(10^25)-7 is divisible by

a) 2 b)3 c)9 d) both 2 & 3

The answer is b)3.

If the last digit of a number is divisible by 2, than the number is divisible by 2 (for example, in 828, the last digit is 8 which is divisible by 2 (8/2 = 4), so 828 is also divisible by 2).

If the sum of all the digits of a number is divisible by 3, then the number is divisible by 3 (for example, in 396, the sum of the digits is 18 (3+9+6=18), so 396 is also divisible by 3). 

Similarly, if the sum of all the digits of a number is divisible by 9, then the number is divisible by 9.

Now, let's check it out for 10²⁵ - 7. Let's first take simple examples:
10² - 7 = 100 - 7 = 93 (The last digit is not divisible by 2. Also, the sum of the digits is 9+3=12 and it is divisible only by 3, but not by 9).
10³ - 7 = 1000 - 7 = 993 (The last digit is not divisible by 2. Also, the sum of the digits is 9+9+3=21 and it is divisible only by 3, but not by 9).
10⁴ - 7 = 10000- 7 = 993 (The last digit is not divisible by 2. Also, the sum of the digits is 9+9+9+3=30 and it is divisible only by 3, but not by 9).

10²⁵-7 = There are many zeros, but based on the previous examples, we can make a conclusion that the last digit is not divisible by 2. Also, the sum of the digitsit is divisible only by 3, but not by 9.

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