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The smallest number that should be subtracted from 2085, so that the new number is completely divisible by 23 is

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For “smallest number to subtract to be divisible by $m$”, compute remainder $r$ of $N$ on division by $m$ and subtract $r$.
Updated On: Jul 15, 2026
  • 9
  • 15
  • 20
  • 19
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The Correct Option is B

Approach Solution - 1

Step 1: Divide 2085 by 23 to find the remainder. Since \( 23 \times 90=2070 \), the remainder is \( 2085-2070=15 \).

Step 2: A number becomes exactly divisible by 23 once its remainder is removed, so subtracting this remainder from 2085 gives a multiple of 23.

Step 3: The smallest number to subtract is exactly the remainder, 15, and \( 2085-15=2070=23 \times 90 \) confirms it.
\[ \boxed{15} \]
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Approach Solution -2

Locate the two multiples of 23 that bracket 2085: since \( 23 \times 90=2070 \) and \( 23 \times 91=2093 \), we have \( 2070<2085<2093 \). The smallest number to subtract must bring 2085 down to the nearer, lower bracketing multiple, 2070. We can check each option by seeing where subtracting it lands us relative to these brackets.

  1. 9: \( 2085-9=2076 \), which falls strictly between the brackets 2070 and 2093 without landing on either, so it is not divisible by 23.
  2. 15: \( 2085-15=2070 \), landing exactly on the lower bracketing multiple of 23.
  3. 20: \( 2085-20=2065 \), which falls below the lower bracket of 2070, overshooting the nearest multiple.
  4. 19: \( 2085-19=2066 \), which again falls between the brackets without landing on either.

The lower bracketing multiple of 23 nearest to 2085 is 2070, reached by subtracting exactly 15.

Therefore, the correct answer is 15.

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