solution: 1) Compute middle n: n = 170 / 5 = 34.2) List the sequence: 32, 33, 34, 35, 36.3) Multiply extremes: 32 * 36 = (32 * 30) + (32 * 6) = 960 + 192 = 1152.4) Optional identity: n^2 - 4 = 34^2 - 4 = 1156 - 4 = 1152. Verification / Alternative check: Confirm the sum: 32 + 33 + 34 + 35 + 36 = 170, ensuring the sequence is correct, thus validating the product. Why Other Options Are Wrong: 1512, 1102, 1210, and 1120 are products from incorrect endpoint choices or arithmetic mistakes (e.g., mixing numbers or using 31/37 by error). Common Pitfalls: Forgetting that consecutive means step 1; miscomputing the average; arithmetic slips in multiplication under time pressure. Final Answer: 1152

Question 7

Consecutive Odd Squares — Difference Property Which number always divides the difference between the squares of two consecutive odd integers?

Accepted Answer

Introduction / Context: This question probes number properties and the algebra of squares. For consecutive odd integers, their separation is 2. Squaring and subtracting reveals a fixed factorization that exposes guaranteed divisibility by a specific integer regardless of which odd numbers are chosen. Given Data / Assumptions: Let the two consecutive odd integers be n and n+2. n is odd, so n+1 is even. We analyze (n+2)^2 - n^2. Concept / Approach: Use the identity a^2 - b^2 = (a - b)(a + b). Here, a - b = 2 and a + b = 2n + 2 = 2(n + 1). Hence the difference equals 2 * 2(n + 1) = 4(n + 1). Since n is odd, n + 1 is even, so write n + 1 = 2k. Then 4(n + 1) = 4 * 2k = 8k, which is divisible by 8 for all choices of n. Step-by-Step Solution: 1) Compute difference: (n+2)^2 - n^2 = 4n + 4.2) Factor: 4n + 4 = 4(n + 1).3) Since n is odd, let n + 1 = 2k, giving 4 * 2k = 8k.4) Therefore the difference is always a multiple of 8. Verification / Alternative check: Try sample pairs: 3 and 5 → 25 - 9 = 16 (divisible by 8). 7 and 9 → 81 - 49 = 32 (divisible by 8). The pattern holds universally. Why Other Options Are Wrong: 7 and 5 do not always divide the result; 3 and 6 divide some cases but not all. Only 8 is guaranteed for every pair of consecutive odd integers. Common Pitfalls: Forgetting that “consecutive odd” means a gap of 2; applying the result to consecutive integers (gap of 1) which yields a different divisibility pattern. Final Answer: 8

Question 8

Arithmetic Simplification — Common Factor Method Evaluate: 9856 × 156 + 9856 × 844. Use factoring to simplify the computation.

Accepted Answer

Introduction / Context: This expression is designed to reward recognition of the distributive property. Rather than multiply two large products separately, factor out the common term to turn the problem into a single, easy multiplication. Given Data / Assumptions: Expression: 9856 × 156 + 9856 × 844. All numbers are integers. Standard arithmetic rules apply. Concept / Approach: Use the distributive property: a*b + a*c = a*(b + c). Here a = 9856, b = 156, c = 844. Adding the small numbers first is simpler and reduces computational effort dramatically. Step-by-Step Solution: 1) Factor: 9856 × 156 + 9856 × 844 = 9856 × (156 + 844).2) Add inside the parentheses: 156 + 844 = 1000.3) Multiply: 9856 × 1000 = 9,856,000.4) Therefore, the value is 9,856,000. Verification / Alternative check: If you computed each product separately and added, you would reach the same total, but with more work. The factorization method is quicker and less error-prone. Why Other Options Are Wrong: Other totals result from arithmetic slips, mis-adding 156 + 844, or misplacing zeros when multiplying by 1000. Common Pitfalls: Ignoring the common factor; performing two large multiplications and making place-value mistakes; forgetting to append three zeros when multiplying by 1000. Final Answer: 9856000

Question 9

Multiplication by 625 — Power-of-10 Trick Compute 6674 × 625 quickly by using 625 = 10^4 / 16.

Accepted Answer

Introduction / Context: Mental math with numbers like 25, 125, and 625 benefits from rewriting them as fractions of powers of 10. Since 625 = 5^4 = 10000 / 16, you can multiply by 10000 (add four zeros) and then divide by 16, which is efficient and accurate. Given Data / Assumptions: We need 6674 × 625. Use 625 = 10000 / 16. All operations are in base-10 decimal arithmetic. Concept / Approach: Compute 6674 × 10000 first (easy place-value shift), then perform an exact division by 16. Division by 16 can be done by successive halving: divide by 2 four times. Step-by-Step Solution: 1) Multiply: 6674 × 10000 = 66,740,000.2) Divide by 2: 33,370,000.3) Divide by 2 again (÷4 total): 16,685,000.4) Divide by 2 again (÷8 total): 8,342,500.5) Divide by 2 again (÷16 total): 4,171,250. Verification / Alternative check: Long multiplication or a calculator yields the same product 4,171,250, confirming the power-of-10 trick is valid. Why Other Options Are Wrong: Other choices stem from halving errors or misplacing zeros during the 10000 multiplication step. Common Pitfalls: Forgetting that four halvings are required to divide by 16; dropping a zero when scaling by 10000; arithmetic slips during repeated division. Final Answer: 4171250

Question 10

Near-1000 Multiplication — Complement Method Evaluate 9998 × 999 efficiently by using (1000 − 2) × (1000 − 1).

Accepted Answer

Introduction / Context: Numbers close to a power of 10 are perfect for the complement method. Computing 9998 × 999 is much easier if we rewrite both factors relative to 1000 and expand carefully, avoiding long multiplication. Given Data / Assumptions: 9998 = 1000 − 2. 999 = 1000 − 1. We want (1000 − 2) × (1000 − 1). Concept / Approach: Use distributive expansion: (a − b)(a − c) = a^2 − a(b + c) + bc. Let a = 1000, b = 2, c = 1. This keeps arithmetic simple and exact while leveraging place-value structure. Step-by-Step Solution: 1) Compute a^2 = 1000^2 = 1,000,000.2) Compute a(b + c) = 1000 × (2 + 1) = 3000.3) Compute bc = 2 × 1 = 2.4) Combine: 1,000,000 − 3000 + 2 = 997,002? Wait—this is for (1000 − 2)(1000 − 3). Re-evaluate directly: (1000 − 2)(1000 − 1) = 1,000,000 − 1000 − 2000 + 2 = 997,002? That still seems off—let’s expand directly:(1000 − 2) × (1000 − 1) = 1000×1000 − 1000 − 2000 + 2 = 1,000,000 − 3000 + 2 = 997,002. But our original factors are 9998 and 999 (not 998 and 997). Recognize a slip: 9998 = 10000 − 2, not 1000 − 2. Correct expansion follows.5) Correct form: 9998 × 999 = (10000 − 2) × (1000 − 1) = 10000×1000 − 10000 − 2000 + 2 = 10,000,000 − 12,000 + 2 = 9,988,002. Verification / Alternative check: Use 9998 × 999 = 9998 × (1000 − 1) = 9,998,000 − 9,998 = 9,988,002, confirming the corrected computation. Why Other Options Are Wrong: 9997001 and 9987012 reflect misapplied complements; 9898012 is far off due to place-value errors; 9990002 ignores the subtraction step. Common Pitfalls: Confusing 9998 with 1000 − 2 instead of 10000 − 2; dropping thousands when subtracting; mixing terms in the distributive expansion. Final Answer: 9988002

Question 11

Complete the Sum — Find the Missing Term Evaluate the missing number “?” if 8888 + 7777 + 666 + ? = 19000.

Accepted Answer

Introduction / Context: Missing-term equations are straightforward arithmetic checks. The idea is to sum the known addends and subtract that total from the required sum to determine the unknown. Accuracy in addition and subtraction under time pressure is the key skill tested here. Given Data / Assumptions: Total required: 19000. Known addends: 8888, 7777, and 666. We must find the fourth addend (?). Concept / Approach: Use the identity: total = sum_known + ?. Hence ? = total − sum_known. Compute the partial sum carefully, then subtract from 19000. Double-check with a quick recomposition to avoid single-digit carry/borrow mistakes. Step-by-Step Solution: 1) Sum known numbers: 8888 + 7777 = 16,665.2) Add 666: 16,665 + 666 = 17,331.3) Compute the missing term: ? = 19,000 − 17,331 = 1,669.4) Verify by recomposition: 8888 + 7777 + 666 + 1669 = 19,000. Verification / Alternative check: Break numbers into thousands and units: (8000 + 7000 + 0) + (888 + 777 + 666) = 15,000 + 2,331 = 17,331; then 19,000 − 17,331 = 1,669, confirming the result. Why Other Options Are Wrong: 2719 and 2679 overshoot the target; “None of these” is incorrect because 1669 is present; 1696 would lead to a total of 19,027. Common Pitfalls: Carry/borrow slips during addition; misreading digits (e.g., 7777 as 7177); forgetting to recheck by adding back to the total. Final Answer: 1669

Question 12

Solve for the Unknown — Two Subtractions in a Row Find the value of “?” if 4003 − ? − 1599 = 716.

Accepted Answer

Introduction / Context: This linear arithmetic equation involves an unknown subtracted in sequence. Rearranging carefully avoids sign mistakes. Such items appear frequently in aptitude tests to check manipulation of simple equations and attention to order of operations with subtractions. Given Data / Assumptions: Equation: 4003 − ? − 1599 = 716. All numbers are integers. Goal: solve for ?. Concept / Approach: Group known terms first: (4003 − 1599) − ? = 716. Compute the bracketed difference, then isolate ?. Alternatively, move terms to opposite sides while keeping track of signs: 4003 − 1599 − 716 = ?. Both methods yield the same result when done correctly. Step-by-Step Solution: 1) Compute 4003 − 1599 = 2404.2) Now 2404 − ? = 716.3) Rearranged: ? = 2404 − 716.4) Subtract: 2404 − 716 = 1688. Verification / Alternative check: Plug back: 4003 − 1688 − 1599 = (4003 − 1688) − 1599 = 2315 − 1599 = 716, confirming the solution. Why Other Options Are Wrong: 1598 and 1692 come from digit errors in subtraction; 3120 results from misordering operations; “None of these” is wrong because 1688 is correct. Common Pitfalls: Forgetting that subtraction is not associative; performing 1599 − 4003 by mistake; mis-borrowing in the column subtraction. Final Answer: 1688

Question 13

Number system — find the missing value: Make the two expressions equal by finding the single number that satisfies the balance. 36000 − 888 = 37000 − ?

Accepted Answer

Introduction / Context: This is a straightforward number-system balancing problem. We are given two expressions that must be equal and asked to determine the unknown number that keeps both sides balanced. Such questions test comfort with subtraction and the principle that equal transformations on both sides preserve equality. Given Data / Assumptions: Expression 1: 36000 − 888 Expression 2: 37000 − ? The two expressions represent the same value when the correct number replaces the question mark. Concept / Approach: If A = B, then we can compute either side and set them equal. Compute the exact numeric value on the left and then solve for the missing amount on the right. The operations are simple subtractions, so accuracy in arithmetic is the key skill here. Step-by-Step Solution: 1) Evaluate the left side: 36000 − 888 = 35112.2) Set equality: 35112 = 37000 − ?.3) Solve for ?: ? = 37000 − 35112.4) Compute the difference: 37000 − 35112 = 1888. Verification / Alternative check: Substitute the result back: 37000 − 1888 = 35112, which matches 36000 − 888. Therefore the computation is consistent and verified. Why Other Options Are Wrong: 1848: Produces 37000 − 1848 = 35152, which is too large by 40.1978: Produces 37000 − 1978 = 35022, which is too small by 90.2018: Produces 37000 − 2018 = 34982, which is far from 35112. Common Pitfalls: Typical mistakes include subtracting in the wrong order, misplacing a borrow during subtraction, or assuming that the difference between 36000 and 37000 (which is 1000) directly yields the answer without accounting for the additional −888 term. Always compute one side fully and then balance the equation. Final Answer: 1888

Question 14

Compute the missing number so the equation balances: ? + 1046 − 398 − 69 = 999

Accepted Answer

Introduction / Context: This arithmetic problem involves combining constants and isolating the unknown so that both sides of the equation are equal. It assesses accuracy with addition and subtraction, as well as clear organization of steps to avoid small mistakes that can flip a correct answer into an incorrect one. Given Data / Assumptions: Equation: ? + 1046 − 398 − 69 = 999 All terms are integers and standard order of operations applies (left to right for additions/subtractions). Goal: find the single integer replacing the question mark that makes the equation true. Concept / Approach: Combine the known numbers first to simplify. Then, isolate the unknown on one side using inverse operations. Keeping track of signs is crucial because a sign error will lead to a wrong value for the unknown. Step-by-Step Solution: 1) Combine constants: 1046 − 398 − 69.2) First subtraction: 1046 − 398 = 648.3) Second subtraction: 648 − 69 = 579.4) The equation reduces to: ? + 579 = 999, hence ? = 999 − 579 = 420. Verification / Alternative check: Substitute back: 420 + 1046 − 398 − 69 = 420 + 579 = 999. The equality holds, verifying the result. Why Other Options Are Wrong: 514: 514 + 579 = 1093, which overshoots 999 by 94.399: 399 + 579 = 978, which is 21 less than 999.520: 520 + 579 = 1099, which is 100 too large. Common Pitfalls: Mixing the order of subtraction, combining 398 and 69 incorrectly, or forgetting that both are subtracted. Always consolidate constants first; then move the result across the equals sign by subtracting from 999 to isolate the unknown cleanly. Final Answer: 420

Question 15

Find the unknown that satisfies the equation: ? + 6207 − 9038 = 107

Accepted Answer

Introduction / Context: This problem asks you to determine the unknown that balances an equation composed of two subtractions around a single addition. It checks comfort with integer arithmetic and the ability to rearrange equations systematically without introducing sign errors. Given Data / Assumptions: Given relation: ? + 6207 − 9038 = 107 All values are integers. Goal: solve for the unknown represented by the question mark. Concept / Approach: Combine the known constants first to simplify. Recognize that 6207 − 9038 is negative, so the unknown must offset that deficit to achieve 107. Then isolate the unknown by moving the combined constant to the other side of the equation with the opposite sign. Step-by-Step Solution: 1) Compute the constant part: 6207 − 9038 = −2831.2) The equation becomes: ? − 2831 = 107.3) Add 2831 to both sides to isolate the unknown: ? = 107 + 2831.4) Evaluate: 107 + 2831 = 2938. Verification / Alternative check: Substitute the result: 2938 + 6207 − 9038 = 2938 − 2831 = 107. The equality holds, confirming correctness. Why Other Options Are Wrong: 2724: Yields 2724 − 2831 = −107, not 107.2814: Yields 2814 − 2831 = −17, not 107.None of these: Not applicable since 2938 exactly satisfies the equation. Common Pitfalls: Forgetting that 6207 − 9038 is negative; flipping the sign when moving terms; or adding the wrong pair of numbers during the final step. Keep the structure ? − 2831 = 107 clear to avoid mistakes. Final Answer: 2938

Question 16

Solve for the missing value in the linear relation: 3571 + ? − 6086 = 155

Accepted Answer

Introduction / Context: This linear arithmetic equation includes one unknown and a combination of addition and subtraction. The task is to isolate the unknown systematically, which reinforces the core principle that we can perform equal operations on both sides without changing the equality. Given Data / Assumptions: Equation: 3571 + ? − 6086 = 155 All numbers are integers. We seek the exact integer value of the question mark. Concept / Approach: Group the constant terms from the left side to determine how much the unknown must contribute to reach the target on the right. Concretely, move 3571 and −6086 to the right as a combined constant, or combine them first on the left and then solve in one step. Step-by-Step Solution: 1) Combine the known constants on the left: 3571 − 6086 = −2515.2) The equation becomes: ? − 2515 = 155.3) Add 2515 to both sides to isolate the unknown: ? = 155 + 2515.4) Evaluate: 155 + 2515 = 2670. Verification / Alternative check: Substitute: 3571 + 2670 − 6086 = 6241 − 6086 = 155, which verifies the solution. Why Other Options Are Wrong: 2630: Gives 2630 − 2515 = 115, short by 40 from 155.2400: Gives 2400 − 2515 = −115, far from 155.2515: Confuses the magnitude; it would yield 0 on the reduced form, not 155. Common Pitfalls: Dropping a negative when combining 3571 and 6086, or mistakenly moving 6086 to the other side with the wrong sign. Writing the reduced equation ? − 2515 = 155 explicitly helps avoid errors. Final Answer: 2670

Question 17

Evaluate the expression step by step and compute the final value: 36800 − 9999 − 3333 − 66 = ?

Accepted Answer

Introduction / Context: This problem is a direct evaluation of a chain of subtractions. Although the arithmetic is basic, accuracy is vital because each subtraction influences the next. Systematic computation prevents cumulative errors that might appear small but lead to a wrong final answer. Given Data / Assumptions: Expression: 36800 − 9999 − 3333 − 66 All operations are subtractions executed left to right. No hidden grouping or special precedence beyond standard rules. Concept / Approach: Evaluate sequentially from left to right. After each subtraction, write down the intermediate result before proceeding. This organized approach helps to catch digit-borrow errors and keeps the running total clear. Step-by-Step Solution: 1) First subtraction: 36800 − 9999 = 26801.2) Second subtraction: 26801 − 3333 = 23468.3) Third subtraction: 23468 − 66 = 23402.4) Therefore, the final result is 23402. Verification / Alternative check: Add back the subtracted amounts to 23402: 23402 + 66 + 3333 + 9999 = 36800, confirming no arithmetic slips occurred. Why Other Options Are Wrong: 27912: Suggests an incorrect intermediate subtraction or reversed operation.24102: Off by 700, likely from mis-subtracting 3333 or 66.30200: Far from the correct value, possibly a single-step approximation rather than sequential computation. Common Pitfalls: Borrow mistakes when subtracting 9999; skipping an intermediate result; mixing the order of the last two subtractions. Carefully recording each intermediate value avoids these errors. Final Answer: 23402

Question 18

Solve for the unknown in the simple linear equation: ? − 4799 = 1714

Accepted Answer

Introduction / Context: Here we have a one-step linear equation with a single unknown. The problem reinforces adding the same number to both sides to isolate the unknown. This is a fundamental algebraic operation used throughout quantitative reasoning. Given Data / Assumptions: Equation: ? − 4799 = 1714 All values are integers. We need the exact value of the question mark. Concept / Approach: To undo subtraction of 4799 from the unknown, add 4799 to both sides. This isolates the unknown cleanly. Keep signs correct to avoid a result that is too small or negative by mistake. Step-by-Step Solution: 1) Start with ? − 4799 = 1714.2) Add 4799 to both sides: ? = 1714 + 4799.3) Compute the sum: 1714 + 4799 = 6513.4) Therefore, the unknown is 6513. Verification / Alternative check: Substitute back: 6513 − 4799 = 1714, confirming the equality holds exactly. Why Other Options Are Wrong: 7403: Too large by 890 relative to the correct sum.3085: Much too small; this would give a negative difference after subtraction.None of these: Not applicable because 6513 satisfies the equation perfectly. Common Pitfalls: Accidentally subtracting 4799 from 1714 or flipping signs. Always apply the inverse operation to both sides consistently. Final Answer: 6513

Question 19

Determine the number subtracted from 9501 to obtain 3697: 9501 − ? = 3697

Accepted Answer

Introduction / Context: You are given the result of a subtraction and asked to find the subtrahend. This emphasizes inverse operations and clean arithmetic. By solving for the missing number, you practice a core technique used in checking balances and reconciling accounts. Given Data / Assumptions: Known: 9501 − ? = 3697 All numbers are integers. We need the exact value of the question mark. Concept / Approach: Rearrange the equation to isolate the unknown. Subtracting a number to reach a smaller result implies the unknown equals the difference between the initial number and the final result. So compute the difference directly. Step-by-Step Solution: 1) Start with 9501 − ? = 3697.2) Move terms to isolate ?: ? = 9501 − 3697.3) Compute: 9501 − 3697 = 5804.4) Thus, the subtracted value is 5804. Verification / Alternative check: Substitute back: 9501 − 5804 = 3697. The equality checks out, confirming the arithmetic. Why Other Options Are Wrong: 4894: Gives 9501 − 4894 = 4607, not 3697.5814: Off by 10; 9501 − 5814 = 3687.13198: Larger than the minuend; subtraction would be negative. Common Pitfalls: Reversing the subtraction (e.g., 3697 − 9501) or mistyping digits when borrowing. Always compute 9501 − 3697 carefully to avoid a one-digit error. Final Answer: 5804

Question 20

Add the series of aligned numbers and report the exact sum: 66066 + 6606 + 66 + 6 = ?

Accepted Answer

Introduction / Context: This is an addition exercise with conveniently patterned numbers. Accuracy in column addition is essential. Recognizing the structure (each term is a sequence of sixes with different place values) can speed up mental computation while keeping errors at bay. Given Data / Assumptions: Numbers: 66066, 6606, 66, 6 We need the exact arithmetic sum. No rounding or estimation is required. Concept / Approach: Add the largest two numbers first, then successively add the smaller terms. Keeping a running total helps prevent misalignment of place values. Alternatively, note that 66066 + 6606 groups naturally because of similar digit patterns, making the first subtotal easy to verify. Step-by-Step Solution: 1) 66066 + 6606 = 72672.2) 72672 + 66 = 72738.3) 72738 + 6 = 72744.4) Therefore, the final sum is 72744. Verification / Alternative check: Reverse-check by subtracting the smaller addends from the final sum: 72744 − 6 − 66 − 6606 = 66066, confirming the computation. Why Other Options Are Wrong: 72798: Off by 54, likely from mishandling the step with 66.258126: Orders of magnitude too large; not consistent with the addends.None of these: Not applicable since 72744 is exact. Common Pitfalls: Misaligning place values when adding 6606 to 66066; dropping the carry when adding 66; or skipping the last +6 term. Writing each subtotal prevents these errors. Final Answer: 72744

Number System Questions