Question

A teacher wrote a 10 digit number on the board and asked her students to subtract the sum of digits of the number from the number. Raju performed the subtraction but accidentally erased one of the digits in the result. The remaining digits were 1,2,3,3,6,6,8,8,9 ( not necessarily in that order). Which digit did Raju erase ? ( A.9 B . 8 C.4)

Answer 1

\[\sum\limits_i d_i10^i - \sum\limits_i d_i = \sum\limits_i d_i(10^i-1)\]

is it easy to show that this is divisible by \(9\) ?

Answer 2

Could you please explain the first part of the equation ? I.e summation di 10 to the power i?

Answer 3

that is the decimal expansion of the number \((d_nd_{n-1}\ldots d_2d_1d_0)_{10}\)

Answer 4

for example the number \(23\) can be expanded as
\[3\times 10^0 + 2\times 10^1 = \sum\limits_{i} d_i10^i\]

here \(d_0=3\) and \(d_1=2\)

Answer 5

Oh ! Wow! Got it ! :)

Answer 6

i think you knew these already, im just using the fancy sum notation in hope of simplifying..

Answer 7

Now we just have to find that number so that the result will be divisible by 9 . that is 8 right ?

Answer 8

recall the geometric series formula
\[x^n-1 = (x-1)(x^{n-1}+x^{n-2}+\cdots +x+1)\]

and rewrite the previous sum

Answer 9

\[\sum\limits_i d_i10^i - \sum\limits_i d_i \\~\\= \sum\limits_i d_i(10^i-1) \\~\\= (10-1)\sum\limits_i d_i(10^{i-1}+10^{i-2}+\cdots + 10+1)\]

which is clearly divisible by \(9\)

Answer 10

Yes, 8 looks good !

Answer 11

Interesting that this problem was "Quant question of the Day" 15/06/14

The T.I.M.E. site looks like a great source of varied problems, numerical and verbal.

https://www.facebook.com/time4education/?fref=nf

http://www.time4education.com/

Answer 12

Thank you so much for all your help @ganeshie8 !

Answer 13

Np :)