Find the sum of divisors of 544 which are perfect squares.

The remainder when 10^10 + 10^100 + ......10^10000000000 is divided by 7 is

Find the last two digits of 15 x 37 x 63 x 51 x 97 x 17.

Find the last two digits of 15 x 37 x 63 x 51 x 97 x 17.

Solution:

15×37×63×51×97×17

= (10+5)*(40-3)*(60+3)*(50+1)*(100-3)*(10+7)

Multiply 5 x 3x 3 x 1 x 3 x 7

Find the remainder when (10^3 + 9^3)^752 is divided by 12^3?

Find the remainder when 7^99 is divided by 2400.

Find the remainder when 7^99 is divided by 2400.

Solution:

Using pattern recognition(cyclicity method).

1) 7^1  mod 2400 = 7
2)  7^2  mod 2400 = 49
3)  7^3 mod 2400 = 343
4)  7^4  mod 2400 = 1
After this the same pattern will be keep on repeating.
So, 

Cyclicty = 4
Power = 99

Power mod Cyclicity = 99 mod 4 = 3

Third value in the above cycle is 343.

Answer: 343

What are the last two digits of 7^2008?

What are the last two digits of 7^2008?

Solution:
7^1 = 07
7^2 = 49  
7^3 = 343   -> 49 times 7 is 343 => last two digits => 43
7^4 = 2401  -> 43 times 7 is 2401 => last  two digits => 01
7^5 = 16807
7^6 = 117649
7^7 = 823543
7^8 = 5764801 -> last two digits again 01

01 -> repeats after every 4 times

Now, 2008 mod 4 = 0

Hence, the last two digits of 7^2008 = 01

When 2^256 is divided by 17 the remainder would be