Number System PYQ 2023 (11-17)

Q.11) What is the unit digit in the expansion of

(57242)^{9×7×5×3×1}           

                                             (UPSC CSAT – 2023)

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

Solution: Ans.(a)

When 9×7×5×3×1 is divided by 4, remainder be 1.

So (57242)⁴ⁿ⁺¹ = (57242)⁴ⁿ × (57242)

                         = (……..6) × (…….2) = (……..2)

---

Q.12) If ABC and DEF are both 3-digit numbers such that A, B, C, D, E and F are distinct non-zero digits such that ABC + DEF = 1111, then what is the value of  A + B + C + D + E + F ?

                                            (UPSC CSAT – 2023)

1. 28
2. 29
3. 30
4. 31

Solution: Ans.(d)

ABC + DEF = 1111

(A×100 + B×10 + C) + (D×100 + E×10 + F) = 1000 + 100 + 11

(A + D)×100 + (B + E)×10 + (C + F) = 10×100 + 10×10 + 11

So (A + D) = 10, (B + E) = 10, (C + F) = 11

A + B + C + D + E + F = 10+10+11 = 31

---

Q.13) D is a 3-digit number such that the ratio of the number to the sum of its digits is least. What is the difference between the digit at the hundred’s place and the digit at the unit’s place of D?

                                            (UPSC CSAT – 2023)

1. 0
2. 7
3. 8
4. 9

Solution: Ans.(c)

199/1+9+9 = Least, D=199

Difference between the digit at the hundred’s place and the digit at the unit’s place = 9 – 1 = 8

---

Q.14) What is the remainder if  2¹⁹²  is divided by 6 ?                                    

                                                  (UPSC CSAT – 2023)

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

Solution: Ans.(d)

 =  

 = Rem. (-1)¹⁹¹ = Rem. (- 1) = Rem. (2)

Final Rem. (2×2) = 4

(Because 2 was the common factor of Dividend and Divisor )

---

Q.15) AB and CD are 2-digit numbers. Multiplying AB with CD results in a 3-digit number DEF. Adding DEF to another 3-digit number GHI results in 975. Further A, B, C, D, E, F, G, H, I are distinct digits. If E = 0, F = 8 then what is A + B + C equal to ?                                  

                                                 (UPSC CSAT – 2023)

1. 6
2. 7
3. 8
4. 9

Solution: Ans.(a)

So D & G should have the value 4 and 5 (not necessary in same order)
The value of D can not be 5, because when we multiply D & B, it can not make 8 (F), So D = 4 & G = 5.

Then B should be 2, which make D×B=8,

All the digits are distinct, therefore  A and C would have any values of  1, 3 or 9.

But as we can see that 12×34 = 408

Therefore A=1 & C=3

Hence A+B+C = 1+2+3 = 6

---

Q.16) How many natural numbers are there which give a remainder of 31 when 1186 is divided by these natural numbers ?    

                                               (UPSC CSAT – 2023)

1. 6
2. 7
3. 8
4. 9

Solution: Ans.(d)

All the factors of 1155 (118 6-31=1155) greater than 31 are the required numbers

How to calculate number of factors of 1155:

1155 = 11¹×7¹×5¹×3¹

Number of factors = 2×2×2×2 = 16

Factors less than 31 are = 1, 3, 5, 7, 11, 15, 21 (7 factors)

Hence the number of natural number which when divides 1186 gives 31 as remainder are = 16 – 7 = 9  

---

Q.17)  Let pp, qq and rr be 2-digit numbers where     p < q < r. If  pp + qq + rr = tt0, where tt0 is a 3-digit number ending with zero, consider the following statements:                     

1. The number of possible values of p is 5.

2. The number of possible values of q is 6.

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.(c)

pp + qq + rr = tt0

---