Repeating Decimals!

Number Theory Level 2

${\dfrac{n}{810} = 0.\overline{9D5} = 0.9D59D59D5...}$

Suppose $n$ and $D$ are integers with $n > 0$ and $0 \leq D \leq 9$ satisfying the above equation. Determine the value of $n+D$ .

The answer is 752.

2 solutions

Kazem Sepehrinia
Aug 19, 2015

$1000\frac{n}{810}-\frac{n}{810}=\overline{9D5} \\ \frac{999}{810} n=\overline{9D5} \\ n=\frac{30}{37} \overline{9D5}$ Now $37|\overline{9D5}$ and since $37\times25= 925$ we get $D=2$ and $n=25\times 30=750$ .

was an AIME question changed it a little bit

Department 8 - 5 years, 6 months ago

Ritabrata Roy
Aug 11, 2018

Quite easy, The expression is just

      n/810=(900+10D+5)/999
 Or, n=(81450+900D)/111


  You can check that among 0≤D≤9
    The optimal case is D=2
    And n=750

So answer is 752

Repeating Decimals!

The answer is 752.

2 solutions

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