Contents

# Q1. If a, b, and c are consecutive positive integers, then the largest number that always divides (a2 + b2 + c2)

1. 14
2. 55
3. None of these

Let the numbers are (x − 1), x, and (x + 1).

Then, (x − 1)2 + x2 + (x + 1)2 = (3x2 + 2)

When x = 2, then 3x2 + 2 = 14 Further, when x = 3, then 3x2 + 2 = 29

Therefore, the largest number that will always divide

(a2+b2+c2) = 1

# Q2. If n^2 is a perfect cube, then which of the following statements is always true?

1. n is odd.
2. n is even.
3. n2 is a perfect square.
4. n is a perfect cube.

Anwer:-

If n is a perfect cube, then n2 will also be a perfect cube. Therefore, the answer is option (4).

# Q3. By multiplying with which of the following numbers, does the product of 8 × 9 × 10 × 11 × 12 become a perfect square?

1. 55
2. 11
3. 165
4. 310

Answer:- Let the number be x. Hence, x × 8 × 9 × 10 × 11 × 12 = x × 26 × 32 × 3 × 5 × 11 Here, we can say that for being a perfect square, x should be 3 × 5 × 11 = 165.

# Q4. If the units digit in the product (47n × 729 × 345 × 343) is 5, what is the maximum number of values that n may take?

1. 9
2. 3
3. 7
4. 5

Answer:- 47n × 729 × 345 × 343 = 47n × 862,66,215  It is given that the units digit of 47n × 862,66,215 or n × 5 is 5.

# Q4. What is the units digit of 21^3 × 21^2 × 34^7 × 46^8 × 77^8 ?

1. 4
2. 8
3. 6
4. 2

Answer:- 21^3 × 21^2 × 34^7 × 46^8 × 77^8 = 1 × 1 × 4 × 6 × 1 = 24  Therefore, units digit = 4

# Q5. p and q are two prime numbers such that p

1. 5
2. 6
3. 7
4. None of these

Answer:- Do it from the actual calculation.

The values of P + Q = 5, 7, 13, 19, 31, 43. Hence, the answer is 6.

# Q6. If p and q are consecutive natural numbers (in increasing order), then which of the following is true?

1. q^2 < p
2. 2p > p^2
3. (q + 1)^2 > p^2
4. (p + 2)^3 < q^3

Answer:- It is given that p and q are consecutive natural numbers,

such that p < q.

Hence, option (a) is incorrect for every possible value of p and q. Option (b) is incorrect for p equals to 1 and q equals to 2. Option (d) is incorrect for every possible value of p and q. Therefore, the answer is option (c)

# Q7. (17^21 + 19^21) is not divisible by

1. 36
2. 8
3. 9
4. 18

Answer:- We know that (an + bn) is divisible by (a + b) if n is an odd number. It means (17^21 + 19^21) is divisible by 36 and all the factors of 36.

Therefore, the answer is 8 because 8 is not a factor of 36.

1. 3
2. 4
3. 1
4. 5

## Q9. I have 77 sweets and I want to distribute them equally among 24 students. After each of the student got maximum integral sweets, how many sweets are left with me?

1. 8
2. 5
3. 1
4. None of these

The question is asking about the remainder when we divide 77 by 24. Remainder is 7.

1. 2
2. 0
3. 6
4. 7

## Q11. MUL has a waiting list of 5005 applicants. The list shows that there are at least 5 males between any two females. The largest number of females in the list could be:

1. 920
2. 835
3. 721
4. 1005

Let the ﬁ rst applicant be female. The remaining applicants = 5005 − 1 = 5004 For maximum female applicants, for every six applicants, there should be a female. Therefore, number of females = 1 + (5004/6) = 1 + 834 = 835

## Q12.What is the units digit of 1! + 2! + 3! +99! + 100!?

1. 3
2. 1
3. 5
4. 6

Units digit of 1! + 2! + 3! + 4! + 5! + 6! …  = 1 + 2 + 6 + 24 + 120 + 0 …= 3

Note: We know that units digit of 5! or for all the numbers greater than 5! is zero

## Q13. A number N has odd number of divisors. Which of the following is true about N?

1. All the divisors of this number will be odd.
2. There will be at least (N − 11) prime divisors.
3. N will be a perfect square.
4. At least one divisor of the number should be odd

If a number has an odd number of divisors, then it is a perfect square.

## Q14. How many zeroes will be there at the end of the expression N = 7 × 14 × 21 × … × 777?

1. 24
2. 25
3. 26
4. None of these

N = 7 × 14 × 21 × … × 777

Method 1 In this expression, every ﬁ fth term is a multiple of 5. Now, there are 111 terms in the expression. Therefore, number of 5s = (111/5) + (111/25) = 22 + 4 = 26

Method 2 N = 7 × 14 × 21 ×…× 777 = (7 × 1) × (7 × 2) × (7 × 3) …× (7 × 111) = 7111 × (1 × 2 × 3 × … × 111) = 7111 × 111! Number of zeroes in 111! = (111/5) + (111/52) = 22 + 4 = 26

1. 0
2. 1
3. 3
4. 6

## Q16. How many divisors of N = 420 will be of the form (4n + 1), where n is a whole number?

1. 3
2. 4
3. 5
4. 8

N = 420 = 22 × 3 × 5 × 7 Odd factors in N = 1, 3, 5, 7, 15, 21, 35, 105 Now, (4n + 1) format → Remainder obtained when divided by 4 is 1. Therefore, (4n + 1) format number = 1, 5, 21, 105

## Q17. Find the units digit of N = 1727!37!

1. 9

71 = 7, 72 = 9, 73 = 3 and 74 = 1 Therefore, the cycle of 7 is 4, and 27!371! is divisible by 4. Therefore, units digit is 1.

## Q18. The number formed by writing any digit 6 times (e.g., 111,111, 444,444, etc.) is always divisible by:

(i) 7       (ii) 11               (iii) 13

1. (i) and (ii)
2. (ii) and (iii)
3. (i) and (iii)
4. (i), (ii) and (iii)

See the divisibility rule of 7, 11, and 13. These types of number will always divisible by 3, 7, 11, 13, and 37

1. 38
2. 39
3. 40
4. 41

## Q20. N = 23 × 53. How many sets of two distinct factors of N are co-prime to each other?

1. 12
2. 24
3. 23
4. 1

For N = 23 × 53 The number of sets of factors co-prime to each other = (x + 1) (y + 1) + xy = (3 + 1) (3 + 1) + 3 × 3 = 25. However, for co-prime set (1, 1), factors are not distinct. Therefore, number of sets = 25 − 1 = 24

1. 1
2. 4

## Q20. N = 23 × 53. How many sets of two distinct factors of N are co-prime to each other?

1. 12
2. 24
3. 23
4. 1

For N = 23 × 53 The number of sets of factors co-prime to each other = (x + 1) (y + 1) + xy = (3 + 1) (3 + 1) + 3 × 3 = 25. However, for co-prime set (1, 1), factors are not distinct. Therefore, number of sets = 25 − 1 = 24

## Q22. A and B are two distinct digits. If the sum of the two-digit numbers formed by using both the digits is a perfect square, what is the value of (A + B)?

1. 9
2. 11
3. 13
4. 17

It is given that (AB + BA) = perfect square (10A + B) + (10B + A) = perfect square 11(A + B) = perfect square. For being a perfect square, (A + B) should be 11.

## iii. 9

1. Either i or ii
2. Either ii or iii
3. Either i or ii or iii
4. None of these

N = 897324P64Q For N divisible by 8, last three digits should be divisible by 8. For N divisible by 9, sum of digits should be divisible by 9. However, 64Q is divisible by 8 when Q equals 0 and 8. Now, if Q = 0, then P should be 2, and if Q = 8, then P should be 3. Then, (P + Q) = 2 and 11

1. 19
2. 38
3. 13
4. 16

## Q25. In the ﬁ ring range, four shooters are ﬁ ring at their respective targets. The ﬁ rst, the second, the third, and the fourth shooter hit the target once every 5 s, 6 s, 7 s, and 8 s, respectively. If all of them hit their target at 10:00 am, when will they hit their target together again?

1. 10:14 am
2. 10:28 am
3. 10:30 am
4. 10:31 am

This question is about the LCM of 5s, 6s, 7s, and 8s. Then, LCM of 5s, 6s, 7s, and 8s = 840 sec = 14 min Hence, the time obtained when they hit target together is = 10:14 am.

## Q26. To celebrate their victory in the World Cup, the Sri Lankans distributed sweets. If the sweets were distributed among 11 players, 2 sweets were left. When the sweets were distributed to 11 players, 3 extra players, and 1 coach, even then 2 sweets were left. What is the minimum number of sweets in the box?

1. 167
2.  334
3.  332
4. 165

This question is about a number which when divided by 11 gives remainder 2 and when divided by 15 gives remainder 2 again. Now, ﬁ nd the number from actual calculation and the number is 167.

## Q27. How many zeroes will be there at the end of N = 18! + 19!?

1. 3
2. Cannot be determined

## Q28. Manish was dividing two numbers by a certain divisor and obtained remainders as 437 and 298, respectively. When he divides the sum of the two numbers by the same divisor, the remainder is 236. What is the divisor?

1. 499
2. 735
3. 971
4.  None of these

Let the number be x. It is given that if we divide the sum of two numbers, then the remainder is 236. Hence, it means when we divide (437 + 298) by x, then the remainder is 236. From here, the number x should be 499

## Q29. How many integers N in the set of integers {1, 2, 3,…, 100} are there such that N^2 + N^3 is a perfect square?

1. 5
2. 11

N2 + N3 = N2 (N + 1) For (N2 + N3) to be a perfect square, (N + 1) should be a perfect square. Further, we know that there are 10 perfect squares till 100. However, we cannot take (N + 1) = (1 → N) = 0

Therefore, there are 9 numbers for which N2 (N + 1) will be a perfect square.

## (B) It is divisible by 13. How many digits are there in N?

1. 6
2. 7
3. 8

Any number of the format abcabc or aaaaaa will be divisible by 7, 11, and 13

## Q31. What is the smallest ﬁ ve-digit number which when divided by 7, 11, and 21 leave a remainder of 3 in each case?

1. 10,019
2. 10,001
3. 10,111
4. 10,16

Number should be like (multiple of LCM of 7, 11, and 21) + 3 Then, ﬁnd the smallest ﬁve-digit multiple of LCM of 7, 11, and 21 and add 3 to that number.

1. 2520
2. 3780
3. 5040
4. 6480

## Q33. Let A, B, and C be digits such that (100A + 10B + C) (A + B + C) = 2005. What is the value of A?

1. 1

Clearly, the two quantities are both integers, and therefore, we check the prime factorization of 2005 = 5 × 401. It can be seen that (A, B, C) = (4, 0, 1) satisﬁ es the relation. Hence, option (1) is the answer.

## Q34. What is the remainder when 9091 is divided by 13?

1. 0
2. 12
3. 1

Hence, the remainder is −1 or 12.

## Q35. A faulty car odometer proceeds from digit 3 to digit 5, always skipping digit 4, regardless of position. If the odometer now reads 002,005, how many miles has the car actually travelled?

1. 1404
2. 1462
3. 1604
4. 1605

Method 1

We ﬁnd the number of numbers with a digit 4 and subtract from 2005. Fast counting tells us that there are 200 numbers with a digit 4 in the hundreds place, 200 numbers

with a digit 4 in the tens place, and 201 numbers with a digit 4 in the units place (counting 2004). There are 20 numbers with a digit 4 in the hundreds and in the tens, and 20 for both the other two intersections. The intersection of all three sets is just 2. Therefore, we get: 2005 − (200 + 200 + 201 − 20 − 20 − 20 + 2) = 1462 Hence, option (2) is the answer.

Method 2

Alternatively, consider that counting without the number 4 is equivalent to counting in base 9; only, in base 9, number 9 is not counted. Since 4 is skipped, the symbol 5 represents 4 miles of travel, and we have travelled 20049 miles. By basic conversion, 20059 = 93(2) + 90(5) = 729(2) + 1(5) = 1458 + 5 = 1463

Hence, option (2) is the answer.

## Q36. If n2 = 123.45654321, which of the following is the exact value of n?

1. 11.1001
2. 11.1101
3. 11.1111
4. 11.1011

n2 = 123.45654321 = 12,345,654,321 × 10−8 Therefore, n2 = (111,111 × 10−4)2. Therefore, n = 11.1111

## Q37. A mule said to a horse, ‘If I take one sack off your back, my load will be double of yours and if you take one off my back, our loads will

1. 12
2. 14

Let the load of mule is x and load of horse is y. Now, from the question

2(x − 1) = (y + 1) 2(x − 2) = (y + 1) → (2x − y) = 3 (i)

and

(x + 1) = (y − 1) (x − y) = −2 (ii) Now, from equation (i) and (ii) x = 5 and y = 7, then (x + y) = 12

## Q38. S is a number formed by writing 8 for 88 times. What will be the remainder of this number when divided by 7

1. 4
2. 1

A number like aaaaaa is divisible by 7. It means 8 written 84 (6 × 14) times is divisible by 7. Now, divide the last four digits of the number by 7 and ﬁnd the remainder. Hence, the remainder obtained = 8888/7 = 5

## Q39. We are writing all the multiples of 3 from 111 to 324. How many times will we write digit 3?

1. 18
2. 19
3. 21
4. 22

Do it from actual counting

1. 3
2. 9

## Q20. N = 23 × 53. How many sets of two distinct factors of N are co-prime to each other?

1. 12
2. 24
3. 23
4. 1

For N = 23 × 53 The number of sets of factors co-prime to each other = (x + 1) (y + 1) + xy = (3 + 1) (3 + 1) + 3 × 3 = 25. However, for co-prime set (1, 1), factors are not distinct. Therefore, number of sets = 25 − 1 = 24

## Q20. N = 23 × 53. How many sets of two distinct factors of N are co-prime to each other?

1. 12
2. 24
3. 23
4. 1

For N = 23 × 53 The number of sets of factors co-prime to each other = (x + 1) (y + 1) + xy = (3 + 1) (3 + 1) + 3 × 3 = 25. However, for co-prime set (1, 1), factors are not distinct. Therefore, number of sets = 25 − 1 = 24

## Q20. N = 23 × 53. How many sets of two distinct factors of N are co-prime to each other?

1. 12
2. 24
3. 23
4. 1

For N = 23 × 53 The number of sets of factors co-prime to each other = (x + 1) (y + 1) + xy = (3 + 1) (3 + 1) + 3 × 3 = 25. However, for co-prime set (1, 1), factors are not distinct. Therefore, number of sets = 25 − 1 = 24

## Q20. N = 23 × 53. How many sets of two distinct factors of N are co-prime to each other?

1. 12
2. 24
3. 23
4. 1

For N = 23 × 53 The number of sets of factors co-prime to each other = (x + 1) (y + 1) + xy = (3 + 1) (3 + 1) + 3 × 3 = 25. However, for co-prime set (1, 1), factors are not distinct. Therefore, number of sets = 25 − 1 = 24

## Q20. N = 23 × 53. How many sets of two distinct factors of N are co-prime to each other?

1. 12
2. 24
3. 23
4. 1

For N = 23 × 53 The number of sets of factors co-prime to each other = (x + 1) (y + 1) + xy = (3 + 1) (3 + 1) + 3 × 3 = 25. However, for co-prime set (1, 1), factors are not distinct. Therefore, number of sets = 25 − 1 = 24

1. 12
2. 24
3. 23
4. 1