Q: Which of the following numbers is divisible by 45?

Given data Test divisibility by 45. Concept / Approach Since 45 = 5 × 9 and gcd(5, 9) = 1, a number is divisible by 45 iff it is divisible by 5 and by 9. Divisible by 5 ⇒ last digit is 0 or 5. Divisible by 9 ⇒ sum of digits is a multiple of 9. Check each option 2700: last digit 0 (✓/5); sum = 2+7+0+0 = 9 (✓/9) ⇒ divisible by 45.1895: last digit 5 (✓/5); sum = 1+8+9+5 = 23 (✗/9) ⇒ not by 45.2235: last digit 5 (✓/5); sum = 2+2+3+5 = 12 (✗/9) ⇒ not by 45.2510: last digit 0 (✓/5); sum = 2+5+1+0 = 8 (✗/9) ⇒ not by 45. Final Answer Only 2700 is divisible by 45.

Q: If 1400 × x = 1050, find x.

Given data & note Interpreting the symbol 'x' between numbers as multiplication (i.e., 1400 × x = 1050), which is the standard reading in such items. Concept / Approach Solve the linear equation by isolating x. Step-by-step calculation x = 1050 / 1400= 105 / 140 = 21 / 28 = 3/4 = 0.75 Verification 1400 × (3/4) = 1400 × 0.75 = 1050 ✔ Common pitfalls Misreading as 1400 ÷ x = 1050; the original uses 'x' as the multiplication sign. Final Answer 3/4

Q: The sum of two numbers is 12 and their product is 35. Find the sum of their reciprocals.

Given data a + b = 12 ab = 35 Concept / Approach Use the identity: 1/a + 1/b = (a + b)/ab. Step-by-step calculation 1/a + 1/b = (a + b) / (ab) = 12 / 35Therefore, the required sum is 12/35. Verification Numbers satisfying a + b = 12 and ab = 35 are 5 and 7. Then 1/5 + 1/7 = (7 + 5)/35 = 12/35 ✔ Common pitfalls Adding reciprocals as 1/(a + b); that is incorrect. Final Answer 12/35

Q: Evaluate: (854^3 − 276^3) ÷ (854^2 + 854×276 + 276^2).

Given data Compute (a^3 − b^3)/(a^2 + ab + b^2) with a = 854, b = 276. Concept / Approach Use the identity: a^3 − b^3 = (a − b)(a^2 + ab + b^2). Step-by-step calculation (a^3 − b^3)/(a^2 + ab + b^2) = a − b= 854 − 276 = 578 Verification Identity guarantees exact cancellation; no large-number expansion needed. Common pitfalls Expanding cubes numerically, which is unnecessary and error-prone. Final Answer 578

Q: Find the sum of the first 45 natural numbers.

Given data We need S = 1 + 2 + … + 45. Concept / Approach Use the standard formula for the sum of the first n natural numbers: S = n(n + 1)/2. Step-by-step calculation S = 45 × 46 / 2= 45 × 23= (40 × 23) + (5 × 23) = 920 + 115 = 1035 Verification Pairing method: (1+45), (2+44), … gives 22 pairs of 46 plus the middle 23 ⇒ 22×46 + 23 = 1012 + 23 = 1035. Common pitfalls Using n(n − 1)/2 by mistake (that is for 1 to n−1). Final Answer 1035

Q: If n is a whole number with remainder 3 upon division by 4, what is the remainder when 2n is divided by 4?

Given data n ≡ 3 (mod 4) Concept / Approach Multiply the congruence by 2 and reduce modulo 4. Step-by-step calculation 2n ≡ 2 × 3 ≡ 6 (mod 4)6 ≡ 2 (mod 4) Verification Example check: n = 3 ⇒ 2n = 6 ⇒ 6 ÷ 4 leaves remainder 2. Common pitfalls Assuming doubling preserves the same remainder; it does not necessarily under modular reduction. Final Answer 2

Q: Find the unit digit of (4137)^754.

Given data Base ends with 7: 4137 → units digit behaves like powers of 7. Concept / Approach The units digit of 7^k cycles with period 4: 7, 9, 3, 1, … Reduce the exponent modulo 4 to select the cycle position. Step-by-step calculation 754 ÷ 4 = 188 remainder 2 ⇒ 754 ≡ 2 (mod 4)Cycle position 2 corresponds to units digit 9 (since 7^1→7, 7^2→9). Verification 7^2 = 49 (units 9), and the cycle repeats every 4 powers; remainder 2 always maps to units 9. Common pitfalls Using the full number instead of just the units digit. Forgetting that the cycle length for 7 is 4, not something else. Final Answer 9

Q: How many 3-digit natural numbers are divisible by 6?

Given data 3-digit numbers range from 100 to 999. Divisible by 6 ⇔ divisible by 2 and 3. Concept / Approach Count multiples of 6 in an inclusive interval using arithmetic progression logic. Step-by-step calculation Smallest 3-digit multiple of 6: 102 (since 100 is not; 102 = 6×17)Largest 3-digit multiple of 6: 996 (since 999 is not; 996 = 6×166)Count = ((996 − 102)/6) + 1 = (894/6) + 1 = 149 + 1 = 150 Alternative formula Count = ⌊999/6⌋ − ⌊99/6⌋ = 166 − 16 = 150. Common pitfalls Accidentally including 100 or 999. Forgetting the +1 when using the AP term-count formula. Final Answer 150

Question 1

Find ?: 7589 − ? = 3434

Accepted Answer

Given data Equation: 7589 − ? = 3434 Concept / Approach Isolate the unknown by transposition: ? = 7589 − 3434. Step-by-step calculation 7589 − 3000 = 45894589 − 400 = 41894189 − 34 = 4155Therefore, ? = 4155 Verification 7589 − 4155 = 3434 ✔ Common pitfalls Reversing the subtraction to compute 3434 − 7589 by mistake. Keep the original order: minuend − subtrahend. Final Answer 4155

Question 2

The 3-digit number 4a3 added to 984 gives 13b7, which is divisible by 11. Find (a + b).

Accepted Answer

Given data 4a3 + 984 = 13b7 (four-digit result) 13b7 is divisible by 11 Concept / Approach Use column-wise addition with carries. Apply divisibility rule for 11: (sum of digits in odd positions) − (sum in even positions) must be a multiple of 11 (i.e., 0, ±11, …). Step-by-step calculation Ones: 3 + 4 = 7 ⇒ matches the ones digit 7, carry = 0.Tens: a + 8 (+0) = b. For now, b = a + 8, with carrytens = 0 (since a ∈ {0,1} to keep sum < 10; see next line).Hundreds: 4 + 9 + carrytens must produce hundreds digit 3 with a carry 1 to thousands. So 4 + 9 + carrytens = 13 ⇒ carrytens = 0, hence a ∈ {0,1} and b = a + 8.Check a = 0 ⇒ 13b7 = 1387. Divisibility by 11: (1 + 8) − (3 + 7) = 9 − 10 = −1 ≠ 0 ⇒ reject.Check a = 1 ⇒ b = 9 ⇒ 13b7 = 1397. Divisibility by 11: (1 + 9) − (3 + 7) = 10 − 10 = 0 ⇒ divisible. Result a = 1, b = 9 ⇒ a + b = 10. Verification 413 + 984 = 1397 ✓ and 1397 is a multiple of 11. Common pitfalls Applying the divisibility-by-11 rule incorrectly (ensure correct positional sums). Forgetting that the tens column must not exceed 9 to keep the carry 0 as required by the hundreds column constraint. Final Answer 10

Question 3

Which of the following numbers is divisible by 45?

Accepted Answer

Given data Test divisibility by 45. Concept / Approach Since 45 = 5 × 9 and gcd(5, 9) = 1, a number is divisible by 45 iff it is divisible by 5 and by 9. Divisible by 5 ⇒ last digit is 0 or 5. Divisible by 9 ⇒ sum of digits is a multiple of 9. Check each option 2700: last digit 0 (✓/5); sum = 2+7+0+0 = 9 (✓/9) ⇒ divisible by 45.1895: last digit 5 (✓/5); sum = 1+8+9+5 = 23 (✗/9) ⇒ not by 45.2235: last digit 5 (✓/5); sum = 2+2+3+5 = 12 (✗/9) ⇒ not by 45.2510: last digit 0 (✓/5); sum = 2+5+1+0 = 8 (✗/9) ⇒ not by 45. Final Answer Only 2700 is divisible by 45.

Question 4

If 1400 × x = 1050, find x.

Accepted Answer

Given data & note Interpreting the symbol 'x' between numbers as multiplication (i.e., 1400 × x = 1050), which is the standard reading in such items. Concept / Approach Solve the linear equation by isolating x. Step-by-step calculation x = 1050 / 1400= 105 / 140 = 21 / 28 = 3/4 = 0.75 Verification 1400 × (3/4) = 1400 × 0.75 = 1050 ✔ Common pitfalls Misreading as 1400 ÷ x = 1050; the original uses 'x' as the multiplication sign. Final Answer 3/4

Question 5

The sum of two numbers is 12 and their product is 35. Find the sum of their reciprocals.

Accepted Answer

Given data a + b = 12 ab = 35 Concept / Approach Use the identity: 1/a + 1/b = (a + b)/ab. Step-by-step calculation 1/a + 1/b = (a + b) / (ab) = 12 / 35Therefore, the required sum is 12/35. Verification Numbers satisfying a + b = 12 and ab = 35 are 5 and 7. Then 1/5 + 1/7 = (7 + 5)/35 = 12/35 ✔ Common pitfalls Adding reciprocals as 1/(a + b); that is incorrect. Final Answer 12/35

Question 6

Evaluate: (854^3 − 276^3) ÷ (854^2 + 854×276 + 276^2).

Accepted Answer

Given data Compute (a^3 − b^3)/(a^2 + ab + b^2) with a = 854, b = 276. Concept / Approach Use the identity: a^3 − b^3 = (a − b)(a^2 + ab + b^2). Step-by-step calculation (a^3 − b^3)/(a^2 + ab + b^2) = a − b= 854 − 276 = 578 Verification Identity guarantees exact cancellation; no large-number expansion needed. Common pitfalls Expanding cubes numerically, which is unnecessary and error-prone. Final Answer 578

Question 7

Find the sum of the first 45 natural numbers.

Accepted Answer

Given data We need S = 1 + 2 + … + 45. Concept / Approach Use the standard formula for the sum of the first n natural numbers: S = n(n + 1)/2. Step-by-step calculation S = 45 × 46 / 2= 45 × 23= (40 × 23) + (5 × 23) = 920 + 115 = 1035 Verification Pairing method: (1+45), (2+44), … gives 22 pairs of 46 plus the middle 23 ⇒ 22×46 + 23 = 1012 + 23 = 1035. Common pitfalls Using n(n − 1)/2 by mistake (that is for 1 to n−1). Final Answer 1035

Question 8

If n is a whole number with remainder 3 upon division by 4, what is the remainder when 2n is divided by 4?

Accepted Answer

Given data n ≡ 3 (mod 4) Concept / Approach Multiply the congruence by 2 and reduce modulo 4. Step-by-step calculation 2n ≡ 2 × 3 ≡ 6 (mod 4)6 ≡ 2 (mod 4) Verification Example check: n = 3 ⇒ 2n = 6 ⇒ 6 ÷ 4 leaves remainder 2. Common pitfalls Assuming doubling preserves the same remainder; it does not necessarily under modular reduction. Final Answer 2

Question 9

Find the unit digit of (4137)^754.

Accepted Answer

Given data Base ends with 7: 4137 → units digit behaves like powers of 7. Concept / Approach The units digit of 7^k cycles with period 4: 7, 9, 3, 1, … Reduce the exponent modulo 4 to select the cycle position. Step-by-step calculation 754 ÷ 4 = 188 remainder 2 ⇒ 754 ≡ 2 (mod 4)Cycle position 2 corresponds to units digit 9 (since 7^1→7, 7^2→9). Verification 7^2 = 49 (units 9), and the cycle repeats every 4 powers; remainder 2 always maps to units 9. Common pitfalls Using the full number instead of just the units digit. Forgetting that the cycle length for 7 is 4, not something else. Final Answer 9

Question 10

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

Accepted Answer

Given data 3-digit numbers range from 100 to 999. Divisible by 6 ⇔ divisible by 2 and 3. Concept / Approach Count multiples of 6 in an inclusive interval using arithmetic progression logic. Step-by-step calculation Smallest 3-digit multiple of 6: 102 (since 100 is not; 102 = 6×17)Largest 3-digit multiple of 6: 996 (since 999 is not; 996 = 6×166)Count = ((996 − 102)/6) + 1 = (894/6) + 1 = 149 + 1 = 150 Alternative formula Count = ⌊999/6⌋ − ⌊99/6⌋ = 166 − 16 = 150. Common pitfalls Accidentally including 100 or 999. Forgetting the +1 when using the AP term-count formula. Final Answer 150

Question 11

Find the unit digit of (6374)^1793 × (625)^317 × (341)^491.

Accepted Answer

Given data We need only the unit digit of (6374)^1793 × (625)^317 × (341)^491. Assumption clarified: the notation (6374)1793 etc. denotes exponents, i.e., (6374)^1793, (625)^317, (341)^491. Concept / Approach The unit digit of a power depends only on the unit digit of the base and the cycle of that unit digit. Unit digit cycles: 4 → (4, 6, 4, 6, …) period 2; 5 → always 5; 1 → always 1. Step-by-step calculation Units((6374)^1793) = Units(4^1793). Since 1793 is odd ⇒ units = 4.Units((625)^317) = 5 (any positive power of a number ending with 5 ends with 5).Units((341)^491) = 1 (any positive power of a number ending with 1 ends with 1).Product's unit digit = Units(4 × 5 × 1) = Units(20) = 0. Verification Because one factor contributes a terminal 5 and another contributes an even digit (4), the product ends with 0 regardless of other digits. Common pitfalls Trying to compute large powers directly instead of using unit-digit cycles. Misreading the notation as multiplication rather than exponentiation. Final Answer 0

Question 12

(x^n − a^n) is divisible by (x − a) when n equals…?

Accepted Answer

Given data Divisibility of x^n − a^n by x − a. Concept / Approach Factorization identity: x^n − a^n = (x − a)(x^{n−1} + x^{n−2}a + ⋯ + xa^{n−2} + a^{n−1}) The identity holds for all integers n ≥ 1 (natural numbers). Step-by-step reasoning Since x^n − a^n has (x − a) as a factor for any n ≥ 1, the quotient is the (n−1)-term geometric sum shown above. Verification Check n = 1: x − a is trivially divisible by x − a.Check n = 2: x^2 − a^2 = (x − a)(x + a). Common pitfalls Thinking it holds only for even or odd n; in fact it holds for every positive integer n. Final Answer Any positive integer n.

Question 13

Evaluate: 2^2 + 4^2 + 6^2 + ⋯ + 20^2.

Accepted Answer

Given data Sum of squares of the first 10 even numbers: 2, 4, …, 20. Concept / Approach Factor out 4 from each square: (2k)^2 = 4k^2. Use the sum of squares formula: 1^2 + 2^2 + ⋯ + n^2 = n(n + 1)(2n + 1)/6. Step-by-step calculation S = 2^2 + 4^2 + ⋯ + 20^2 = 4(1^2 + 2^2 + ⋯ + 10^2)= 4 × [10 × 11 × 21 / 6]= 4 × 385 = 1540 Verification Compute a few terms: 4 + 16 + 36 + ⋯ + 400; the formula-based result matches known tabulated sums. Common pitfalls Accidentally summing 2 + 4 + ⋯ + 20 (not squared). Using n = 20 in the squares formula directly without halving to 10 even indices. Final Answer 1540

Question 14

Compute: 8796 × 223 + 8796 × 77.

Accepted Answer

Given data Expression: 8796 × 223 + 8796 × 77. Concept / Approach Use distributive property: a·b + a·c = a(b + c). Step-by-step calculation 8796 × (223 + 77) = 8796 × 300= 8796 × 3 × 100 = 26,388 × 100 = 2,638,800 Verification Direct multiplication of each term and addition gives the same total. Common pitfalls Arithmetic slips when multiplying by 300; remember ×300 = ×3 then append two zeros. Final Answer 2638800

Question 15

How many of these are divisible by 3 but not by 9: 2133, 2343, 3474, 4131, 5286, 5340, 6336, 7347, 8115, 9276?

Accepted Answer

Given data We must count numbers divisible by 3 but not by 9. Concept / Approach Divisible by 3 ⇔ digit sum is a multiple of 3. Divisible by 9 ⇔ digit sum is a multiple of 9. So we want digit sum ≡ 0 (mod 3) but ≠ 0 (mod 9). Check each number 2133: sum = 9 ⇒ divisible by 9 ⇒ exclude.2343: sum = 12 ⇒ /3 yes, /9 no ⇒ include.3474: sum = 18 ⇒ /9 yes ⇒ exclude.4131: sum = 9 ⇒ /9 yes ⇒ exclude.5286: sum = 21 ⇒ /3 yes, /9 no ⇒ include.5340: sum = 12 ⇒ /3 yes, /9 no ⇒ include.6336: sum = 18 ⇒ /9 yes ⇒ exclude.7347: sum = 21 ⇒ /3 yes, /9 no ⇒ include.8115: sum = 15 ⇒ /3 yes, /9 no ⇒ include.9276: sum = 24 ⇒ /3 yes, /9 no ⇒ include. Total Included: {2343, 5286, 5340, 7347, 8115, 9276} ⇒ count = 6. Common pitfalls Assuming “divisible by 3” automatically means “divisible by 9.” It does not. Final Answer 6

Question 16

Evaluate: 12345679 × 72.

Accepted Answer

Given data Compute 12345679 × 72. Concept / Approach Use the classic pattern: 12345679 × 9 = 111111111. Then scale appropriately. Since 72 = 8 × 9, we can compute sequentially. Step-by-step calculation 12345679 × 9 = 111111111111111111 × 8 = 888888888 Verification Direct multiplication (calculator or long multiplication) matches the patterned result. Common pitfalls Multiplying by 72 directly and making arithmetic errors; using the 9-multiple pattern simplifies. Final Answer 888888888

Question 17

If 97215*6 is divisible by 11, find the smallest digit that can replace .

Accepted Answer

Given data Number: 9 7 2 1 5 * 6 must be divisible by 11. Concept / Approach Divisibility by 11: (sum of digits in odd positions) − (sum in even positions) ≡ 0 (mod 11). Positions from left: 1:9, 2:7, 3:2, 4:1, 5:5, 6:, 7:6. Step-by-step calculation Odd sum = 9 + 2 + 5 + 6 = 22Even sum = 7 + 1 + * = 8 + *Difference = 22 − (8 + *) = 14 − *Set ≡ 0 (mod 11) ⇒ 14 − * ≡ 0 ⇒ * ≡ 14 ≡ 3 (mod 11)Smallest digit solution is 3. Verification With * = 3: difference = 14 − 3 = 11 ⇒ divisible by 11. Common pitfalls Summing positions incorrectly (ensure consistent left-to-right odd/even indexing). Final Answer 3

Question 18

What is the smallest number to add to 4456 so the sum is divisible by 6?

Accepted Answer

Given data We want the least k ≥ 0 such that 4456 + k is divisible by 6. Concept / Approach Divisible by 6 ⇔ divisible by 2 and 3. 4456 is even ⇒ divisible by 2 already; adjust k to satisfy divisibility by 3. Step-by-step calculation Digit sum of 4456 = 4 + 4 + 5 + 6 = 19We require 19 + k ≡ 0 (mod 3)Next multiple of 3 after 19 is 21 ⇒ k = 2Check parity: 4456 + 2 = 4458 (even). Both conditions met.Therefore, k = 2. Verification 4458 ÷ 6 = 743 exactly. Common pitfalls Making the sum divisible by 3 but turning it odd; verify both divisibility-by-2 and by-3. Final Answer 2

Question 19

Find the difference between the place value and the face value of 7 in 32675149.

Accepted Answer

Given data Number: 32675149; target digit: 7. Concept / Approach Face value of a digit is the digit itself. Place value = digit × (place weight). Identify the position of 7. Step-by-step identification From right, digits and weights: 9(10^0), 4(10^1), 1(10^2), 5(10^3), 7(10^4), 6(10^5), 2(10^6), 3(10^7).Thus, place value of 7 = 7 × 10^4 = 70,000.Face value of 7 = 7.Difference = 70,000 − 7 = 69,993. Verification Position check confirms 7 is in the ten-thousands place. Common pitfalls Confusing “local/place value” with “face value,” or miscounting the position from the right. Final Answer 69993

Question 20

Evaluate: 11^2 + 12^2 + 13^2 + ⋯ + 20^2.

Accepted Answer

Given data Sum of squares from 11 to 20. Concept / Approach Use the closed form for 1^2 + 2^2 + ⋯ + n^2, then subtract 1^2 + ⋯ + 10^2. Formula: 1^2 + ⋯ + n^2 = n(n + 1)(2n + 1)/6. Step-by-step calculation Sum(1→20) = 20 × 21 × 41 / 620 × 21 = 420; 420 × 41 = 17,220; divide by 6 ⇒ 2,870Sum(1→10) = 10 × 11 × 21 / 6 = 385Required sum = 2,870 − 385 = 2,485 Verification Spot-check: 20^2 = 400, 19^2 = 361, …; partial sums align with the computed total. Common pitfalls Forgetting to subtract the first 10 squares. Arithmetic errors when multiplying 420 × 41. Final Answer 2485

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