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 745 ?
The last two digits of 71 are 07
The last two digits of 72are 49
The last two digits of 73 are 43
The last two digits of 74 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 390 + 590 by 34?
390 + 590 can be written as (32)45 + ( 52)45
= (9)45 + (25)45
Any number of the form an + bn 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 (32)45 + ( 52)45 by 34 is equal to 0
3. N2 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 N2 also has a remainder of 1. we can also see that if the remainder when we divide N by 24 is -1, then N2 a remainder of 1
When remainder when we divide N by 24 is ±3, then N2 has a remainder of 9 When remainder when we divide N by 24 is ±5, then N2 has a remainder of 1 When remainder when we divide N by 24 is ±7, then N2 has a remainder of 1 When remainder when we divide N by 24 is ±9, then N2 has a remainder of 9 When remainder when we divide N by 24 is ±11, then N2 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