Question

Q. What is the remainder when expression
2^2+22^2+222^2+2222^2+....+22222...48 times^2 divided by 9.

Answer 1

Given :   2²+22²+222²+........+(222.....48times)² divided by 9​

To find :  Remainder

Solution:

Any number divided by 9 or Sum of Digits of that number  Divided by 9

give the same remainder

22/9  - remainder = 4

(2 + 2)  = 4 divided by 9  - remainder  4

(22)² = 484   divided by 9  remainder = 7

4 + 8 + 4 = 16 divided by 9 remainder = 7

Hence

2² + 22² + 222² +.....................................+(222......48times)²

Equivalent to  ( for remainder purpose after dividing by 9)

2²  + 4²  + 6²  +.................................+ 96²

= 2²(1² + 2² +....................+ 48²)

∑n² = n(n+1)(2n + 1)/6

= 4 * 48(49)(97)/6

= 152096

Sum of Digits = 1 + 5 + 2 + 0 + 9 + 6  =  23

23  divided by 9 leaves remainder 5

Hence the remainder when 2²+22²+222²+........+(222.....48times)² divided by 9​    is  5

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