Number System PYQ 2023 (1-10)

Q.1) Let x be a positive integer such that 7x + 96 is divisible by x. How many values of x are possible ?

                                                (UPSC CSAT – 2023)

1. 10
2. 11
3. 12
4. Infinitely many

Solution: Ans.(c)

If (7x + 96) is divisible by x, then 96 is also divisible by x. Therefore the possible values of x are the factors of 96.

Number of factors of 96 :

96 = 3¹×2⁵

Number of factors of 96 = (1+1)×(5+1) = 2×6 = 12

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Q.2) If p, q, r and s are distinct single digit positive numbers, then what is the greatest value of (p+q)(r+s) ?

                                                  (UPSC CSAT – 2023)

1. 230
2. 225
3. 224
4. 221

Solution: Ans.(b)

To be (p+q)(r+s) greatest, (p+q) and (r+s) both should be greatest

So p, q, r and s must be 9, 8, 7 and 6 not necessary in same order.

As we know the sum of all 4 numbers remain constant, so to make product maximum the two number’s difference should be minimum

So (9+6)(8+7) would be greatest

(9+6)(8+7) = 15×15 = 225

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Q.3) A number N is formed by writing 9 for 99 times. What is the remainder if N is divided by 13?

                                                 (UPSC CSAT – 2023)

1. 11
2. 9
3. 7
4. 1

Solution: Ans.(a)

As we know 1001 = 7×11×13, 1001 is divisible by 13

1001×999 = 999999 (6 times 9s) is also divisible by 13

Therefore number formed by writing 96 times 9s is also divisible by 13

Now, when remaining 3 times 9s (999) is divided by 13, the remainder is 11.

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Q.4) Each digit of a 9-digit number is 1. It is multiplied by itself. What is the sum of the digits of the resulting number?

                                                (UPSC CSAT – 2023)

1. 64
2. 80
3. 81
4. 100

Solution: Ans.(c)

111111111×111111111 = 12345678987654321

Sum of digits = 81

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Q.5) What is the sum of all digits which appear in all the integers from 10 to 100 ? 

                                                (UPSC CSAT – 2023)

1. 855
2. 856
3. 910
4. 911

Solution: Ans.(b)

Every non-zero digit (1, 2, 3, 4, 5, 6, 7, 8, 9) appears 20 times in all the integers from 1 to 100.

19 times in all the integers from 10 to 99.

Sum of all digits appear in the integers from 10 to 99 is = (1×19 + 2×19 + 3×19 + 4×19 + 5×19 + 6×19 + 6×19 + 7×19 + 8×19 + 9×19) = 45 × 19 = 855

Sum of all digits appear in the integers from 10 to 100 is = 855+1 = 856

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Q.6) Three of the five positive integers p, q, r, s, t are even and two of them are odd (not necessarily in order). Consider the following:

1. p + q + r – s – t  is definitely even.

2. 2p + q + 2r – 2s + t  is definitely odd.

Which of the above statements is/are correct?

                                           (UPSC CSAT – 2023)

1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2

Solution: Ans.(a)

Statement 1.  If we add or subtract three even integers it results even number always. And if we add or subtract two odd integers it results even number always.

Therefore ( p + q + r – s – t ) is definitely even.

Statement 2.  In (2p + q + 2r – 2s + t), (2p + 2r – 2s) is always even number and (q + t) may be even or odd as per the value of q and t (even or odd)

Therefore (2p + q + 2r – 2s + t) may be even or odd.

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Q.7) Consider the following in respect of prime number p and composite number c.

1.     can be even.

2.  2p + c  can be odd.

3.  pc  can be odd.

Which of the statements given above are correct?

                                        (UPSC CSAT – 2023)

1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3

Solution: (d)

Statement 1.   For p = 11, c = 9,  can be even.

Statement 2.   For p =11, c = 9, 2p + c can be odd.

Statement 3.   For p = 11, c = 9, pc  can be odd.

Hence all the statements are correct.

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Q.8) A 3-digit number ABC, on multiplication with

D gives 37DD where A, B, C and D are different non-zero digits. What is the value of A + B + C ?

                                           (UPSC CSAT – 2023)

1. 18
2. 16
3. 15
4. Cannot be determined due to insufficient data.

Solution: Ans.(a)

ABC × D = 37DD,

When D multiply by C, it should be D as unit digit, therefore D is a single digit number that can give itself by multiplying any other number, so it can be 2, 4, 5 or 8.

If D is 2, then C should be 6 which gives unit digit as 2 and 1 carry forward, but it can not make again 2 by multiplying it to B.

If D is 5, then C should be any odd number, which gives unit digit 5 and some carry forward, but next it also can not make again 5.

If D is 8, then C should be 6 which gives 8 as unit digit and 4 carry forward. And B should be 3 which gives 28 (=24+4) but next any value of A can not make 37.

If D is 4, then C should be 6 which gives unit digit 4 and 2 carry forward. And B should be 3 which gives 14 (=12+2), and next A should be 9 which gives 37 (=36+1).

Hence A=9, B=3, C=6 and D=4

And A + B + C = 9+3+6 = 18.

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Q.9) For any choices of values of X, Y and Z, the 6-digit number of the form XYZXYZ is divisible by

                                           (UPSC CSAT – 2023)

1. 7 and 11 only
2. 11 and 13 only
3. 7 and 13 only
4. 7,11 and 13

Solution: Ans.(d)

XYZXYZ = XYZ×1000 + XYZ

                  = XYZ × (1000 + 1)

                  = XYZ × 1001

                  = XYZ × 7×11×13

Hence XYZXYZ is divisible by 7, 11 and 13 all.

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Q.10) What is the remainder when

85 × 87 × 89 × 91 × 95 × 96  is divided by 100?

                                              (UPSC CSAT – 2023)

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

Solution: Ans.(a)

100 = 5×5×4

The expression 85 × 87 × 89 × 91 × 95 × 96 have all the factors 5×5×4, therefore all the factors cancelled out and remainder will be 0.

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