Discussion
A number when successively divided by 4, 5 and 6 leaves remainders 2, 3 and 4 respectively. The least such number is:
Answer: A
Correct answer: 94
In successive division each quotient becomes the next dividend, so work backwards from the last step.
Let the final quotient be \(c\). Then, reading the divisions in reverse:
\[b = 6c + 4, \qquad a = 5b + 3, \qquad N = 4a + 2\]
The least number comes from \(c = 0\), which gives \(b = 4\), then \(a = 5(4) + 3 = 23\), and
\[N = 4(23) + 2 = 94\]
Verifying the chain: \(94 \div 4\) gives quotient 23 remainder 2; \(23 \div 5\) gives quotient 4 remainder 3; \(4 \div 6\) gives quotient 0 remainder 4.
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