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.

Viewing 2 posts - 1 through 2 (of 2 total)
  • Author
    Posts
  • #29182

    Gouri Basak
    Participant

    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

    Nitin Prasad
    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.
Viewing 2 posts - 1 through 2 (of 2 total)

You must be logged in to reply to this topic.