Have given below the solutions to the questions on remainders. The questions can be found here.

1. What are the last two digits of the number 7^{45 }?

The last two digits of 7^{1} are 07

The last two digits of 7^{2 }are 49

The last two digits of 7^{3} are 43

The last two digits of 7^{4} are 01

The last two digits of powers of 7 go in a cycle – 07,49,43,01

So, the last two digits of 745 are 07

2. What is the remainder when we divide 3^{90} + 5^{90} by 34?

3^{90} + 5^{90 }can be written as (3^{2})^{45} + ( 5^{2})^{45}= (9)^{45} + (25)^{45}

Any number of the form a^{n} + b^{n} is a multiple of (a + b) whenever n is odd

So (9)^{45} + (25)^{45} is a multiple of 9 +25 = 34

So, the remainder when we divide (3^{2})^{45} + ( 5^{2})^{45 } by 34 is equal to 0

3. N^{2 }leaves a remainder of 1 when divided by 24. What are the possible remainders we can get if we divide N by 12?

This again is a question that we need to solve by trial and error. Clearly, N is an odd number. So, the remainder when we divide N by 24 has to be odd.

If the remainder when we divide N by 24 = 1, then N^{2 }also has a remainder of 1. we can also see that if the remainder when we divide N by 24 is -1, then N^{2 }a remainder of 1

When remainder when we divide N by 24 is ±3, N^{2 }has a remainder of 9

When remainder when we divide N by 24 is ±5, N^{2 } has a remainder of 1

When remainder when we divide N by 24 is ±7, N^{2 } has a remainder of 1

When remainder when we divide N by 24 is ±9, N^{2 } has a remainder of 9

When remainder when we divide N by 24 is ±11, N^{2 } has a remainder of 1

So, the remainder when we divide N by 24 could be ±1, ±5, ±7, or ±11

Or, the possible remainders when we divide N by 24 are 1, 5, 7, 11, 13, 17, 19, 23

Or, the possible remainders when we divide N by 12 are 1, 5, 7, 11

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