The number of natural numbers lying between 1012 and 23421 that can be formed using the digits 2,3,4,5,6 (repetition of digits is not allowed) and divisible by 55 is _________.

Solution

Maths sir · 5 steps

Divisible by 55 means divisible by 5 and by 11, so fix the last digit as 5 and test the 11-rule.

1. Split 55. 55 = 5 × 11, so the number must pass both tests. No 0 is available, so divisibility by 5 forces the last digit to be 5.
2. Which lengths. Every 4-digit number from these digits is at least 2345, so all of them lie between 1012 and 23421. A 5-digit number must use all five digits, start with 2, and stay below 23421.
3. Four-digit case. Write the number as d₁ d₂ d₃ 5. The 11-rule needs (d₁ + d₃) - (d₂ + 5) to be a multiple of 11. With d₁, d₂, d₃ from {2,3,4,6}, this difference lies between -6 and 3, so it must be 0. That gives d₁ + d₃ = d₂ + 5.

   d₂	Need d₁ + d₃	Pair from the rest	Numbers
   2	7	3, 4	3245, 4235
   3	8	2, 6	2365, 6325
   4	9	3, 6	3465, 6435
   6	11	none from 2, 3, 4	none

4. Five-digit case. The digits add to 20, so the odd-place sum is 10 and d₁ + d₃ = 5, which means {d₁, d₃} = {2, 3}. With d₁ = 2 the only numbers are 24365 and 26345. Both are above 23421, so none count.
5. So. Only the six 4-digit numbers work, and the answer is 6.

The common slip: 10. Forgetting the upper limit 23421 also counts 24365, 26345, 34265 and 36245, which gives 10 instead of 6.

The idea to keep. For divisibility by a composite like 55, split it into coprime factors. Fix the easy digit first (last digit for 5), then use the alternating-sum rule for 11, and check the range at the end.

Answer from the NTA final answer key. Solution by PrepWiser.