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
$1-2799 \to[\frac{2799}{33}]=84$
$1-999 \to[\frac{999}{33}]=30$
$1000-2799 \to 54$
$\therefore n(3)+n(11)-n(33)$
$600+164-54=710$
