Welcome 2016! Part 35

Number Theory Level 3

The repeating decimals 0. a b a b a b ‾ 0.abab\overline{ab} and 0. a b c a b c a b c ‾ 0.abcabc\overline{abc} satisfy

0. a b a b a b ‾ + 0. a b c a b c a b c ‾ = 33 37 0.abab\overline{ab} + 0.abcabc\overline{abc}=\dfrac{33}{37}

where a, b, c \text{a, b, c} are (not necessarily distinct) digits. Find the three digit number a b c ‾ \overline{abc} .


The answer is 447.

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Anish Harsha by Anish Harsha , 20, India

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1 solution

Anish Harsha
Jan 9, 2016

Note that 33 37 = 891 999 = 0. 891 ‾ \dfrac{33}{37}=\dfrac{891}{999}=0.\overline{891} . Also note that the period of 0. a b a b a b ‾ + 0. a b c a b c a b c ‾ 0.abab\overline{ab}+0.abcabc\overline{abc} is at most 6. Therefore, we need to only worry about the sum 0. a b a b a b ‾ + 0. a b c a b c a b c ‾ 0.abab\overline{ab}+ 0.abcabc\overline{abc} . Adding the two, we get

From this we can see that a = 4 a=4 , b = 4 b=4 and c = 1 c=1 , so our desired answer is 447.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nice idea for your solution! (Though I'm sure you mean that c=7 :) )

Michael Ng - 5 years, 5 months ago

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why 441 is your answer to this?....above the answer is 447..

jonathan dapadap - 5 years, 5 months ago

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