JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Permutations and Combinations

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:

Answer: 6

Solution

Divisibility by

55

needs divisibility by both

5

and

11

, so the last digit must be

5

. Four-digit numbers of the form

d_{1} d_{2} d_{3} 5

draw

d_{1}, d_{2}, d_{3}

from

{2, 3, 4, 6}

without repetition, and the alternating-sum rule for

11

requires

(d_{1} + d_{3}) - (d_{2} + 5)

to be a multiple of

11

; with these digits that is possible only when

(d_{1} + d_{3}) - d_{2} = 5

. The qualifying choices give

3245, 4235, 2365, 6325, 3465

and

6435

, each genuinely a multiple of

55

. A five-digit number would use all five digits, but the smallest such,

23456

, already exceeds the upper limit

23421

, so none contribute, leaving exactly six numbers.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer