The Calculator Problem Home › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › Number Theory › The Calculator Problem Tagged: Fun with Numbers 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 June 12, 2019 at 10:06 am #29182 Gouri BasakParticipant 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 . June 12, 2019 at 10:55 am #29185 Nitin PrasadParticipant 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. Author Posts Viewing 2 posts - 1 through 2 (of 2 total) You must be logged in to reply to this topic.