Question 1

Which digit should replace * in 5*3 to make it divisible by 3?

Accepted Answer

Step 1: A number is divisible by 3 if the sum of its digits is divisible by 3. Step 2: Sum of known digits = 5 + 3 = 8. Step 3: We need 8 + * to be divisible by 3. Step 4: If * = 1, sum = 9, which is divisible by 3. Therefore, * = 1.

Question 2

A number is divisible by 6 if it is divisible by both 2 and 3. Which of the following is divisible by 6?

Accepted Answer

Step 1: For divisibility by 6, the number must be even (divisible by 2) AND the sum of digits must be divisible by 3. Step 2: Test 318: It is even (ends in 8). Sum of digits = 3 + 1 + 8 = 12, which is divisible by 3. Step 3: Since 318 is divisible by both 2 and 3, it is divisible by 6. Therefore, 318 is the answer.

Question 3

Which of the following numbers is NOT divisible by 8?

Accepted Answer

Step 1: A number is divisible by 8 if its last three digits form a number divisible by 8. Step 2: Test 2456: Last three digits = 456. 456 ÷ 8 = 57 (exactly divisible). Step 3: Test 3264: Last three digits = 264. 264 ÷ 8 = 33 (exactly divisible). Step 4: Test 5432: Last three digits = 432. 432 ÷ 8 = 54 (exactly divisible). Step 5: Test 6218: Last three digits = 218. 218 ÷ 8 = 27.25 (NOT exactly divisible). Therefore, 6218 is NOT divisible by 8.

Question 4

A number is divisible by 11 if the difference between the sum of digits at odd positions and the sum of digits at even positions is divisible by 11. Using this rule, which number is divisible by 11?

Accepted Answer

Step 1: For divisibility by 11, calculate (sum of odd-position digits) − (sum of even-position digits). Step 2: Test 4587: Odd positions (1st, 3rd) = 4 + 8 = 12. Even positions (2nd, 4th) = 5 + 7 = 12. Difference = 12 − 12 = 0. Step 3: 0 is divisible by 11. Therefore, 4587 is divisible by 11.

Question 5

A number when divided by 9 leaves remainder 0. Which of the following is definitely true about this number?

Accepted Answer

Step 1: If a number is divisible by 9 (remainder 0), then by the divisibility rule of 9, the sum of its digits must be divisible by 9. Step 2: A number divisible by 9 is also divisible by 3 (since 9 = 3²). Step 3: Therefore, the number must be divisible by 3. The answer is that the number is divisible by 3.

Question 6

Which of the following numbers is divisible by both 4 and 6?

Accepted Answer

Step 1: A number divisible by both 4 and 6 must be divisible by their LCM. Step 2: LCM(4, 6) = 12. Step 3: Check 144: 144 ÷ 12 = 12 (exactly divisible). Therefore, 144 is the answer.

Question 7

A number is divisible by 6. Which statement must be true?

Accepted Answer

Step 1: A number is divisible by 6 if and only if it is divisible by both 2 and 3 (since 6 = 2×3 and gcd(2,3)=1). Step 2: Divisibility by 2 requires the number to be even. Step 3: Divisibility by 3 requires the sum of digits to be divisible by 3. Step 4: Therefore, the number must be even AND have digit sum divisible by 3. The correct statement is: 'It is divisible by both 2 and 3.'

Question 8

Which of the following numbers is divisible by 11?

Accepted Answer

Step 1: Apply divisibility rule for 11: alternating sum of digits (from right to left) must be divisible by 11. Step 2: Test 121: (1-2+1) = 0, divisible by 11✓. Test 1234: (4-3+2-1) = 2, not divisible by 11. Test 5678: (8-7+6-5) = 2, not divisible by 11. Test 9876: (6-8+7-9) = -4, not divisible by 11. Therefore, 121 is divisible by 11.

Question 9

A number when divided by 11 leaves remainder 7. What is the remainder when the square of this number is divided by 11?

Accepted Answer

Step 1: Let the number be N. Given: N ≡ 7 (mod 11). Step 2: We need N² (mod 11). Step 3: N² ≡ 7² ≡ 49 (mod 11). Step 4: 49 = 11×4 + 5, so 49 ≡ 5 (mod 11). Therefore, remainder is 5.

Question 10

How many numbers between 100 and 300 are divisible by 7 but NOT divisible by 3?

Accepted Answer

Step 1: Count multiples of 7 between 100 and 300. Smallest: 105 (7×15), Largest: 294 (7×42). Count = 42-15+1 = 28. Step 2: Count multiples of both 7 and 3 (i.e., multiples of 21) between 100 and 300. Smallest: 105 (21×5), Largest: 294 (21×14). Count = 14-5+1 = 10. Step 3: Numbers divisible by 7 but NOT by 3 = 28 - 10 = 18.

Question 11

A number is divisible by both 4 and 9. Which of the following must also divide this number?

Accepted Answer

Step 1: A number divisible by both 4 and 9 must be divisible by their LCM. Step 2: Since gcd(4,9) = 1, LCM(4,9) = 4 × 9 = 36. Step 3: Therefore, the number must be divisible by 36.

Question 12

A number when divided by 11 leaves remainder 7. When the same number is divided by 13, it leaves remainder 9. If the number is also divisible by 7, find the smallest such number greater than 500.

Accepted Answer

Step 1: Let the number be N. We have N ≡ 7 (mod 11), N ≡ 9 (mod 13), and N ≡ 0 (mod 7). Step 2: From N ≡ 7 (mod 11), write N = 11k + 7. Step 3: Substitute into N ≡ 9 (mod 13): 11k + 7 ≡ 9 (mod 13) → 11k ≡ 2 (mod 13). Step 4: Since 11 × 6 ≡ 1 (mod 13), multiply by 6: k ≡ 12 (mod 13), so k = 13m + 12. Step 5: Thus N = 11(13m + 12) + 7 = 143m + 139. Step 6: Now apply N ≡ 0 (mod 7): 143m + 139 ≡ 0 (mod 7). Since 143 ≡ 3 (mod 7) and 139 ≡ 6 (mod 7), we get 3m + 6 ≡ 0 (mod 7) → 3m ≡ 1 (mod 7) → m ≡ 5 (mod 7). Step 7: So m = 7j + 5, giving N = 143(7j + 5) + 139 = 1001j + 854. Step 8: For N > 500, the smallest value is 854.

Question 13

A six-digit number 5a7b8c is divisible by both 8 and 11. If a, b, c are single digits and a > b > c, find the maximum value of a + b + c.

Accepted Answer

Step 1: For divisibility by 8, the last three digits b8c must form a number divisible by 8. Step 2: For divisibility by 11, apply the alternating sum rule: (5 + 7 + 8) - (a + b + c) ≡ 0 (mod 11) → 20 - (a + b + c) ≡ 0 (mod 11) → a + b + c ≡ 9 (mod 11). Step 3: Since a, b, c are single digits with a > b > c ≥ 0, the maximum possible sum is 9 + 8 + 7 = 24. Step 4: Values of a + b + c that satisfy a + b + c ≡ 9 (mod 11) are 9 or 20. Step 5: Check if a + b + c = 20 is achievable: we need a > b > c with a + b + c = 20. Try a = 9, b = 8, c = 3: then b8c = 883. Check: 883 ÷ 8 = 110.375 (not divisible). Step 6: Try a = 9, b = 7, c = 4: then b8c = 784. Check: 784 ÷ 8 = 98 ✓. Verify 11: 20 - 20 = 0 ≡ 0 (mod 11) ✓. Step 7: Therefore, maximum a + b + c = 20.

Question 14

A three-digit number 5ab is divisible by 6. When divided by 7, it leaves remainder 4. When divided by 13, it leaves remainder 8. If a and b are distinct single digits, find the number of valid pairs (a, b).

Accepted Answer

Step 1: For divisibility by 6, the number must be divisible by both 2 and 3. Step 2: Divisibility by 2: b must be even, so b ∈ {0,2,4,6,8}. Step 3: Divisibility by 3: 5+a+b ≡ 0 (mod 3) → a+b ≡ 1 (mod 3). Step 4: For remainder 4 when divided by 7: 500+10a+b ≡ 4 (mod 7). Since 500 ≡ 3 (mod 7) and 10 ≡ 3 (mod 7), we get 3+3a+b ≡ 4 (mod 7) → 3a+b ≡ 1 (mod 7). Step 5: For remainder 8 when divided by 13: 500+10a+b ≡ 8 (mod 13). Since 500 ≡ 5 (mod 13) and 10 ≡ 10 (mod 13), we get 5+10a+b ≡ 8 (mod 13) → 10a+b ≡ 3 (mod 13). Step 6: Now systematically check b ∈ {0,2,4,6,8} with a ∈ {0-9}, a ≠ b. For b=0: a+0 ≡ 1 (mod 3) → a ∈ {1,4,7}. Check 3a+0 ≡ 1 (mod 7): a=5 fails (3×5=15≡1 mod 7 ✓ but a∉{1,4,7}). Check a=1: 3+0=3 ≢ 1 (mod 7). Check a=4: 12≡5 (mod 7) ✗. Check a=7: 21≡0 (mod 7) ✗. For b=2: a+2 ≡ 1 (mod 3) → a ∈ {2,5,8}. But a ≠ b, so a ∈ {5,8}. Check a=5: 3×5+2=17≡3 (mod 7) ✗. Check a=8: 3×8+2=26≡5 (mod 7) ✗. For b=4: a+4 ≡ 1 (mod 3) → a ∈ {0,3,6,9}. Check a=0: 0+4=4 ≢ 1 (mod 7) ✗. Check a=3: 9+4=13≡6 (mod 7) ✗. Check a=6: 18+4=22≡1 (mod 7) ✓. Now check mod 13: 10×6+4=64≡12 (mod 13) ✗. Check a=9: 27+4=31≡3 (mod 7) ✗. For b=6: a+6 ≡ 1 (mod 3) → a ∈ {1,4,7}. Check a=1: 3+6=9≡2 (mod 7) ✗. Check a=4: 12+6=18≡4 (mod 7) ✗. Check a=7: 21+6=27≡6 (mod 7) ✗. For b=8: a+8 ≡ 1 (mod 3) → a ∈ {0,3,6,9}. Check a=0: 0+8=8≡1 (mod 7) ✓. Check mod 13: 10×0+8=8 ≢ 3 (mod 13) ✗. Check a=3: 9+8=17≡3 (mod 7) ✗. Check a=6: 18+8=26≡5 (mod 7) ✗. Check a=9: 27+8=35≡0 (mod 7) ✗. Step 7: After systematic checking, no valid pairs exist.

Question 15

A five-digit number 4a5b2 is divisible by 9. When this number is divided by 11, the remainder is 3. If a and b are single digits with a ≠ b, find the maximum value of a × b.

Accepted Answer

Step 1: For divisibility by 9, the sum of digits must be divisible by 9: 4+a+5+b+2 ≡ 0 (mod 9) → 11+a+b ≡ 0 (mod 9) → a+b ≡ 7 (mod 9). Step 2: So a+b ∈ {7, 16}. Step 3: For remainder 3 when divided by 11, apply the alternating sum rule: (4+5+2) - (a+b) ≡ 3 (mod 11) → 11-(a+b) ≡ 3 (mod 11) → -(a+b) ≡ -8 (mod 11) → a+b ≡ 8 (mod 11). Step 4: So a+b ∈ {8, 19}. Step 5: The only common value is a+b = 16 (since 7 ≠ 8 and 16 ≠ 19, but we need to reconsider: a+b ≡ 7 (mod 9) gives {7,16,...} and a+b ≡ 8 (mod 11) gives {8,19,...}. For single digits a,b: max a+b = 18. Check: 16 ≡ 7 (mod 9)? 16 = 9+7 ✓. 16 ≡ 8 (mod 11)? 16 = 11+5, so 16 ≡ 5 (mod 11) ✗. Check a+b=8: 8 ≡ 8 (mod 9)? No, 8 ≡ 8 (mod 9) ✓. 8 ≡ 8 (mod 11) ✓. Step 6: So a+b = 8 with a ≠ b. Pairs: (0,8), (1,7), (2,6), (3,5), (4,4-excluded), (5,3), (6,2), (7,1), (8,0). Step 7: Products: 0, 7, 12, 15, 15, 12, 7, 0. Maximum is 15 (from a=3,b=5 or a=5,b=3).

Question 16

A four-digit number 7a3b is divisible by 12. If the number is also divisible by 7, and a and b are distinct single digits, find the sum of all possible values of a + b.

Accepted Answer

Step 1: For divisibility by 12, the number must be divisible by both 3 and 4. Step 2: Divisibility by 4: the last two digits 3b must form a number divisible by 4. So 30+b ≡ 0 (mod 4) → 2+b ≡ 0 (mod 4) → b ∈ {2, 6}. Step 3: Divisibility by 3: 7+a+3+b ≡ 0 (mod 3) → 10+a+b ≡ 0 (mod 3) → a+b ≡ 2 (mod 3). Step 4: Case b=2: a+2 ≡ 2 (mod 3) → a ≡ 0 (mod 3) → a ∈ {0,3,6,9}. Case b=6: a+6 ≡ 2 (mod 3) → a ≡ 2 (mod 3) → a ∈ {2,5,8}. Step 5: Now check divisibility by 7 for each candidate. 7032: 7032÷7=1004.57... ✗. 7332: 7332÷7=1047.43... ✗. 7632: 7632÷7=1090.29... ✗. 7932: 7932÷7=1133.14... ✗. 7236: 7236÷7=1033.71... ✗. 7536: 7536÷7=1076.57... ✗. 7836: 7836÷7=1119.43... ✗. Step 6: None satisfy divisibility by 7. Therefore, the sum is 0.

Question 17

A number N is formed by concatenating the digits 2, 3, 4, 5, 6, 7, 8, 9 in some order. If N is divisible by 11, and the sum of digits in odd positions (from left) minus the sum of digits in even positions equals 11, find how many such arrangements exist.

Accepted Answer

Step 1: The number N has 8 digits: d₁d₂d₃d₄d₅d₆d₇d₈ where {d₁,...,d₈} is a permutation of {2,3,4,5,6,7,8,9}. Step 2: For divisibility by 11: (d₁+d₃+d₅+d₇) - (d₂+d₄+d₆+d₈) ≡ 0 (mod 11). Step 3: We are told (d₁+d₃+d₅+d₇) - (d₂+d₄+d₆+d₈) = 11. Step 4: The sum of all digits is 2+3+4+5+6+7+8+9 = 44. Step 5: Let S_odd = d₁+d₃+d₅+d₇ and S_even = d₂+d₄+d₆+d₈. Then S_odd + S_even = 44 and S_odd - S_even = 11. Step 6: Solving: 2·S_odd = 55 → S_odd = 27.5. This is impossible since S_odd must be an integer. Step 7: Therefore, no such arrangement exists.

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