Let-x-be-a-positive-integer-multiple-of-17-that-satisfies-the-inequality-0-lt-5-x-120-x-lt-1-Find-the-value-of-x-

Question Number 182461 by HeferH last updated on 09/Dec/22

L e t x b e a p o s i t i v e i n t e g e r m u l t i p l e o f 17

t h a t s a t i s f i e s t h e i n e q u a l i t y :

0 < \frac{5 (x - 120)}{x} < 1

F i n d t h e v a l u e o f x .

Answered by mr W last updated on 09/Dec/22

x = 17 k

0 < \frac{5 (17 k - 120)}{17 k} < 1

0 < 85 k - 600 < 17 k

600 < 85 k \Rightarrow k > \frac{600}{85} \Rightarrow k ⩾ 8

68 k < 600 \Rightarrow k < \frac{600}{68} \Rightarrow k ⩽ 8

\Rightarrow k = 8 \Rightarrow x = 17 \times 8 = 136 ✓

Answered by MJS_new last updated on 09/Dec/22

x = 17 n

0 < \frac{85 n - 600}{17 n} < 1

0 < 85 n - 600 < 17 n

600 < 85 n < 17 n + 600

\frac{120}{17} < n < \frac{n}{5} + \frac{120}{17}

7.05 \dots < n < \frac{n}{5} + 7.05 \dots

n_{\min} = 8 and because of 9 > \frac{9}{5} + 7.05 \dots {it}^{'} s the

only solution

\Rightarrow x = 8 \times 17 = 136

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