Question

the remainder when 2²+22²+222²+........+(222.....49times)² divided by 9​

Answer 1

Given :   2²+22²+222²+........+(222.....49times)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......49times)²

Equivalent to  ( for remainder purpose after dividing by 9)

2²  + 4²  + 6²  +.................................+ 98²

= 2²(1² + 2² +....................+ 49²)

∑n² = n(n+1)(2n + 1)/6

= 4 * 49(50)(99)/6

= 161700

Sum of Digits = 1 + 6 + 1 + 7 + 0 + 0  =  15

15 divided by 9 leaves remainder 6

Hence the remainder when 2²+22²+222²+........+(222.....49times)² divided by 9​    is  6

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