Competition problem — Number theory

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Number theory Difficulty 5.1 AIME, harder Find the answer

7. Set
$A=\left\{x \left\lvert\, x=\left[\frac{5 k}{6}\right]\right., k \in \mathbf{Z}, 100 \leqslant k \leqslant 999\right\},$

where $[x]$ denotes the greatest integer not exceeding the real number $x$ . Then the number of elements in set $A$ is $\qquad$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

When $k=100$ , $\left[\frac{5 k}{6}\right]=83$ ;
When $k=999$ , $\left[\frac{5 k}{6}\right]=832$ .
It is also easy to see that for $100 \leqslant k \leqslant 999$ , we have
$0 \leqslant\left[\frac{5(k+1)}{6}\right]-\left[\frac{5 k}{6}\right] \leqslant 1 \text {. }$

Therefore, the elements in $A$ can cover all integers from 83 to 832, a total of 750 elements.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.