Question
A transaction code 92419 is not divisible by 13. To make it divisible by 13, the least digit should be appended at the end of the number. What is that digit?
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RRB NTPC 2025 Graduate CBT 2 Question Paper PDF (10-Jul-2026) (Shift 1)
The correct answer is
6
Appending a digit \(d\) to 92419 gives the number \(924190 + d\), and we need this to be divisible by 13.
Divide 924190 by 13: \(13 \times 71091 = 924183\), leaving \(924190 - 924183 = 7\). So 924190 leaves remainder 7.
To reach the next multiple of 13, add \(13 - 7 = 6\). Then \(924190 + 6 = 924196 = 13 \times 71092\), which is divisible by 13.
Since \(d = 6\) is a single digit, it is a valid append, and it is the least such digit.
Hence, the digit to be appended is 6.
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