If n is a natural number what is the remainder when (23*35*57*79)^n is divided by 2?

Answer: 1

Explanation:

23, 35, 57, 79 are all odd numbers. The product of any number of odd numbers will also be an odd number and an odd number raised to any power will again be an odd number. When an odd number is divided by 2 the remainder is 1.

If n is a natural number what is the remainder when (23*35*57*79)^n is divided by 10?

Answer: 5

Explanation:

The remainder of any number divided by 10 is nothing but its units digit. The product 23*35*57*79 will end in ‘5’. And a number that ends in 5 when raised to any power will also end in 5. Hence the units digit of (23*35*57*79)^n will be 5.

If n is a natural number what is the remainder when (23*35*57*79)^n is divided by 100?

Answer: 25 when n is even and 75 when n is odd (except 1) and 15 when n = 1.

Explanation:

Here again, the remainder when divided by 100 is nothing but the last two digits.

Please remember that you will have an on-screen calculator in your GRE exam. You can use that to compute the product. Hence (23*35*57*79)^n is nothing but (3624915)^n. If you compute (15)^n for n = 2 and n = 3 it would be evident that for n = 2 the last two digits are 25 and for n = 3 the last two digits are 75.

As a rule,

When the last digit of a number is 5 and the previous digit is even i.e., say 345, then this number raised to any power greater than 1 will have its last two digits as 25.

When the last digit of a number is 5 and the previous digit is odd i.e., say 315, then this number raised to any even power will have its last two digits as 25 and raised to any odd power greater than 1 will have its last two digits as 75.

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