Here's the question you clicked on:
Ujji
(10^25)-7 is divisible by a) 2 b)3 c)9 d) both 2 & 3
10^25=10000000000000000000000000 the number is not so important u know that the last digit will be 0 and is u substract 7 u will have tha last digit 3 si it is divisible by 3
10-7=3 therefore 10^25-7 is divisible by 3.
its not last digit but all preceeding digits are 9. therefore sum of 23 9s+3 is a multiple of 3.
if u have 10 or 10^2 meaning 100 or 10^x it doesn't matter cause by substracting 7 u will have the last digit 3 and 3 is divisible only with 3
133 for instance is not divisible by 3.
anyways,alternately, we know, 10^n = 3q + 1 and 7 = 3p + 1 subtracting both, 10^n - 7 = 3(q-p) = 3r ==> divisible by 3