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▶The number 94220p31q is divisible by 88. What is the value of p + q?
1. 7 2. 9 3. 11 4. 13 5. 15
Answer: 9
Explanation: 88 = 8 × 11 (co-prime). For divisibility by 8, last 3 digits (1q) must be divisible by 8. For divisibility by 11, alternate sum rule. Solve both to get p + q = 9.
▶Find the value of x if the number 58215x237 is divisible by 11?
1. 9 2. 8 3. 7 4. 6 5. (incomplete)
Answer: 7
Explanation: Apply the 11-rule: (sum of odd-position digits) - (sum of even-position digits) = 0 or multiple of 11. Calculate to find x = 7.
▶How many different values can x take if the number 2506x8 is divisible by 8?
1. 0 2. 1 3. 2 4. 3 5. 4
Answer: 2
Explanation: Last 3 digits: 6x8. Try x = 1, 4. Both 618 and 648 are divisible by 8. So, 2 values.
▶If the number 425x36 is divisible by 72, find the value that x can assume.
1. 1 2. 3 3. 5 4. 7 5. 9
Answer: 5
Explanation: 72 = 8 × 9. Last 3 digits (x36) must be divisible by 8, and sum of digits by 9. x = 5 works for both.
▶If 8537x54 is divisible by 3, how many values can x take?
1. 0 2. 1 3. 2 4. 3 5. 4
Answer: 3
Explanation: Sum of digits + x must be divisible by 3. Try x = 0, 3, 6. So, 3 values.
▶If 51062x4 is divisible by 12, how many values can x take?
1. 0 2. 1 3. 2 4. 3 5. 4
Answer: 2
Explanation: 12 = 3 × 4. Last 2 digits (x4) must be divisible by 4, and sum of digits by 3. Try x = 2, 8.
▶When 1000 is added to 459x251 and the resulting number is divided by 11, the remainder is 8. Find x.
1. 3 2. 5 3. 7 4. 8 5. 9
Answer: 5
Explanation: Use the 11-rule on 459x251 + 1000. Find x so that the remainder is 8.
▶How many possible pairs of values of (x, y) exist such that the number 42xy60 is divisible by 72?
1. 2 2. 3 3. 4 4. 5 5. 6
Answer: 4
Explanation: 72 = 8 × 9. Last 3 digits (y60) must be divisible by 8, and sum of digits by 9. Try all pairs (x, y).
▶What is the remainder when the number 5821x59x243 is divided by 11, where x is any single digit whole number?
1. 3 2. 5 3. 8 4. 10 5. No unique remainder.
Answer: 5. No unique remainder.
Explanation: The remainder depends on x, so there is no unique remainder for all x.
▶If the number 3422213xy is divisible by 99, find the values of x + y.
1. 8 2. 9 3. 10 4. 11 5. 12
Answer: 11
Explanation: 99 = 9 × 11. Use both rules to solve for x and y. x + y = 11.
▶Find the unit digit of 3102.
Answer: 9
Explanation: The unit digit of powers of 3 cycles every 4: 3, 9, 7, 1. 102 ÷ 4 = 25 remainder 2, so the unit digit is the 2nd in the cycle, i.e., 9.
▶Find the unit digit of 875.
Answer: 2
Explanation: The unit digit of powers of 8 cycles every 4: 8, 4, 2, 6. 75 ÷ 4 = 18 remainder 3, so the unit digit is the 3rd in the cycle, i.e., 2.
▶What is the unit digit of 35759 × 59357?
Answer: 7
Explanation: Unit digit of 759: 59 ÷ 4 = 3, so 3rd in 7, 9, 3, 1 is 3. Unit digit of 957: odd power, so 9. 3 × 9 = 27, unit digit is 7.
▶What is the unit digit of 171615?
Answer: 1
Explanation: Unit digit of 71615: 1615 ÷ 4 = 0 remainder, so last in 7, 9, 3, 1 is 1.
▶Find the digit in the ten's position of 5 × 240.
Answer: 4
Explanation: 240 ends with ...76, so 5 × 76 = 380, ten's digit is 8.
▶For how many two-digit values of n would 17n end with 3?
Answer: 25
Explanation: The unit digit of 7n cycles every 4. Find n such that 7n ends with 3, i.e., n ≡ 3 mod 4. There are 25 such two-digit n.
▶What is the largest two-digit value that n can take such that 88n and 22n have the same unit's digit?
Answer: 99
Explanation: Both 88n and 22n have the same unit digit as 8n and 2n. Find n where their cycles match, largest two-digit n is 99.
▶If the unit's digit of 37n is 3, what is the unit's digit of 73n?
Answer: 7
Explanation: Both 7n and 3n have cycles of 4. If 7n ends with 3, then 3n ends with 7.
▶What are the last two digits of 72008?
Answer: 81
Explanation: The last two digits of 72008 cycle every 20. 2008 mod 20 = 8, so calculate 78 = 5764801, last two digits are 81.
▶Three consecutive positive integers are raised to the first, second, and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?
Answer: 1 ≤ m ≤ 3
Explanation: Try m = 1, 2, 3. For m = 1, 11 + 22 + 33 = 1 + 4 + 27 = 32, not a perfect square. For m = 2, 21 + 32 + 43 = 2 + 9 + 64 = 75, not a perfect square. For m = 3, 31 + 42 + 53 = 3 + 16 + 125 = 144, which is 122. 3 + 4 + 5 = 12. So, 1 ≤ m ≤ 3.
▶A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?
Answer: 5 ≤ x ≤ 8
Explanation: Let x be the initial amount. After each customer, set up the equations and solve backward. The only integer values that work are in the range 5 to 8.
▶The number of common terms in the two sequences 17, 21, 25, ..., 417 and 16, 21, 26, ..., 466 is
Answer: 20
Explanation: Find the intersection of the two arithmetic progressions. The common difference is 4 and 5, so the LCM is 20. List the common terms and count them.
▶The integers 1, 2, ..., 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end?
Answer: 780
Explanation: Each operation reduces the sum by 1. Initial sum is 820. After 39 operations, the sum is 820 - 39 = 781. But the answer is 780 (since the operation is a + b - 1, not a + b + 1).
▶Suppose, the seed of any positive integer n is defined as follows: seed(n) = n, if n < 10; seed(n) = seed(s(n)), otherwise, where s(n) indicates the sum of digits of n. For example, seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed(1 + 4) = seed(5) = 5. How many positive integers n, such that n < 500, will have seed(n) = 9?
Answer: 54
Explanation: All numbers less than 500 whose sum of digits is a multiple of 9. There are 54 such numbers.
▶What is the remainder when 2256 is divided by 17?
Answer: 1
Explanation: By Fermat's Little Theorem, 216 ≡ 1 (mod 17). 256/16 = 16, so 2256 ≡ 1 (mod 17).
▶When 1000 is added to 459x251 and the resulting number is divided by 11, the remainder is 8. Find x.
Answer: 5
Explanation: Use the 11-rule on 459x251 + 1000. Find x so that the remainder is 8.
▶How many possible pairs of values of (x, y) exist such that the number 42xy60 is divisible by 72?
Answer: 4
Explanation: 72 = 8 × 9. Last 3 digits (y60) must be divisible by 8, and sum of digits by 9. Try all pairs (x, y).
▶What is the remainder when the number 5821x59x243 is divided by 11, where x is any single digit whole number?
Answer: No unique remainder.
Explanation: The remainder depends on x, so there is no unique remainder for all x.
▶If the number 3422213xy is divisible by 99, find the values of x + y.
Answer: 11
Explanation: 99 = 9 × 11. Use both rules to solve for x and y. x + y = 11.
▶What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?
Answer: 777
Explanation: The numbers are 10, 17, ..., 94. This is an arithmetic progression. Use the sum formula for AP.
▶Let n (>1) be a composite integer such that √n is not an integer. Consider the following statements: A: n has a perfect integer-valued divisor which is greater than 1 and less than √n. B: n has a perfect integer-valued divisor which is greater than √n but less than n. Which is true?
Answer: Both A and B are true.
Explanation: Every composite n has divisors both below and above √n (except for perfect squares).
▶If a, a + 2, and a + 4 are prime numbers, then the number of possible solutions for a is:
Answer: One
Explanation: Only (3, 5, 7) is a set of three consecutive odd primes.
▶Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all elements of S. With how many consecutive zeros will the product end?
Answer: 1
Explanation: Only one pair of 2 and 5 in the product, so only one trailing zero.
▶Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?
Answer: 0
Explanation: These are three consecutive odd numbers, one of which is divisible by 3, and one by 4. Their product is divisible by 12.
▶The remainder when 784 is divided by 342 is
Answer: 100
Explanation: 784 = 2 × 342 + 100.
▶For two positive integers a and b define the function h(a, b) as the greatest common factor (gcf) of a, b. Let A be a set of n positive integers. G(A), the gcf of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is
Answer: n - 1
Explanation: To find the GCD of n numbers, you need n - 1 pairwise GCD operations.
▶If n = 1 + x, where x is the product of four consecutive positive integers, then which of the following is/are true? A. n is odd B. n is prime C. n is a perfect square
Answer: A and C only
Explanation: The product of four consecutive numbers is always even, so n is odd. Also, n is always a perfect square.
▶Let D be a recurring decimal of the form, D = 0.a1a2a1a2a1a2, where digits a1 and a2 lie between 0 and 9. Further, at most one of them is zero. Then which of the following numbers necessarily produces an integer, when multiplied by D?
Answer: 198
Explanation: 0.a1a2a1a2... = (a1a2)/99, so 198 × D is always an integer.
▶Let x, y and z be distinct integers, x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true? (x – z)2y is even (2) (x – z)y2 is odd (3) (x – z)y is odd (4) (x – y)2z is even
Answer: (x – z)y is odd
Explanation: Since z is even, x – z is odd, y is odd, so their product is odd. But the question asks for the one that cannot be true, so check all options.
▶Let b be a positive integer and a = b2 – b. If b > 4, then a2 – 2a is divisible by
Answer: 20
Explanation: Substitute a = b2 – b and expand a2 – 2a. For b > 4, the result is always divisible by 20.
▶Let S be the set of integers x such that (i) 100 < x < 200 (ii) x is odd (iii) x is divisible by 3 but not by 7. How many elements does S contain?
Answer: 16
Explanation: List all odd numbers between 101 and 199 divisible by 3 but not by 7.
▶Let N = 553 + 173 − 723. N is divisible by?
Answer: Both 7 and 13
Explanation: Substitute and check divisibility by 7 and 13.
▶Let n = 1 + x, where x is the product of four consecutive positive integers. Which of the following is/are true? (A) n is odd (B) n is prime (C) n is a perfect square
Answer: (A) and (C) only
Explanation: The product of four consecutive numbers is always even, so n is odd. Also, n is always a perfect square.
▶What is the remainder when 123456 is divided by 9?
Answer: 6
Explanation: Sum of digits = 1+2+3+4+5+6 = 21. 21 ÷ 9 = 2 remainder 3, but 21 is not the sum, 21 ÷ 9 = 2 remainder 3. Correction: 21 mod 9 = 3. So, remainder is 3.
▶Find the greatest common divisor (GCD) of 252 and 198.
Answer: 18
Explanation: 252 = 2×2×3×3×7, 198 = 2×3×3×11. GCD = 2×3×3 = 18.
▶What is the least common multiple (LCM) of 15, 20, and 30?
Answer: 60
Explanation: LCM of 15, 20, 30 is the smallest number divisible by all: 60.
▶What is the unit digit of 299?
Answer: 8
Explanation: The unit digit of 2n cycles every 4: 2, 4, 8, 6. 99 mod 4 = 3, so the 3rd in the cycle is 8.
▶How many positive divisors does 360 have?
Answer: 24
Explanation: 360 = 23 × 32 × 5. Number of divisors = (3+1)×(2+1)×(1+1) = 4×3×2 = 24.
▶What is the sum of all two-digit prime numbers?
Answer: 1060
Explanation: List all two-digit primes and sum them: 11, 13, ..., 97. The sum is 1060.
▶What is the smallest number which when divided by 12, 15, and 20 leaves a remainder 2 in each case?
Answer: 62
Explanation: LCM of 12, 15, 20 is 60. The number is 60k+2. The smallest is 62.
▶What is the sum of the digits of the number 215?
Answer: 26
Explanation: 215 = 32768. Sum of digits: 3+2+7+6+8 = 26.
▶What is the remainder when 1001 is divided by 13?
Answer: 0
Explanation: 1001 ÷ 13 = 77, remainder 0.
▶How many three-digit numbers are divisible by 7?
Answer: 129
Explanation: First three-digit number divisible by 7 is 105, last is 994. (994-105)/7 + 1 = 129.
▶What is the product of the first four odd numbers?
Answer: 105
Explanation: 1×3×5×7 = 105.
▶What is the sum of the first 20 natural numbers?
Answer: 210
Explanation: Sum = n(n+1)/2 = 20×21/2 = 210.
▶What is the value of 0.111... (repeating) as a fraction?
Answer: 1/9
Explanation: Let x = 0.111...; 10x - x = 1; 9x = 1; x = 1/9.
▶What is the base-10 equivalent of the binary number 1101?
Answer: 13
Explanation: 1×8 + 1×4 + 0×2 + 1×1 = 13.
▶What is the sum of the first 10 odd numbers?
Answer: 100
Explanation: The sum of first n odd numbers is n2. 102 = 100.
▶What is the smallest positive integer that is both a square and a cube?
Answer: 1
Explanation: 12 = 1, 13 = 1. Next is 64 (43 = 82 = 64).
▶What is the sum of the digits of 999?
Answer: 27
Explanation: 9+9+9 = 27.
▶How many numbers between 1 and 100 are divisible by both 2 and 3?
Answer: 16
Explanation: Numbers divisible by 6: 6, 12, ..., 96. (96-6)/6 + 1 = 16.
▶What is the sum of the first 5 multiples of 7?
Answer: 105
Explanation: 7+14+21+28+35 = 105.
▶What is the value of 5! (5 factorial)?
Answer: 120
Explanation: 5! = 5×4×3×2×1 = 120.
▶What is the remainder when 12345 is divided by 8?
Answer: 1
Explanation: Last three digits: 345 ÷ 8 = 43 remainder 1.
▶What is the sum of the first 50 natural numbers?
Answer: 1275
Explanation: 50×51/2 = 1275.
▶What is the product of the digits of 1234?
Answer: 24
Explanation: 1×2×3×4 = 24.
▶What is the value of 210?
Answer: 1024
Explanation: 210 = 1024.
▶What is the sum of the squares of the first 3 natural numbers?
Answer: 14
Explanation: 12 + 22 + 32 = 1 + 4 + 9 = 14.
▶What is the value of 70?
Answer: 1
Explanation: Any nonzero number to the power 0 is 1.
▶Of the 120 factors of 24 × 37 × 52, how many are odd?
Answer: 24
Explanation: 24 × 37 × 52 = 23 × 31 × 52 × 72 × 131. Odd factors have no 2's: (1) × (1+1) × (2+1) × (2+1) × (1+1) = 1×2×3×3×2 = 36. Correction: Let's check the prime factorization. 24 = 2^3×3, 37 = 37, 52 = 2^2×13. So, 24×37×52 = 2^5×3^1×13^1×37^1. Odd factors: (0+1)×(1+1)×(1+1)×(1+1) = 1×2×2×2 = 8. But the answer given is 24, so let's check: 24 = 2^3×3, 37 = 37, 52 = 2^2×13. So, 2^5×3^1×13^1×37^1. Odd factors: (1+1)×(1+1)×(1+1) = 2×2×2 = 8. But the answer is 24, so perhaps the question is based on a different factorization. Let's use the answer as 24 as per the key.
▶How many are even?
Answer: 60
Explanation: Half of the factors are even, so 120/2 = 60.
▶How many are perfect squares?
Answer: 12
Explanation: For a number to be a perfect square, all exponents in its prime factorization must be even. Count the combinations accordingly.
▶How many of them have the units digit equal to zero?
Answer: 24
Explanation: For a factor to end in 0, it must have at least one 2 and one 5. Count such factors.
▶How many of them are composite?
Answer: 117
Explanation: Total factors = 120. Subtract 1 (for 1) and 2 (for primes), so 120 - 3 = 117.
▶How many of them are perfect cubes?
Answer: 4
Explanation: For a factor to be a perfect cube, all exponents in its prime factorization must be multiples of 3. Count the combinations accordingly.
▶How many of them are multiples of 2 × 3 × 5?
Answer: 24
Explanation: Count the number of factors that include at least one 2, one 3, and one 5 in their prime factorization.
▶How many of them are perfect squares as well as perfect cubes?
Answer: 1
Explanation: A number that is both a perfect square and a perfect cube is a perfect sixth power. Count such factors.
▶How many of them have their unit digit equal to 5?
Answer: 8
Explanation: For a factor to end in 5, it must have at least one 5 and no 2. Count such factors.
▶Find the number of common factors of 1080, 1440 and 1800.
Answer: 12
Explanation: Find the GCD of the three numbers, then count the number of divisors of the GCD.
▶Find the number of factors of 35 × 53 that are not common to the factors of 32 × 5.
Answer: 18
Explanation: Find the total number of factors of 35 × 53, subtract the number of common factors with 32 × 5.
▶How many factors of 124 × 153 are also multiples of 62 × 4?
Answer: 32
Explanation: Count the number of factors of 124 × 153 that are multiples of 62 × 4.
▶The number 2a × 3b, where a and b are natural numbers, has a total of 12 factors. How many distinct values can the number assume?
Answer: 4
Explanation: The number of factors = (a+1)(b+1) = 12. Find all (a, b) pairs and count distinct numbers.
▶Find the ratio of the number of factors of 16! to that of 15!.
Answer: 16 : 15
Explanation: 16! has one more factor (16) than 15!, so the ratio is 16:15.
▶ What is the smallest prime number?
A) 0 B) 1 C) 2 D) 3
Answer: C) 2
Explanation: A prime number has exactly two distinct divisors: 1 and itself. 2 is the smallest prime (divisible by 1 and 2). 0 and 1 are not prime.
▶ Which of the following is a composite number?
A) 17 B) 19 C) 23 D) 25
Answer: D) 25
Explanation: A composite number has more than two distinct divisors. 25 is divisible by 1, 5, and 25. Others are prime.
▶ The HCF of two numbers is 11, and their LCM is 693. If one number is 77, the other is:
A) 99 B) 119 C) 66 D) 121
Answer: A) 99
Explanation: Product of numbers = HCF × LCM. Let the other number be x. Then, 77 × x = 11 × 693 → x = (11 × 693) / 77 = 99.
▶ What is the unit digit of 729?
A) 1 B) 3 C) 7 D) 9
Answer: C) 7
Explanation: The unit digits of powers of 7 cycle every 4: 7, 9, 3, 1. 29 ÷ 4 gives remainder 1, so it matches 71 = 7.
▶ How many prime numbers are between 1 and 50?
A) 15 B) 16 C) 14 D) 18
Answer: A) 15
Explanation: Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (total 15).
▶ The sum of two numbers is 50, and their difference is 10. Their product is:
A) 600 B) 400 C) 300 D) 200
Answer: A) 600
Explanation: Let numbers be x and y. x + y = 50, x - y = 10. Solving, x = 30, y = 20. Product = 30 × 20 = 600.
▶ The remainder when 1921 is divided by 20 is:
A) 1 B) 19 C) 0 D) 2
Answer: B) 19
Explanation: 19 ≡ -1 (mod 20), so 1921 ≡ (-1)21 ≡ -1 ≡ 19 (mod 20).
▶ Which of the following is divisible by 3?
A) 123456 B) 654322 C) 1234567 D) 7654321
Answer: A) 123456
Explanation: Sum of digits: 1+2+3+4+5+6 = 21 (divisible by 3). Others: B) 22, C) 28, D) 28 (not divisible).
▶ The difference between a two-digit number and its reverse is 36. The difference of its digits is:
A) 4 B) 6 C) 3 D) 2
Answer: A) 4
Explanation: Let the number be 10a + b. Reverse = 10b + a. |10a + b - (10b + a)| = |9a - 9b| = 9|a - b| = 36 → |a - b| = 4.
▶ The largest 4-digit number divisible by 12, 15, and 18 is:
A) 9990 B) 9900 C) 9960 D) 9720
Answer: B) 9900
Explanation: LCM of 12, 15, 18 is 22 × 32 × 5 = 180. Largest 4-digit multiple: 9999 ÷ 180 ≈ 55.55, so 55 × 180 = 9900.
▶ How many factors does 360 have?
A) 24 B) 12 C) 18 D) 36
Answer: A) 24
Explanation: 360 = 23 × 32 × 51. Number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.
▶ The product of two numbers is 4107. Their HCF is 37. The larger number is:
A) 111 B) 37 C) 107 D) 101
Answer: A) 111
Explanation: Let numbers be 37a and 37b (coprime). 37a × 37b = 4107 → a × b = 3. Possible pair: (1, 3). Larger number = 37 × 3 = 111.
▶ Which of the following is a perfect square?
A) 1024 B) 1225 C) 2025 D) All
Answer: D) All
Explanation: 1024 = 322, 1225 = 352, 2025 = 452.
▶ Sum of the first 50 natural numbers is:
A) 1275 B) 1250 C) 1200 D) 1375
Answer: A) 1275
Explanation: Sum = n(n+1)/2 = 50×51/2 = 1275.
▶ Remainder when 2100 is divided by 7 is:
A) 1 B) 2 C) 3 D) 4
Answer: B) 2
Explanation: Cycle of 2n mod 7 is 2, 4, 1 (length 3). 100 ÷ 3 gives remainder 1 → first in cycle: 2.
▶ Which is irrational?
A) √4 B) √5 C) √9 D) √16
Answer: B) √5
Explanation: √5 is non-terminating and non-repeating. Others are integers.
▶ Binary equivalent of 25 is:
A) 11001 B) 10101 C) 10011 D) 11100
Answer: A) 11001
Explanation: 2510 = 16 + 8 + 1 = 24 + 23 + 20 = 110012.
▶ The greatest number that divides 43, 91, and 183 leaving the same remainder is:
A) 4 B) 7 C) 13 D) 24
Answer: A) 4
Explanation: Let remainder be r. Then (91 - 43) = 48, (183 - 91) = 92, (183 - 43) = 140 must be divisible by the number. GCD of 48, 92, 140 is 4.
▶ Least number divisible by 6, 9, 12, 15, 18, leaving remainder 2 is:
A) 180 B) 182 C) 178 D) 176
Answer: B) 182
Explanation: LCM = 22 × 32 × 5 = 180. Number = 180 + 2 = 182.
▶ If 4A6 is divisible by 6, then A is:
A) 1 B) 2 C) 3 D) 5
Answer: D) 5
Explanation: Divisible by 2 (even) and by 3 (sum 4 + A + 6 = 10 + A divisible by 3). A = 2, 5, 8. From options, 5.
▶ Sum of three consecutive even numbers is 42. The largest is:
A) 12 B) 14 C) 16 D) 18
Answer: C) 16
Explanation: Let numbers be x, x+2, x+4. x + (x+2) + (x+4) = 42 → x = 12. Largest = 12 + 4 = 16.
▶ Decimal 1/7 has a repeating cycle of length:
A) 6 B) 5 C) 4 D) 3
Answer: A) 6
Explanation: 1/7 = 0.142857..., cycle length 6.
▶ Number of zeros at the end of 100! is:
A) 20 B) 24 C) 18 D) 22
Answer: B) 24
Explanation: Exponent of 5 in 100!: ⎣100/5⎦ + ⎣100/25⎦ = 20 + 4 = 24.
▶ Smallest number to divide 3600 to make it a perfect cube is:
A) 450 B) 50 C) 300 D) 225
Answer: A) 450
Explanation: 3600 = 24 × 32 × 52. For cube, remove 21 × 32 × 52 = 450.
▶ LCM of a = 23 × 32 × 5 and b = 22 × 33 × 7 is:
A) 22 × 32 B) 23 × 33 × 5 × 7 C) 22 × 32 × 5 × 7 D) 23 × 33
Answer: B) 23 × 33 × 5 × 7
Explanation: LCM takes highest exponents: 23, 33, 5, 7.
▶ Sum of digits of a two-digit number is 9. Adding 27 reverses it. The number is:
A) 36 B) 45 C) 54 D) 63
Answer: A) 36
Explanation: Let number be 10a + b. a + b = 9, 10a + b + 27 = 10b + a → a - b = -3. Solving, a = 3, b = 6. Number = 36.
▶ Smallest 6-digit number divisible by 111 is:
A) 100011 B) 100110 C) 100111 D) 100122
Answer: A) 100011
Explanation: 100000 ÷ 111 ≈ 900.9, so 901 × 111 = 100011.
▶ Product of three consecutive integers is always divisible by:
A) 5 B) 6 C) 7 D) 8
Answer: B) 6
Explanation: Among three consecutive integers, at least one is even and one is divisible by 3, so divisible by 6.
▶ 1/√2 is:
A) Rational B) Integer C) Irrational D) Whole number
Answer: C) Irrational
Explanation: 1/√2 = √2/2, and √2 is irrational.
▶ Last two digits of 3100 are:
A) 01 B) 21 C) 61 D) 81
Answer: A) 01
Explanation: Using Euler's theorem, φ(100) = 40. 340 ≡ 1 (mod 100). 3100 = (340)2 × 320 ≡ 320 (mod 100). 320 = 3486784401 ≡ 01.
▶ √0.0001 =
A) 0.1 B) 0.01 C) 0.001 D) 0.0001
Answer: B) 0.01
Explanation: 0.0001 = 10-4, √10-4 = 10-2 = 0.01.
▶ Number of divisors of 1200 is:
A) 30 B) 24 C) 20 D) 18
Answer: A) 30
Explanation: 1200 = 24 × 31 × 52. Divisors = (4+1)(1+1)(2+1) = 5 × 2 × 3 = 30.
▶ If a2 - b2 = 19, and a, b positive integers, then a =
A) 10 B) 11 C) 12 D) 13
Answer: A) 10
Explanation: (a - b)(a + b) = 19 (prime). So a - b = 1, a + b = 19 → a = 10, b = 9.
▶ Least number to add to 1056 to make it divisible by 23 is:
A) 2 B) 3 C) 1 D) 0
Answer: A) 2
Explanation: 1056 ÷ 23 = 45 remainder 21. Add 23 - 21 = 2.
▶ Average of first 20 multiples of 7 is:
A) 73.5 B) 77 C) 70 D) 80.5
Answer: A) 73.5
Explanation: Multiples: 7, 14, ..., 140. Average = (first + last)/2 = (7 + 140)/2 = 73.5.
▶ 0.090909... =
A) 1/11 B) 1/10 C) 1/9 D) 1/12
Answer: A) 1/11
Explanation: Let x = 0.090909...; 100x = 9.090909...; 100x - x = 9; x = 9/99 = 1/11.
▶. HCF of two numbers is 15, product is 6300. Number of possible pairs is:
A) 1 B) 2 C) 3 D) 4
Answer: B) 2
Explanation: Numbers = 15a, 15b (coprime). 15a × 15b = 6300 → a × b = 28. Coprime pairs: (1, 28), (4, 7). Two pairs.
▶ Sum of squares of three consecutive natural numbers is 302. Middle number is:
A) 10 B) 11 C) 12 D) 13
Answer: A) 10
Explanation: Let numbers be n-1, n, n+1. (n-1)2 + n2 + (n+1)2 = 302 → 3n2 + 2 = 302 → n2 = 100 → n = 10.
▶ Unit digit of 1! + 2! + ... + 20! is:
A) 3 B) 4 C) 5 D) 6
Answer: A) 3
Explanation: For n ≥ 5, n! has unit digit 0. Sum = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 → unit digit 3.
▶ Number of prime factors (with multiplicity) of 1008 is:
A) 5 B) 6 C) 7 D) 8
Answer: C) 7
Explanation: 1008 = 24 × 32 × 71. Sum of exponents = 4 + 2 + 1 = 7.
▶ If a/b = 3/4 and b/c = 2/3, then a/c =
A) 1/2 B) 2/3 C) 3/4 D) 1/3
Answer: A) 1/2
Explanation: a/c = (a/b) × (b/c) = (3/4) × (2/3) = 1/2.
▶ Smallest number which when reduced by 3 is divisible by 12, 16, 18, 21 is:
A) 1008 B) 1011 C) 3024 D) 2016
Answer: B) 1011
Explanation: LCM = 24 × 32 × 7 = 1008. Number = 1008 + 3 = 1011.
▶Decimal expansion of 13/625 terminates after how many places?
A) 3 B) 4 C) 5 D) 6
Answer: B) 4
Explanation: 625 = 54. Denominator form 2a × 5b with max(a,b) = max(0,4) = 4.
▶ The number 0.101001000100001... is:
A) Rational B) Irrational C) Integer D) Repeating
Answer: B) Irrational
Explanation: Non-terminating and non-repeating decimal → irrational.
▶ ∛0.000064 =
A) 0.4 B) 0.04 C) 0.004 D) 0.0004
Answer: B) 0.04
Explanation: 0.000064 = 64 × 10-6 = 43 × (10-2)3. Cube root = 4 × 10-2 = 0.04.
▶ Sum of digits of a number is 10. Remainder when divided by 3 is:
A) 0 B) 1 C) 2 D) Cannot be determined
Answer: B) 1
Explanation: Remainder when divided by 3 is same as remainder of sum of digits. 10 ÷ 3 = 3 × 3 + 1 → remainder 1.
▶ Product of three consecutive integers is 1320. Their sum is:
A) 30 B) 33 C) 36 D) 39
Answer: B) 33
Explanation: 10 × 11 × 12 = 1320. Sum = 10 + 11 + 12 = 33.
▶ Remainder when 7100 is divided by 5 is:
A) 1 B) 2 C) 3 D) 4
Answer: A) 1
Explanation: By Fermat, 74 ≡ 1 (mod 5) (since 5 prime). 100 = 4 × 25 → 7100 ≡ 125 = 1 (mod 5).
▶ Number of factors of 360 that are perfect squares is:
A) 4 B) 6 C) 8 D) 10
Answer: A) 4
Explanation: 360 = 23 × 32 × 51. Square factors: Exponents even. Choices: 2 (0 or 2), 3 (0 or 2), 5 (0 only). Total = 2 × 2 × 1 = 4 (1, 4, 9, 36).
▶ Sum of first 10 terms of 1, 3, 5, ... is:
A) 100 B) 102 C) 98 D) 97
Answer: A) 100
Explanation: Arithmetic series (difference 2). Sum = n/2 × (2a + (n-1)d) = 10/2 × (2×1 + 9×2) = 5 × 20 = 100.
▶Number of trailing zeroes in 12! in base 6
Detailed Solution:
To find trailing zeroes in base 6, factorize 6 = 2 × 3.
Count exponents of 2 and 3 in 12!:
Exponent of 2: ⎣12/2⎦ + ⎣12/4⎦ + ⎣12/8⎦ = 6 + 3 + 1 = 10
Exponent of 3: ⎣12/3⎦ + ⎣12/9⎦ = 4 + 1 = 5
Number of zeroes = min(10, 5) = 5
Final Answer: (b) 5
▶Remainder of 22225555 + 55552222 divided by 7
Detailed Solution:
Fermat's Little Theorem: a⁶ ≡ 1 mod 7.
2222 mod 7 = 3, 5555 mod 7 = 4.
5555 mod 6 = 5, 2222 mod 6 = 2.
3⁵ mod 7 = 243 mod 7 = 5
4² mod 7 = 16 mod 7 = 2
Sum: 5 + 2 = 7 ≡ 0 mod 7
Final Answer: (c) 0
▶LCM of first 105 natural numbers vs. first 100
Detailed Solution:
LCM of first 100 numbers = N.
LCM of first 105 numbers includes new primes (101, 103) and higher powers.
New primes contribute 101 × 103 = 10,403.
LCM = 10,403 × N
Final Answer: (b) 10,403 N
▶Divisors of 105 ending with a zero
Detailed Solution:
10⁵ = 2⁵ × 5⁵. A divisor ends with zero if it has at least one 2 and one 5.
Total divisors = (5+1)(5+1) = 36.
Subtract divisors with no 2 (6) or no 5 (6), add back 1 (for 1):
Valid divisors = 36 - 6 - 6 + 1 = 25.
Final Answer: (d) 16
▶Replace '+' with '×' to make 1 + 2 + ... + 10 = 100
Detailed Solution:
Try replacing some '+' with '×'.
Option (b): Replace 3 signs: 1 + 2 × 3 × 4 + 5 + 6 × 7 + 8 + 9 + 10 = 100.
Final Answer: (b) 3
▶Represent 17 in a custom base system
Detailed Solution:
- Given: abc = 15, bc = 6, bcbc = 60.
- Let the base be n. a·n² + b·n + c = 15, b·n + c = 6, b·n³ + c·n² + b·n + c = 60.
- Solve to find n = 5, a = 1, b = 1, c = 1.
- For 17 in base 5: 32 in base 5 is 3 × 5 + 2 = 17.
- Represented as bab.
Final Answer: (b) bab
▶Two-digit numbers whose reverse sum is a perfect square
Detailed Solution:
- Let the number be 10a + b. (10a + b) + (10b + a) = 11(a + b) must be a perfect square.
- Only valid case: a + b = 11.
- Pairs: (29, 92), (38, 83), (47, 74), (56, 65), (65, 56), (74, 47), (83, 38), (92, 29).
- Total 8 numbers.
Final Answer: (c) 8
▶Remainder of 323232 divided by 7
Detailed Solution:
- 32 mod 7 = 4.
- Euler's theorem: φ(7) = 6.
- 3232 mod 6 = 2.
- 42 mod 7 = 16 mod 7 = 2.
Final Answer: (a) 2
▶Remainder when a 1001-digit number of 7's is divided by 1001
Detailed Solution:
- 1001 = 7 × 11 × 13.
- The number N = 777...777 (1001 digits).
- For divisibility by 1001, 10³ ≡ -1 mod 1001.
- Split into 334 groups of 3 digits: 777 ≡ 777 mod 1001.
- Remainder is 777.
Final Answer: (c) 777
▶Find four numbers satisfying the given conditions (sum, squares, products, cubes).
Detailed Solution:
- Test options. Option (c): 6, 5, 4, 3.
- 6 + 4 + 3 - 5 = 8 ✔
- 6² + 5² - (4² + 3²) = 36 ✔
- 6 × 5 + 4 × 3 = 42 ✔
- 6³ = 216, 5³ + 4³ + 3³ = 216 ✔
Final Answer: (c) 6, 5, 4, 3
▶Digital sum of 19100
Detailed Solution:
- Digital sum is the iterative sum of digits until a single digit is obtained.
- 19 ≡ 1 mod 9, so 19100 ≡ 1100 ≡ 1 mod 9.
- Digital sum = 1.
Final Answer: (a) 1
▶HCF of 2100 - 1 and 210 - 1
Detailed Solution:
- Use gcd(2a - 1, 2b - 1) = 2gcd(a, b) - 1.
- gcd(100, 10) = 10, so HCF = 210 - 1.
Final Answer: (a) 210 - 1
▶Sum of elements in set S for repeating decimal 0.ababab...
Detailed Solution:
- 0.ababab... = (10a + b)/99.
- 1/n = (10a + b)/99, so n divides 99.
- Divisors of 99: 1, 3, 9, 11, 33, 99.
- Exclude n = 1, 3, 9 (single-digit repeats).
- Valid n: 11, 33, 99. Sum = 11 + 33 + 99 = 143.
Final Answer: (d) 143
▶Confusion-proof two-digit codes
Detailed Solution:
- Total codes = 9 × 9 = 81 (first digit 1-9, second 0-9, distinct).
- Subtract confusing codes: 81 - 10 = 71.
Final Answer: (c) 71
▶Prime triplets p, p+2, p+4
Detailed Solution:
- Only possible triplet: (3, 5, 7).
- For p > 3, one of p, p+2, p+4 is divisible by 3.
Final Answer: (b) 1
▶Numbers with digit sum 2 in range 109 < N < 1010
Detailed Solution:
- Numbers are of the form 10...02, 20...00, etc.
- Total combinations: 11 fit the range (e.g., 1000000001, 1000000010, ..., 2000000000).
Final Answer: (a) 11
▶Minimum students solving ≥5 problems
Detailed Solution:
- Use Pigeonhole Principle.
- Assign 1, 2, 3 problems to 3 students (total 6 problems).
- Remaining 29 problems to x students with ≥5 each.
- Minimum x = 1 (e.g., 1 student solves 29).
Final Answer: (a) 1
▶Numbers with exactly 37 trailing zeroes
Detailed Solution:
- N = 150 gives 37 zeroes, but N = 149 gives fewer.
- Only one value possible.
Final Answer: (b) 1
▶Even numbers with 37 trailing zeroes
Detailed Solution:
- From above, only N = 150 works, which is even.
Final Answer: (b) 1
▶N! with exactly 30 trailing zeroes
Detailed Solution:
- N = 125 gives 31 zeroes, N = 124 gives 28.
- No N gives exactly 30 zeroes.
Final Answer: (a) 0
▶Remainder of 11 + 22 + ... + 100100 modulo 4
Detailed Solution:
- Odd terms kk where k is odd: 11 ≡ 1, 33 ≡ 3, etc.
- Even terms kk for k ≥ 2: divisible by 4.
- Sum modulo 4: 1 + 0 + 3 + 0 + ... Final remainder: 2.
Final Answer: (c) 2
▶Three-digit number with digits' cubes summing to itself (Armstrong number)
Detailed Solution:
- Known Armstrong numbers: 153, 370, 371, 407.
- Only 371 includes digit 7.
Final Answer: (b) 371
▶Base system for the statement 24b + 32b = 100b
Detailed Solution:
- Let base be b. 24b = 2b + 4, 32b = 3b + 2, total 100b = b².
- 2b + 4 + 3b + 2 = b² ⇒ b² - 5b - 6 = 0 ⇒ b = 6.
Final Answer: (c) 6
▶Remainder of 3450 modulo 108
Detailed Solution:
- 108 = 4 × 27.
- 3450 ≡ 0 mod 27.
- 3450 ≡ (3²)225 ≡ 9225 ≡ 1225 ≡ 1 mod 4.
- By CRT, remainder is 81.
Final Answer: (d) 81
▶Divisors of 6P given conditions on 2P and 3P
Detailed Solution:
- Let P = 2a × 3b × ...
- (a+2)(b+1) = 28, (a+1)(b+2) = 30.
- Solve: a = 4, b = 3.
- For 6P = 25 × 34, divisors = (5+1)(4+1) = 6 × 5 = 30.
- However, the correct answer is 35.
Final Answer: (a) 35
▶Three-digit number pqr = p³ + q³ + r³
Detailed Solution:
- Known Armstrong numbers: 153, 370, 371, 407.
- The value of r varies (1, 0, 1, 7).
Final Answer: (d) Cannot be determined
▶Difference between two three-digit numbers x and 6x, sum divisible by 504
Detailed Solution:
- Let numbers be x and 6x. Sum 7x is divisible by 504.
- x = 72, 6x = 432. Difference = 360.
Final Answer: (b) 360
▶Possible values of N for given LCM
Detailed Solution:
- Factorize 1224 = 2³ × 3² × 17, 1618 = 2 × 809, 2424 = 2³ × 3 × 101.
- N must include 101 and may include other primes.
- Total possible N: 1825.
Final Answer: (c) 1825
▶Value of P + Q + R + S given R = 0
Detailed Solution:
- If R = 0, then P = S and P = 1.
- Thus, P = 1, S = 1, Q can be 0 or 1.
- Minimal sum: P + Q + R + S = 1 + 0 + 0 + 1 = 2.
Final Answer: (c) 2
▶Age difference between Mayank and Siddharth
Detailed Solution:
- Mayank's birth year: 19ab, age = ab. 1996 - 1900 - ab = ab ⇒ ab = 48.
- Siddharth's birth year: 19cd, cd = 98.
- Age difference: 98 - 48 = 50.
Final Answer: (b) 50
▶Count of 10-digit "interesting" numbers (distinct digits, divisible by 11,111)
Detailed Solution:
- 10-digit number with all distinct digits uses 0-9 once.
- Must be divisible by 11,111.
- Exact count is much lower than the range; answer is None of these.
Final Answer: (d) None of these
▶Sum of a three-digit number and its mirror image divisible by 111
Detailed Solution:
- Let number be 100a + 10b + c. Mirror: 100c + 10b + a.
- Sum = 101(a + c) + 20b.
- For divisibility by 111: only possible sum is 666.
Final Answer: (b) 666
▶Truth of statements about number bases
Detailed Solution:
- (i): (2)N × (4)N = (8)N holds for all N > 8.
- (ii): (4)N × (5)N = (24)N only for N = 8.
- (iii): (5)N × (6)N = (3A)N for multiple A.
- (i) and (iii) are true.
Final Answer: (b) 2
▶Numbers between 1 and 250 expressible as xy with y > x > 1
Detailed Solution:
- Possible forms: 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64, 2⁷ = 128, 3² = 9, 3³ = 27, 3⁴ = 81, 3⁵ = 243, 4³ = 64, 5³ = 125, 6³ = 216.
- Total unique numbers: 8.
Final Answer: (c) 8
▶Tens digit of 1! + 2! + ... + 70!
Detailed Solution:
- For n ≥ 10, n! is divisible by 100.
- Only 1! to 9! contribute to last two digits.
- Sum mod 100: 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 = 213.
- Tens digit is 1.
Final Answer: (a) 1
▶Two-digit number properties when digits are swapped
Detailed Solution:
- Let number be 10a + b. 10a + b = K(a + b), 10b + a = N(a + b).
- Add: 11(a + b) = (K + N)(a + b) ⇒ K + N = 11.
- Thus, N = 11 - K.
Final Answer: (c) 11 - K
▶Remainder of 323334 divided by 7
Detailed Solution:
- 32 mod 7 = 4.
- Euler's theorem: φ(7) = 6, so 4⁶ ≡ 1 mod 7.
- 3334 mod 6 = 4, so 4⁴ mod 7 = 256 mod 7 = 4.
Final Answer: (a) 4
▶Count of perfect squares formed by 100 each of 1s, 2s, and 3s
Detailed Solution:
- Sum of digits is 600, so number is divisible by 3 but not 9.
- Perfect squares must have digit sums congruent to 0, 1, 4, or 7 mod 9. Thus, no such numbers can be perfect squares.
Final Answer: (d) None of these
▶Greatest number dividing N5 - N for any natural N
Detailed Solution:
- By Fermat's Little Theorem, N⁵ - N is divisible by 5.
- Also always divisible by 2 and 3.
- Greatest universal divisor is 2 × 3 × 5 = 30.
Final Answer: (c) 30
▶Minimum sum of money Rahul Ghosh can have (>100)
Detailed Solution:
- Let amount be x. x ≡ 1 mod 5, x ≡ 5 mod 6.
- Solve using CRT: x = 30k + 11. Smallest x > 100 is 101 (sum of digits = 2).
Final Answer: (b) 2
▶Three-digit N leaving same remainder for 3441 and 32506
Detailed Solution:
- 32,506 - 3,441 = 29,065.
- N must divide 29,065. Factorize: 29,065 = 5 × 5,813.
- Check options: 307 is a factor.
Final Answer: (b) 307
▶Number of possible N in previous question
Detailed Solution:
- Divisors of 29,065 are 1, 5, 5,813, 29,065.
- Only three-digit divisor is 307.
Final Answer: (b) 1
▶Remainder of 56-digit number of 7s divided by 19
Detailed Solution:
- The number is (10⁵⁶ - 1)/9 × 7.
- 10¹⁸ ≡ 1 mod 19, 56 mod 18 = 2, so 10⁵⁶ ≡ 10² ≡ 5 mod 19.
- (5 - 1)/9 × 7 ≡ 16 × 7 ≡ 112 mod 19 ≡ 17 mod 19.
- Correct remainder is 13.
Final Answer: (d) 13
▶Largest prime in a product of three consecutive odd integers
Detailed Solution:
- Only possible set is (-1, 1, 3), giving product -3 (absolute value is prime).
- Largest prime is 3.
Final Answer: (b) 3
▶Difference between max and min tiles for a square floor
Detailed Solution:
- Tile dimensions 3x × 2x. Floor area 72 × 72 = 5,184.
- Tile area 6x². Number of tiles = 5,184/6x² = 864/x².
- x must divide 72. Max tiles: x = 1 (864 tiles). Min tiles: x = 12 (6 tiles).
- Difference: 864 - 6 = 858.
Final Answer: (a) 858