Permutation Combination Question 15

Question 15 - 31 January - Shift 1

Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to

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Answer: (710)

Solution:

Formula: Principle of counting

$1000-2799$

Divisible by 3

$1002+(n-1) 3=2799$

$n=\mathbf{6 0 0}$

Divisible by 11

$1-2799 \to[\frac{2799}{11}]=[254]=254$

$1-999=[\frac{999}{11}]=90$

$1000-2799=254-90=164$

Divisible by 33

$1-2799 \to[\frac{2799}{33}]=84$

$1-999 \to[\frac{999}{33}]=30$

$1000-2799 \to 54$

$\therefore n(3)+n(11)-n(33)$

$600+164-54=710$