📘 Quiz

Solution: Step 1: Recall the divisibility rule for 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Step 2: In the given number 89715938*, the last two digits form the number `8*`. Step 3: We need to find a non-zero digit `*` such that `8*` is divisible by 4. Step 4: Test non-zero digits from 1 to 9: - If `* = 1`, 81 is not divisible by 4. - If `* = 2`, 82 is not divisible by 4. - If `* = 3`, 83 is not divisible by 4. - If `* = 4`, 84 is divisible by 4 (`84 = 4 * 21`). Step 5: Therefore, the unknown non-zero digit `*` must be 4.

Solution: Step 1: To find the remainder of 4^96 divided by 6, let's observe the pattern of the last digits (or remainders when divided by 6) of powers of 4. Step 2: Calculate the first few powers of 4 and their remainders when divided by 6: 4^1 = 4 => 4 mod 6 = 4 4^2 = 16 => 16 mod 6 = 4 4^3 = 64 => 64 mod 6 = 4 Step 3: We observe that for any positive integer exponent n, 4^n always leaves a remainder of 4 when divided by 6. This is because 4^n = 4 * 4^(n-1). Since 4 leaves a remainder of 4 when divided by 6, multiplying it by itself will continue to yield a remainder of 4 (e.g., (6k+4)(6j+4) = 36kj + 24k + 24j + 16 = 6(...) + 16 = 6(...) + 6*2 + 4 = 6(...) + 4). Step 4: Since the exponent is 96 (a positive integer), 4^96 will also leave a remainder of 4 when divided by 6. Step 5: The remainder is 4.

Solution: Step 1: Let the number be N. The successive divisions can be expressed as: N = 9 × Q1 + 8 Q1 = 11 × Q2 + 9 Q2 = 13 × Q3 + 8 Step 2: To find the smallest such number N, assume the last quotient Q3 is 1. Calculate Q2: Q2 = 13 × 1 + 8 = 21. Calculate Q1: Q1 = 11 × 21 + 9 = 231 + 9 = 240. Calculate N: N = 9 × 240 + 8 = 2160 + 8 = 2168. So, the number is 2168. Step 3: Now, perform successive division of 2168 by the reversed order of divisors: 13, then 11, then 9. Divide 2168 by 13: 2168 ÷ 13 = 166 with a remainder of 10. (2168 = 13 × 166 + 10) The first remainder is 10. Step 4: Divide the quotient (166) by the next divisor, 11: 166 ÷ 11 = 15 with a remainder of 1. (166 = 11 × 15 + 1) The second remainder is 1. Step 5: Divide the quotient (15) by the final divisor, 9: 15 ÷ 9 = 1 with a remainder of 6. (15 = 9 × 1 + 6) The third remainder is 6. Step 6: The remainders when the order of divisors is reversed are 10, 1, and 6.

Solution: Step 1: Adjust numbers to account for remainders: 17 - 4 = 13, 42 - 3 = 39, 93 - 15 = 78. Step 2: Find GCD of 13, 39, and 78 using Euclidean algorithm. Step 3: GCD(13, 39) = 13, GCD(13, 78) = 13. Step 4: The greatest number satisfying the condition is 13.

Solution: Step 1: To find the least number to subtract, divide 13601 by 87 to determine the remainder. Step 2: Perform the long division: 87 | 13601 (156 - 87 ---- 490 - 435 ---- 551 - 522 ---- 29 Step 3: The remainder when 13601 is divided by 87 is 29. Step 4: To make the number completely divisible by 87, the remainder must be subtracted from the original number. Step 5: Therefore, the required least number to subtract is 29.

Solution: Step 1: Let the number be 'x'. According to the problem, when x is divided by 357, the remainder is 39. This can be expressed using the division algorithm: x = 357q + 39, where q is the quotient. Step 2: We need to find the remainder when this same number 'x' is divided by 17. Step 3: First, check if the original divisor (357) is a multiple of the new divisor (17). Divide 357 by 17: 357 ÷ 17 = 21. So, 357 = 17 × 21. Step 4: Substitute this into the equation for x: x = (17 × 21)q + 39. Step 5: Now, focus on the remainder term, 39, and divide it by 17. 39 ÷ 17 = 2 with a remainder of 5 (since 39 = 17 × 2 + 5). Step 6: Substitute this back into the equation for x: x = 17 × 21q + (17 × 2) + 5. Step 7: Factor out 17 from the terms that are multiples of 17: x = 17(21q + 2) + 5. Step 8: This form shows that when 'x' is divided by 17, the remainder is 5.

Solution: Step 1: Divide 13851 by 87 to find the remainder. 13851 ÷ 87 = 159 with a remainder. 87 × 159 = 13833. Remainder = 13851 - 13833 = 18. Step 2: To make the number exactly divisible by 87, the amount to be added is the difference between the divisor and the remainder. Step 3: Number to be added = Divisor - Remainder = 87 - 18 = 69. Step 4: Adding 69 to 13851 (13851 + 69 = 13920) makes the sum divisible by 87.

Solution: Step 1: Let the unknown number be X. We are looking for the smallest X such that 7 × X equals a number composed solely of the digit 9 (e.g., 9, 99, 999, ...). Step 2: This implies that a number consisting of only nines must be perfectly divisible by 7. Step 3: Start checking numbers consisting of repeating nines for divisibility by 7: - 9 ÷ 7 = 1 remainder 2 - 99 ÷ 7 = 14 remainder 1 - 999 ÷ 7 = 142 remainder 5 - 9999 ÷ 7 = 1428 remainder 3 - 99999 ÷ 7 = 14285 remainder 4 - 999999 ÷ 7 = 142857 remainder 0 Step 4: The smallest number consisting of only nines that is perfectly divisible by 7 is 999999. Step 5: The number of digits in 999999 is 6. Step 6: Therefore, the smallest number X (which is 142857) has 6 digits.

Solution: Step 1: Express the given number in a suitable form: (2^48 - 1). Step 2: Rewrite 2^48 as (2^6)^8. So the expression becomes ( (2^6)^8 - 1^8 ). Step 3: Let x = 2^6 = 64. The expression is (x^8 - 1^8). Step 4: Recall the algebraic divisibility rule: For any even integer n, (a^n - b^n) is exactly divisible by both (a - b) and (a + b). Step 5: Here, a = x = 64, b = 1, and n = 8 (which is an even number). Step 6: Therefore, (64^8 - 1^8) is divisible by: - (64 - 1) = 63 - (64 + 1) = 65 Step 7: Both 63 and 65 are numbers between 60 and 70. Step 8: The numbers are 63 and 65.

Solution: Step 1: Divide 1056 by 23 to find the remainder. Step 2: 1056 ÷ 23. Performing the division: 1056 = 23 × 45 + 21. Step 3: The quotient is 45 and the remainder is 21. Step 4: To make 1056 completely divisible by 23, the remainder must be 0. Step 5: The amount that needs to be added is the difference between the divisor and the current remainder. Step 6: Required number to add = Divisor - Remainder = 23 - 21 = 2. Step 7: Adding 2 to 1056 gives 1058, which is 23 × 46, thus completely divisible by 23.

Solution: Step 1: Let the number be N. According to the division algorithm: N = 729 * Q + 56 (where Q is the quotient) Step 2: Check if the original divisor (729) is a multiple of the new divisor (27). 729 / 27 = 27. Yes, 729 is a multiple of 27. Step 3: If the original divisor is a multiple of the new divisor, the remainder when the number is divided by the new divisor can be found by dividing the original remainder by the new divisor. Divide 56 by 27: 56 = 27 * 2 + 2 Step 4: The remainder is 2. Alternatively, since N = 729Q + 56, and 729 = 27 × 27: N = (27 × 27)Q + 56 N = 27 × (27Q) + 56 Now, divide 56 by 27: N = 27 × (27Q) + (27 × 2 + 2) N = 27 × (27Q + 2) + 2 So, when N is divided by 27, the remainder is 2.

Solution: Step 1: Identify the smallest 6-digit number, which is 100,000. Step 2: Divide the smallest 6-digit number (100,000) by 111 to find the remainder. 100,000 ÷ 111 = 900 with a remainder. 111 × 900 = 99,900. Remainder = 100,000 - 99,900 = 100. So, 100,000 = (111 × 900) + 100. Step 3: To find the smallest 6-digit number exactly divisible by 111, we need to add the difference between the divisor and the remainder to the original number. Required number = Original number + (Divisor - Remainder) Required number = 100,000 + (111 - 100). Step 4: Calculate the required number: Required number = 100,000 + 11 = 100,011. Step 5: Therefore, the smallest 6-digit number exactly divisible by 111 is 100,011.

Solution: Step 1: Observe the relationship between the base (7) and the divisor (342). Calculate powers of 7: 7^1 = 7, 7^2 = 49, 7^3 = 343. Notice that 343 is very close to 342. Step 2: Rewrite the expression 7^84 using 7^3. 7^84 = (7^3)^28 = (343)^28. Step 3: Apply the divisibility rule: (x^n - a^n) is divisible by (x - a) for all integer values of n. Let x = 343 and a = 1, and n = 28. So, (343^28 - 1^28) is divisible by (343 - 1). This means (343^28 - 1) is divisible by 342. Step 4: If (343^28 - 1) is divisible by 342, it implies that (343^28 - 1) = 342 × k for some integer k. Therefore, 343^28 = 342 × k + 1. Step 5: When 343^28 (which is 7^84) is divided by 342, the remainder is 1.

Solution: Step 1: Identify the largest 4-digit number, which is 9999. Step 2: Divide 9999 by 88 to determine the remainder. Step 3: Performing the division: 9999 = 88 × 113 + 55. Step 4: The remainder of this division is 55. Step 5: To find the largest 4-digit number exactly divisible by 88, subtract this remainder from 9999. Step 6: Required number = 9999 - 55 = 9944. Step 7: This means 9944 is the largest 4-digit number that is a multiple of 88 (9944 = 88 × 113).

Solution: Step 1: The greatest four-digit number is 9999. Step 2: We need to find the remainder when 9999 is divided by 307. Step 3: Divide 9999 by 307: 9999 = 307 × 32 + 175. The remainder is 175. Step 4: To make 9999 exactly divisible by 307, we need to add the difference between the divisor (307) and the remainder (175). Step 5: Number to be added = 307 - 175 = 132. Step 6: Therefore, 132 must be added to 9999 to make it divisible by 307.

Solution: Step 1: Recall the divisibility rule for sums of powers: The expression (x^n + a^n) is divisible by (x + a) if and only if 'n' is an odd positive integer. Step 2: Compare this rule to the given expression (12^n + 1). Here, x = 12 and a = 1 (since 1 can be written as 1^n for any n). The divisor is (x + a) = (12 + 1) = 13. Step 3: According to the rule from Step 1, (12^n + 1^n) is divisible by (12 + 1) = 13 if 'n' is an odd integer. Step 4: Therefore, for (12^n + 1) to be divisible by 13, 'n' must be any odd integer.

Solution: Step 1: Let the four-digit number be 37X3. For it to be divisible by 7, we can apply the divisibility rule for 7. Step 2: The divisibility rule for 7 states: subtract twice the last digit from the number formed by the remaining digits. If the result is divisible by 7, then the original number is divisible by 7. Step 3: For 37X3: - Remaining digits: 37X - Last digit: 3 - Twice the last digit: 2 × 3 = 6 - New number to check: 37X - 6. This must be divisible by 7. Step 4: Express 37X as 370 + X (since X is in the tens place, or 3700 + 10X if X is hundreds). The problem implies X is the hundreds digit: 37X3 = 3000 + 700 + 10X + 3 = 3703 + 10X. *Self-correction: The provided solution's derivation implies 37X3 means 3700 + 100*X + 3. Let's re-interpret `37 X 3` as `3700 + 100*X + 3`. This seems more likely given the solution provided.* `3703 + 100X` must be divisible by 7. `3703 = 7 * 529`. So, `3703` is already divisible by 7. For `3703 + 100X` to be divisible by 7, `100X` must be divisible by 7. Since 7 is a prime number, either 7 divides 100 or 7 divides X. 7 does not divide 100. So, 7 must divide X. The only digit X (0-9) that is divisible by 7 is 0 or 7. Step 5: Now, check the provided options. The options are '0', '3', '5', '9'. Step 6: Among the options, only 0 satisfies the condition (3703 is divisible by 7, and 3773 is also divisible by 7, but only 0 is an option). Step 7: So, X must be 0.

Solution: Step 1: Let the three consecutive natural numbers be n, n + 1, and n + 2. Step 2: Calculate their sum. Sum = n + (n + 1) + (n + 2) = 3n + 3. Step 3: Factor out the common term. Sum = 3(n + 1). Step 4: Conclude divisibility. Since the sum can be expressed as 3 multiplied by an integer (n + 1), it is always divisible by 3.

Solution: Step 1: Recall the algebraic identity: If 'n' is an odd integer, then (a^n + b^n) is always divisible by (a + b). Step 2: In the given expression, a = 15, b = 23, and n = 23. Since 23 is an odd integer, the property applies. Step 3: Therefore, (15^23 + 23^23) is divisible by (15 + 23). Step 4: Calculate the sum (a + b): 15 + 23 = 38. Step 5: Since the expression is divisible by 38, and 38 is a multiple of 19 (38 = 2 × 19), the expression (15^23 + 23^23) must also be divisible by 19. Step 6: When a number is perfectly divisible by another number, the remainder is 0. Step 7: The remainder is 0.