NCERT Exemplar Solutions
Class 6 - Mathematics - Unit 1: Number System - Problems and Solutions

Question 192

Goal: smallest 4-digit number using 1, 2, 3, 4 once, divisible by 4.

1. Rule: A number is divisible by 4 if its last two digits form a multiple of 4.
2. Form all possible last-two-digit pairs from 1, 2, 3, 4.
3. Check which pairs are multiples of 4:
   \(12\) is divisible by \(4\).
   \(24\) is divisible by \(4\).
   \(32\) is divisible by \(4\).
4. So possible endings are:
   \(\_\_12\), \(\_\_24\), \(\_\_32\).
5. Make the smallest thousands and hundreds digits with the remaining numbers for each ending:
   Ending \(12\): use \(3,4\) in front → \(3412\).
   Ending \(24\): use \(1,3\) in front → \(1324\).
   Ending \(32\): use \(1,4\) in front → \(1432\).
6. Compare:
   \(1324 < 1432 < 3412\).
7. Smallest number is \(\boxed{1324}\).