IPMPrepWhen x2 + 2x + 6 + k is divided by x + 1, the remainder is −2. Find k.
Solution
The remainder equals the value at x = −1: −5 + k = −2.
k = 3.
Answer: 3
Quantitative Ability · Numbers
Number System: IPMAT practice questions
24 questions, 12 of them short answer (type the number, as in IPMAT Indore). Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
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What is the smallest positive integer by which 525 must be multiplied to make it a perfect square?
Solution
525 = 3 × 52 × 7.
Primes with an odd power need one more copy: multiply by 21.
525 × 21 = 11025 = 1052.
Answer: (B) 21
Find the remainder when 9170 is divided by 10.
Solution
Work with powers of 9 modulo 10.
92 ≡ 1 (mod 10), so powers repeat every 2.
170 = 2 × 85 + 0, so 9170 ≡ 90 ≡ 1 (mod 10).
Answer: 1
What is the units digit of 1415 × 16133?
Solution
The units digit of 1415 is 4 (units digits of powers of 4 repeat in a cycle of at most 4).
The units digit of 16133 is 6.
So the answer is the units digit of 4 × 6: 4.
Answer: 4
Find the remainder when x3 + 5x2 − 3x is divided by x + 3.
Solution
By the remainder theorem, the remainder is the value at x = −3.
(−3)3 + 5(−3)2 − 3(−3) + 0 = 27.
Answer: (A) 27
What is the largest integer m such that 12m divides 168!?
Solution
12 = 22 × 3.
Power of 2 in 168! = 165, giving 82 copies of 22.
Power of 3 in 168! = 82.
The limiting count is 82, so m = 82.
Answer: 82
How many positive factors does 2200 have?
Solution
2200 = 23 × 52 × 11.
Number of factors = (3 + 1) × (2 + 1) × (1 + 1) = 24.
Answer: (B) 24
What is the largest integer m such that 7m divides 163!?
Solution
7 is prime; count the 7s in 163!.
Power of 7 in 163! = 26.
The limiting count is 26, so m = 26.
Answer: (D) 26
Find the remainder when 3414 is divided by 5.
Solution
Work with powers of 3 modulo 5.
34 ≡ 1 (mod 5), so powers repeat every 4.
414 = 4 × 103 + 2, so 3414 ≡ 32 ≡ 4 (mod 5).
Answer: (D) 4
What is the units digit of 69329 × 85205?
Solution
The units digit of 69329 is 9 (units digits of powers of 9 repeat in a cycle of at most 4).
The units digit of 85205 is 5.
So the answer is the units digit of 9 × 5: 5.
Answer: 5
What is the units digit of 3135 + 38101?
Solution
The units digit of 3135 is 1 (units digits of powers of 1 repeat in a cycle of at most 4).
The units digit of 38101 is 8.
So the answer is the units digit of 1 + 8: 9.
Answer: 9
Find the remainder when x3 + 2x2 + 2x − 1 is divided by x + 2.
Solution
By the remainder theorem, the remainder is the value at x = −2.
(−2)3 + 2(−2)2 + 2(−2) − 1 = −5.
Answer: (D) −5
How many zeros are there at the end of 733!?
Solution
Each zero needs a factor 10 = 2 × 5, and 5s are scarcer than 2s.
Count of 5s = [7335] + [73325] + [733125] + [733625] = 146 + 29 + 5 + 1 = 181.
Answer: 181
How many perfect-square factors does 540 have?
Solution
540 = 22 × 33 × 5.
A square factor uses even powers only: (1 + 1) × (1 + 1) × (0 + 1) = 4.
Answer: 4
How many zeros are there at the end of 881!?
Solution
Each zero needs a factor 10 = 2 × 5, and 5s are scarcer than 2s.
Count of 5s = [8815] + [88125] + [881125] + [881625] = 176 + 35 + 7 + 1 = 219.
Answer: 219
What is the largest integer m such that 3m divides 95!?
Solution
3 is prime; count the 3s in 95!.
Power of 3 in 95! = 45.
The limiting count is 45, so m = 45.
Answer: 45
What is the smallest positive integer by which 560 must be multiplied to make it a perfect square?
Solution
560 = 24 × 5 × 7.
Primes with an odd power need one more copy: multiply by 35.
560 × 35 = 19600 = 1402.
Answer: (B) 35
How many perfect squares lie between 1212 and 2027 (inclusive)?
Solution
The smallest square ≥ 1212 is 352 = 1225; the largest ≤ 2027 is 452 = 2025.
Count = 45 − 35 + 1 = 11.
Answer: (C) 11
The number 5x3113 is divisible by 11, where x is a single digit. Find x.
Solution
A number is divisible by 11 when the alternating sum of its digits is divisible by 11.
Only x = 5 works: 553113 ÷ 11 = 50283.
Answer: (C) 5
The number 2x245 is divisible by 11, where x is a single digit. Find x.
Solution
A number is divisible by 11 when the alternating sum of its digits is divisible by 11.
Only x = 5 works: 25245 ÷ 11 = 2295.
Answer: 5
Find the remainder when 3312 is divided by 5.
Solution
Work with powers of 3 modulo 5.
34 ≡ 1 (mod 5), so powers repeat every 4.
312 = 4 × 78 + 0, so 3312 ≡ 30 ≡ 1 (mod 5).
Answer: (C) 1
How many zeros are there at the end of 798!?
Solution
Each zero needs a factor 10 = 2 × 5, and 5s are scarcer than 2s.
Count of 5s = [7985] + [79825] + [798125] + [798625] = 159 + 31 + 6 + 1 = 197.
Answer: 197
The number 88666x is divisible by 8, where x is a single digit. Find x.
Solution
A number is divisible by 8 when its last three digits form a multiple of 8.
Only x = 4 works: 886664 ÷ 8 = 110833.
Answer: (C) 4
How many positive factors does 90 have?
Solution
90 = 2 × 32 × 5.
Number of factors = (1 + 1) × (2 + 1) × (1 + 1) = 12.
Answer: (B) 12
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.