A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible remainder?
- (a) 9
- (b) 27
- (c) 36
- (d) 45 ✓ UPSC's answer
Why the answer is (d)
• For a number and its reverse, the larger divided by the smaller gives the biggest remainder when the quotient is 1, and then the remainder equals their difference, 9 × (difference of the digits).
• The quotient stays 1 only if the larger is less than twice the smaller. 94 and 49 work: 94 = 1 × 49 + 45.
• A digit difference of 6 or more forces a quotient of at least 2 (for example 93 = 2 × 39 + 15), so no pair does better.
• The largest possible remainder is 45, option (d).
Why the other options are wrong
- (a) 9
- A remainder of 9 occurs for several pairs like 98÷89 or 81÷18, but these are not the maximum possible remainder.
- (b) 27
- A remainder of 27 occurs when dividing 96 by 69, but this is not the maximum; larger remainders exist for other digit pairs.
- (c) 36
- A remainder of 36 occurs for pairs like 95÷59 or 84÷48, which is close to the maximum but not the largest possible value.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2017, held on 18 June 2017. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.