IAS 2017 — CSAT Maths

2017mcqmedium

A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible remainder?

Official previous-year question

Verified 2 Aug 2026.

Options

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Solution

Correct Option (D)

Let the 2-digit number be represented as $10a + b$ , where $a$ is the tens digit and $b$ is the units digit. For both the original number and its reverse to be valid 2-digit numbers, both $a$ and $b$ must be integers from 1 to 9 (i.e., $a, b \in \{1, 2, ..., 9\}$ ).

The reversed number will be $10b + a$ .

The problem requires dividing the larger of these two numbers by the smaller one. Let $D$ be the smaller number (divisor) and $R$ be the remainder. According to the properties of division, the remainder $R$ must satisfy $0 \le R < D$ .

To identify the largest possible remainder, we should consider scenarios where the divisor $D$ is relatively large, allowing for a larger potential remainder $R$ that is just below $D$ . Given the options, the maximum possible remainder is likely to be 45. If the remainder is 45, the divisor $D$ must be at least 46.

We systematically examine pairs of numbers $(10a+b, 10b+a)$ where the smaller number (divisor $D$ ) is 46 or greater:

Consider the number 49 ( $a=4, b=9$ $a = 4, b = 9$ ). Its reverse is 94.
- The smaller number is 49 (divisor).
- The larger number is 94 (dividend).
- Dividing 94 by 49 yields: $94 = 1 \times 49 + 45$ . The remainder is 45.

Further examination of other relevant pairs confirms that 45 is the maximum remainder:

For numbers 46 ( $a=4, b=6$ ) and 64 ( $10b+a$ ): $64 \div 46 \implies R = 18$ .
For numbers 47 ( $a=4, b=7$ ) and 74 ( $10b+a$ ): $74 \div 47 \implies R = 27$ .
For numbers 48 ( $a=4, b=8$ ) and 84 ( $10b+a$ ): $84 \div 48 \implies R = 36$ .
For numbers 59 ( $a=5, b=9$ ) and 95 ( $10b+a$ ): $95 \div 59 \implies R = 36$ .
For numbers 69 ( $a=6, b=9$ ) and 96 ( $10b+a$ ): $96 \div 69 \implies R = 27$ .
For numbers 79 ( $a=7, b=9$ ) and 97 ( $10b+a$ ): $97 \div 79 \implies R = 18$ .
For numbers 89 ( $a=8, b=9$ ) and 98 ( $10b+a$ ): $98 \div 89 \implies R = 9$ .

The largest remainder obtained through this systematic analysis is 45.

Incorrect Options:

Option A (9): A remainder of 9 is a possible outcome (e.g., 98 divided by 89 yields a remainder of 9). However, this is not the largest possible remainder.
Option B (27): A remainder of 27 is a possible outcome (e.g., 96 divided by 69 yields a remainder of 27). This is not the largest possible remainder.
Option C (36): A remainder of 36 is a possible outcome (e.g., 95 divided by 59 yields a remainder of 36). This is not the largest possible remainder.

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Solution

Correct Option (D)

The reversed number will be $10b + a$ .

We systematically examine pairs of numbers $(10a+b, 10b+a)$ where the smaller number (divisor $D$ ) is 46 or greater:

Consider the number 49 ( $a=4, b=9$ $a = 4, b = 9$ ). Its reverse is 94.
- The smaller number is 49 (divisor).
- The larger number is 94 (dividend).
- Dividing 94 by 49 yields: $94 = 1 \times 49 + 45$ . The remainder is 45.

Further examination of other relevant pairs confirms that 45 is the maximum remainder:

For numbers 46 ( $a=4, b=6$ ) and 64 ( $10b+a$ ): $64 \div 46 \implies R = 18$ .
For numbers 47 ( $a=4, b=7$ ) and 74 ( $10b+a$ ): $74 \div 47 \implies R = 27$ .
For numbers 48 ( $a=4, b=8$ ) and 84 ( $10b+a$ ): $84 \div 48 \implies R = 36$ .
For numbers 59 ( $a=5, b=9$ ) and 95 ( $10b+a$ ): $95 \div 59 \implies R = 36$ .
For numbers 69 ( $a=6, b=9$ ) and 96 ( $10b+a$ ): $96 \div 69 \implies R = 27$ .
For numbers 79 ( $a=7, b=9$ ) and 97 ( $10b+a$ ): $97 \div 79 \implies R = 18$ .
For numbers 89 ( $a=8, b=9$ ) and 98 ( $10b+a$ ): $98 \div 89 \implies R = 9$ .

The largest remainder obtained through this systematic analysis is 45.

Incorrect Options:

Option A (9): A remainder of 9 is a possible outcome (e.g., 98 divided by 89 yields a remainder of 9). However, this is not the largest possible remainder.
Option B (27): A remainder of 27 is a possible outcome (e.g., 96 divided by 69 yields a remainder of 27). This is not the largest possible remainder.
Option C (36): A remainder of 36 is a possible outcome (e.g., 95 divided by 59 yields a remainder of 36). This is not the largest possible remainder.

IAS 2017 — CSAT Maths

Options

Did you get this right?

Solution

Correct Option (D)

Incorrect Options:

Your note

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Comprehension

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IAS 2017 — CSAT Maths

Options

Did you get this right?

Solution

Correct Option (D)

Incorrect Options:

Your note

IAS Maths in other years

More IAS Maths questions

Other CSAT topics for IAS

Comprehension

Reasoning

Keep practicing Maths