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# The Calculator Problem

Home Forums Math Olympiad, I.S.I., C.M.I. Entrance Number Theory The Calculator Problem

This topic contains 1 reply, has 2 voices, and was last updated by  Nitin Prasad 2 months, 1 week ago.

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• #29182

Gouri Basak
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Take all the 9 numbers 1,2,3,….,9 in a calculator and make three 3 – digit numbers out of it. Add those numbers . Prove that the number is divisible by 9 .

#29185

Participant

Let (abc) denote the number 100a+10b+c where a,b,c are digits

Now let (abc), (def), (ghi) be the three 3-digit numbers formed using the digits from 1 to 9.

Now observe that-

$$(abc)+(def)+(ghi) =100a+10b+c+100d+10e+f+100g+10h+i =99a+9b+99d+9e+99g+9h+a+b+c+d+e+f+g+h+i$$

Since $$a+b+c+d+e+f+g+h+i= 1+2+3+\cdots +9=45$$

Hence (abc)+(def)+(ghi) is divisible by 9

• This reply was modified 2 months, 1 week ago by  Nitin Prasad.
• This reply was modified 2 months, 1 week ago by  Nitin Prasad.
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