Question
A four-digit number 3P2Q is divisible by both 2 and 3. Find the smallest value of P + Q.
This question was previously asked in
RRB NTPC 2025 Under Graduate CBT 1 Question Paper PDF (20-Jun-2026) (Shift 3)
The correct answer is
1
For divisibility by 2, Q must be even.
For divisibility by 3, the digit sum \(3+P+2+Q = 5+P+Q\) must be a multiple of 3.
Testing the smallest possible values with Q even: P=1, Q=0 gives \(P+Q=1\) and digit sum \(5+1=6\), which is divisible by 3.
Hence, the smallest value of P + Q is 1.
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