Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is
- (a) 2
- (b) 5
- (c) 7
- (d) 8 ✓ UPSC's answer
Why the answer is (d)
• The equation is 136 + 5B7 = 7A3, which simplifies to 5B7 = 7A3 - 136.
• Performing the subtraction: 7A3 - 136 results in a number ending in 7 (since 13 - 6 = 7), confirming the units digit of 5B7 is 7.
• The tens digit of the result is determined by (A - 3) or (A + 10 - 3) if borrowing occurs from the hundreds place.
• Since 7A3 is divisible by 3, the sum of its digits (7 + A + 3) must be a multiple of 3, so 10 + A is divisible by 3.
• Possible values for A are 2 (10+2=12) or 5 (10+5=15). If A=2, 723 - 136 = 587, so B=8. If A=5, 753 - 136 = 617, which does not match the form 5B7 (hundreds digit is 6, not 5).
• Therefore, the only valid case is A=2 and B=8.
Why the other options are wrong
- (a) 2
- If B were 2, the number would be 527, and 527 + 136 = 663, which does not match the form 7A3.
- (b) 5
- If B were 5, the number would be 557, and 557 + 136 = 693, which does not match the form 7A3.
- (c) 7
- If B were 7, the number would be 577, and 577 + 136 = 713, but 713 is not divisible by 3 (7+1+3=11).
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2019, held on 2 June 2019. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.