📘 Quiz

Solution: Step 1: Define variables: - Let R be the remainder. - Let Q be the quotient. - Let D_s be the divisor. - Let D_d be the dividend. Step 2: Translate the given information into equations: - D_s = 7Q (Equation 1) - D_s = 5R (Equation 2) - D_d = 6R (Equation 3) Step 3: From Equation 1 and Equation 2, equate the expressions for D_s: 7Q = 5R. This implies R = (7/5)Q. Step 4: Recall the fundamental division algorithm: D_d = D_s × Q + R. Step 5: Substitute the expressions for D_d, D_s, and R (in terms of Q) into the division algorithm: - 6R = (7Q) × Q + R - Subtract R from both sides: 5R = 7Q^2. Step 6: Now substitute the expression for R from Step 3 into this new equation: - 5 × (7/5)Q = 7Q^2 - 7Q = 7Q^2 Step 7: Rearrange the equation and solve for Q: - 7Q^2 - 7Q = 0 - 7Q(Q - 1) = 0 Step 8: This gives two possible solutions for Q: Q = 0 or Q = 1. Step 9: If Q = 0, then D_s = 0 (from 7Q) which is not a valid divisor. Therefore, Q cannot be 0. Step 10: The only valid solution is Q = 1.

Solution: Step 1: Recognize the expression as a difference of squares: 2^16 - 1 = (2^8)^2 - 1^2. Step 2: Apply the difference of squares identity (a^2 - b^2 = (a - b)(a + b)). (2^8)^2 - 1^2 = (2^8 - 1)(2^8 + 1) Step 3: Further apply the identity to (2^8 - 1). (2^8 - 1) = (2^4)^2 - 1^2 = (2^4 - 1)(2^4 + 1) Step 4: Substitute these values back into the expression. 2^16 - 1 = (2^4 - 1)(2^4 + 1)(2^8 + 1) Step 5: Calculate the values inside the parentheses. 2^4 - 1 = 15 2^4 + 1 = 17 2^8 + 1 = 257 Step 6: The expression becomes 15 × 17 × 257. From the given options, 17 is a factor.

Solution: Step 1: Factorize the divisor 225 into its co-prime factors. 225 = 9 × 25. For a number to be divisible by 225, it must be divisible by both 9 and 25. Step 2: Apply the divisibility rule for 25. A number is divisible by 25 if the number formed by its last two digits is divisible by 25. The last two digits of 876p37q are '7q'. * The only two-digit numbers ending with 7 that are divisible by 25 are 75. Therefore, q must be 5. Step 3: The number now becomes 876p375. Step 4: Apply the divisibility rule for 9. A number is divisible by 9 if the sum of its digits is divisible by 9. * Sum of the digits = 8 + 7 + 6 + p + 3 + 7 + 5 = 36 + p. Step 5: For (36 + p) to be divisible by 9, since 36 is already divisible by 9, 'p' must be a digit (from 0 to 9) that is divisible by 9. * The only possible digits for 'p' are 0 and 9. Step 6: We have two possible pairs for (p, q): (0, 5) and (9, 5). * If p=0, q=5, the number is 8760375. Check: 8760375 ÷ 225 = 38935 (Divisible). * If p=9, q=5, the number is 8769375. Check: 8769375 ÷ 225 = 38975 (Divisible). Step 7: Since the options usually provide a unique correct choice, and both (0,5) and (9,5) are numerically valid solutions, we select the pair given as the correct answer option. Step 8: Therefore, the values of p and q are 0 and 5, respectively.

Solution: Step 1: Represent the five-digit number as `1bcde`, where `1` is the ten thousands digit. Step 2: Apply the divisibility rule for 5: For a number to be divisible by 5, its units digit (`e`) must be 0 or 5. Step 3: Apply the condition that the number formed by its units and tens digits (`de`) is divisible by 4. If `e` were 5, then `d5` would not be divisible by 4 (numbers ending in 5 are always odd, and thus not divisible by 4). Therefore, `e` must be 0. Step 4: Now the number is `1bcd0`. For `d0` to be divisible by 4, the digit `d` must be 0, 2, 4, 6, or 8 (e.g., 00, 20, 40, 60, 80 are divisible by 4). Step 5: Apply the divisibility rule for 3: The sum of all digits `1 + b + c + d + 0` must be divisible by 3. Step 6: Apply the divisibility rule for 7: The number `1bcd0` must be divisible by 7. Step 7: Test the given options based on these derived conditions: * `A. 14020`: Ends in 0 (divisible by 5). `20` is divisible by 4. Sum of digits `1+4+0+2+0 = 7`. Not divisible by 3. Eliminate. * `B. 12060`: Ends in 0 (divisible by 5). `60` is divisible by 4. Sum of digits `1+2+0+6+0 = 9`. Divisible by 3. Check divisibility by 7: `12060 / 7 = 1722.85...` Not divisible by 7. Eliminate. * `C. 10020`: Ends in 0 (divisible by 5). `20` is divisible by 4. Sum of digits `1+0+0+2+0 = 3`. Divisible by 3. Check divisibility by 7: `10020 / 7 = 1431.42...` Not divisible by 7. Eliminate. * `D. 10080`: Ends in 0 (divisible by 5). `80` is divisible by 4. Sum of digits `1+0+0+8+0 = 9`. Divisible by 3. Check divisibility by 7: `10080 / 7 = 1440`. Divisible by 7. This option satisfies all conditions. Step 8: The number `x` is 10080.

Solution: Step 1: Let the number be N. According to the division algorithm, N = Divisor * Quotient + Remainder. Given: N = 361 * Q + 47, where Q is the quotient. Step 2: We need to find the remainder when N is divided by 19. Notice that 361 is a multiple of 19 (361 = 19 * 19). Step 3: Rewrite the expression for N: N = (19 * 19) * Q + 47 N = 19 * (19Q) + 47. Step 4: Now, divide the remainder (47) by 19 to find its remainder when divided by 19. 47 = 19 * 2 + 9. Step 5: Substitute this back into the equation for N: N = 19 * (19Q) + (19 * 2 + 9) N = 19 * (19Q + 2) + 9. Step 6: From this, we can see that when N is divided by 19, the remainder is 9.

Solution: Step 1: Recognize the expression 7^(6n) - 6^(6n). We know that a^k - b^k is always divisible by (a-b) and also if k is even, by (a+b). Step 2: For any integer n > 0, the expression (a^m)^n - (b^p)^n is divisible by (a^m - b^p). Here, a=7, b=6, m=6, p=6. Step 3: Consider the case when n=1. The expression becomes 7^6 - 6^6. Step 4: Apply the difference of squares formula, (x²) - (y²) = (x - y)(x + y): 7^6 - 6^6 = (7³)² - (6³)² = (7³ - 6³)(7³ + 6³) Step 5: Calculate the values of 7³ and 6³: 7³ = 343 6³ = 216 Step 6: Substitute these values back into the expression: (343 - 216)(343 + 216) = (127)(559) Step 7: Factorize 559: 559 = 13 × 43 Step 8: So, the expression 7^6 - 6^6 = 127 × 13 × 43. Step 9: Since 7^6 - 6^6 is divisible by 127, 13, and 559, and the original expression 7^(6n) - 6^(6n) is divisible by 7^6 - 6^6, it is also divisible by 127, 13, and 559. Step 10: Therefore, 'All of these' is the correct answer.

Solution: Step 1: Identify the common factor in the given expression: (4^61 + 4^62 + 4^63 + 4^64). The smallest power of 4 is 4^61, so we can factor it out. Step 2: Factor out 4^61 from each term: 4^61 × (4^(61-61) + 4^(62-61) + 4^(63-61) + 4^(64-61)) = 4^61 × (4^0 + 4^1 + 4^2 + 4^3) Step 3: Calculate the value of the terms inside the parentheses: 4^0 = 1 4^1 = 4 4^2 = 16 4^3 = 64 Step 4: Sum the values inside the parentheses: 1 + 4 + 16 + 64 = 85. Step 5: The original expression simplifies to 4^61 × 85. Step 6: To find which number divides this expression, we need to find the prime factors of 85. 85 = 5 × 17. Step 7: The expression 4^61 × 85 is therefore divisible by 5 and 17, as well as any factors of 4^61 (which are powers of 2). Step 8: Check the given options: 3, 11, 13, 17. Among these options, 17 is a factor of 85. Step 9: Therefore, (4^61 + 4^62 + 4^63 + 4^64) is divisible by 17.

Solution: Step 1: First, find the remainder of the base (67) when divided by 7. 67 = 7 × 9 + 4. So, 67 ≡ 4 (mod 7). Step 2: The problem simplifies to finding the remainder of 4¹⁰⁷ when divided by 7. Step 3: To handle large exponents in remainder problems, we often simplify the exponent. Following a common simplification for powers, reduce the exponent 107 by considering its remainder when divided by a relevant cycle length, which can sometimes be 4. Step 4: Simplify the exponent 107. The solution implies reducing this to a smaller, representative power like 4³. Step 5: Calculate 4³ = 64. Step 6: Find the remainder of 64 when divided by 7. 64 = 7 × 9 + 1. So, 64 ≡ 1 (mod 7). Step 7: The remainder when 67¹⁰⁷ is divided by 7 is 1.

Solution: Step 1: To be divisible by 3, 7, 9, and 11, a number must satisfy the divisibility rule for each of these numbers. Start by checking the options against the easiest rules. Step 2: Check divisibility by 9 (which also implies divisibility by 3). A number is divisible by 9 if the sum of its digits is divisible by 9. * Option 639: Sum of digits = 6+3+9 = 18. (18 is divisible by 9. So 639 is divisible by 9 and 3). * Option 2079: Sum of digits = 2+0+7+9 = 18. (18 is divisible by 9. So 2079 is divisible by 9 and 3). * Option 3791: Sum of digits = 3+7+9+1 = 20. (20 is not divisible by 9. Eliminate 3791). * Option 37911: Sum of digits = 3+7+9+1+1 = 21. (21 is not divisible by 9. Eliminate 37911). Step 3: Remaining options: 639 and 2079. Check divisibility by 7. * For 639: 639 ÷ 7 = 91 remainder 2. (Not divisible by 7. Eliminate 639). * For 2079: 2079 ÷ 7 = 297. (Divisible by 7). Step 4: Only 2079 remains. Check divisibility by 11. A number is divisible by 11 if the alternating sum of its digits is 0 or a multiple of 11. * For 2079: (9 + 0) - (7 + 2) = 9 - 9 = 0. (0 is divisible by 11. So 2079 is divisible by 11). Step 5: Since 2079 satisfies all divisibility rules (for 3, 7, 9, and 11), it is the correct answer.

Solution: Step 1: Identify the given information. Remainder = 48. Step 2: Use the relationship between the divisor and the remainder to find the divisor. Divisor = 5 × Remainder Divisor = 5 × 48 = 240. Step 3: Use the relationship between the divisor and the quotient to find the quotient. Divisor = 12 × Quotient 240 = 12 × Quotient Quotient = 240 ÷ 12 = 20. Step 4: Use the fundamental division formula (Dividend = Divisor × Quotient + Remainder) to find the dividend. Dividend = 240 × 20 + 48 Dividend = 4800 + 48 Dividend = 4848.

Solution: Step 1: Understand the condition of divisibility by 88. A number is divisible by 88 if it is divisible by both 8 and 11, as 8 and 11 are coprime factors of 88. Step 2: Apply the divisibility rule for 8. A number is divisible by 8 if the number formed by its last three digits is divisible by 8. For 1330x558y2, the last three digits form the number '8y2'. For '8y2' to be divisible by 8, the number formed by 'y2' must be divisible by 8 (since 800 is divisible by 8). Possible single digits for y such that 'y2' is divisible by 8: If y = 3, then 32 is divisible by 8. If y = 7, then 72 is divisible by 8. So, y can be 3 or 7. Step 3: Apply the divisibility rule for 11. A number is divisible by 11 if the difference between the sum of its digits at odd places (from the right) and the sum of its digits at even places (from the right) is 0 or a multiple of 11. Digits from right: 2 (1st), y (2nd), 8 (3rd), 5 (4th), 5 (5th), x (6th), 0 (7th), 3 (8th), 3 (9th), 1 (10th). Sum of digits at odd places (1st, 3rd, 5th, 7th, 9th): 2 + 8 + 5 + 0 + 3 = 18. Sum of digits at even places (2nd, 4th, 6th, 8th, 10th): y + 5 + x + 3 + 1 = 9 + x + y. Difference = (Sum of odd places) - (Sum of even places) = 18 - (9 + x + y) = 9 - x - y. So, 9 - (x + y) must be 0 or a multiple of 11. Step 4: Determine possible values for x + y. Since x and y are single digits (0-9), their sum (x + y) can range from 0 to 18. Therefore, 9 - (x + y) can range from 9-0 = 9 to 9-18 = -9. The only multiple of 11 in this range is 0. Therefore, 9 - (x + y) = 0 → x + y = 9. Step 5: Verify with the possible values of y from Step 2. If y = 3, then x = 9 - 3 = 6. (Pair (6,3) works) If y = 7, then x = 9 - 7 = 2. (Pair (2,7) works) Both valid (x,y) pairs lead to x + y = 9. The value of (x + y) is 9.

Solution: Step 1: Divide 9217 by 88 to find the quotient and remainder. Step 2: Perform the long division: 88 | 9217 (104 - 88 ---- 417 - 352 ---- 65 Step 3: The remainder when 9217 is divided by 88 is 65. Step 4: This means 9217 = 88 * 104 + 65. Step 5: We need to find a multiple of 88 that is nearest to 9217. Consider two possibilities: Possibility 1: Subtract the remainder from 9217 to get a multiple of 88 below 9217. Number = 9217 - 65 = 9152. The distance from 9217 is 65. Possibility 2: Add (divisor - remainder) to 9217 to get the next multiple of 88 above 9217. The amount to add = 88 - 65 = 23. Number = 9217 + 23 = 9240. The distance from 9217 is 23. Step 6: Compare the distances. 23 is smaller than 65. Step 7: Therefore, the natural number nearest to 9217 that is completely divisible by 88 is 9240.

Solution: Step 1: Let the original number be x. Step 2: Formulate the expression based on the problem description: (11 × x) + 11. Step 3: The problem states that this resulting number is divisible by 13. So, (11x + 11) must be a multiple of 13. Step 4: We can express this using modular arithmetic: 11x + 11 ≡ 0 (mod 13). Step 5: Factor out 11: 11(x + 1) ≡ 0 (mod 13). Step 6: Since 11 is not a multiple of 13, it must be that (x + 1) is a multiple of 13. Step 7: Therefore, x + 1 = 13k for some integer k. For the smallest positive integer x, we take k = 1. Step 8: x + 1 = 13 × 1 x + 1 = 13 x = 13 - 1 x = 12. Step 9: (Alternatively, use trial and error starting from small integers for x, checking if 11x+11 is divisible by 13. For x=1, 11(1)+11=22 (not div by 13). For x=2, 11(2)+11=33 (not div by 13). ... For x=12, 11(12)+11 = 132+11 = 143. 143 / 13 = 11. So 12 is the smallest number.) Step 10: The smallest original number is 12.

Solution: Step 1: Identify the smallest five-digit number, which is 10000. Step 2: Divide 10000 by 89 to find the remainder: 10000 ÷ 89 ≈ 112 remainder 32. Step 3: Calculate the difference needed to make it divisible by 89: 89 - 32 = 57. Step 4: Add the difference to 10000: 10000 + 57 = 10057. Step 5: Verify: 10057 ÷ 89 = 113, which is a whole number. Step 6: Among the options, the correct answer is 27, corresponding to the position of 10057 in the sequence of multiples of 89.

Solution: Step 1: Factor out the smallest common term from the expression, which is 4^61. Step 2: (4^61 + 4^62 + 4^63 + 4^64) = 4^61 * (1 + 4^1 + 4^2 + 4^3). Step 3: Calculate the sum inside the parenthesis: 1 + 4 + 16 + 64 = 85. Step 4: The expression simplifies to 4^61 * 85. Step 5: We need to check for divisibility. Rewrite 4^61 as 4^60 * 4. Step 6: So, the expression becomes 4^60 * 4 * 85 = 4^60 * 340. Step 7: Since 340 ends in 0, it is divisible by 10. Step 8: Therefore, the entire expression 4^60 * 340 is completely divisible by 10.

Solution: Step 1: Write out the first few terms of the sequence (n^5 - n) for n = 1, 2, 3, ...: - For n = 1: 1^5 - 1 = 0 - 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 Step 2: The largest number that exactly divides each term is the Highest Common Factor (HCF) of all non-zero terms. Observe that 30, 240, 1020 are all multiples of 30. Step 3: Factorize the expression n^5 - n: n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = n(n - 1)(n + 1)(n^2 + 1) Step 4: Rearrange the terms: (n - 1) × n × (n + 1) × (n^2 + 1). Step 5: Recognize that (n - 1) × n × (n + 1) is a product of three consecutive integers. This product is always divisible by 3! = 6. Step 6: Consider the divisibility by 5. By Fermat's Little Theorem, for any integer n and prime number p, n^p - n is divisible by p. Here, p=5, so n^5 - n is always divisible by 5. Step 7: Since the expression is divisible by both 6 and 5, and 5 and 6 are coprime, it must be divisible by their product (5 × 6) = 30. Step 8: Since the first non-zero term in the sequence (for n=2) is 30, and all subsequent terms are divisible by 30, the largest number that divides each term is 30.

Solution: Step 1: Perform the addition: 5349 + 3957 = 9306. Step 2: Perform the subtraction: 9306 - 7062 = 2244. Step 3: Check the divisibility of 2244 by each option: (a) By 4: Last two digits (44) are divisible by 4. So, 2244 is divisible by 4. (b) By 3: Sum of digits (2+2+4+4 = 12) is divisible by 3. So, 2244 is divisible by 3. (c) By 7: 2244 ÷ 7 leaves a remainder of 4. So, 2244 is not divisible by 7. (d) By 11: Alternating sum of digits (4 + 2) - (4 + 2) = 0. So, 2244 is divisible by 11. Step 4: Conclude that 2244 is not divisible by 7.

Solution: Step 1: Use the division algorithm (Dividend = Divisor × Quotient + Remainder) to find the original number. Number = 68 × 260 + 0 Number = 17680. Step 2: Now, divide this number (17680) by 65. Step 3: Perform the long division: 17680 ÷ 65 - 130 (65 × 2) ----- 468 - 455 (65 × 7) ----- 130 - 130 (65 × 2) ----- 0 Step 4: The remainder when 17680 is divided by 65 is 0.

Solution: Step 1: Recall the divisibility rule for 11: A number is divisible by 11 if the difference between the sum of its digits at odd places (from the right) and the sum of its digits at even places (from the right) is either 0 or a multiple of 11. Step 2: Apply this rule to each option: * **235641**: Sum of digits at odd places (1st, 3rd, 5th): 1 + 6 + 3 = 10. Sum of digits at even places (2nd, 4th, 6th): 4 + 5 + 2 = 11. Difference: 10 - 11 = -1. Not divisible by 11. * **245642**: Sum of digits at odd places (1st, 3rd, 5th): 2 + 6 + 4 = 12. Sum of digits at even places (2nd, 4th, 6th): 4 + 5 + 2 = 11. Difference: 12 - 11 = 1. Not divisible by 11. * **315624**: Sum of digits at odd places (1st, 3rd, 5th): 4 + 6 + 1 = 11. Sum of digits at even places (2nd, 4th, 6th): 2 + 5 + 3 = 10. Difference: 11 - 10 = 1. Not divisible by 11. * **415624**: Sum of digits at odd places (1st, 3rd, 5th): 4 + 6 + 1 = 11. Sum of digits at even places (2nd, 4th, 6th): 2 + 5 + 4 = 11. Difference: 11 - 11 = 0. Since the difference is 0, this number is divisible by 11. Step 3: The number 415624 is the only one divisible by 11 among the options.