Question

If the integers m and n are chosen at random between 1 and 100, then atmost distinct numbers of the form $7^m+7^n$ is divisible by 5 equals to:

• A. 1250
• B. 10000
• C. 2500
• D. none of these

Answer 1

The correct option is A 1250

We know that if a number is divisible by 5, then its unit digit must be either 0 or 5.
Again, the unit digits of $7^1=7$
$7^2$ = 9
$7^3$ = 3 and $7^4$ =1
i.e., we can get only 7, 9, 3, 1 as unit digits.
Now, the combinations of 7, 9, 3, 1 never gives unit digit 5, but 7 + 3 = 9 +1 = 10, these two combinations gives us unit digit zero.
So $7^m+7^n$ is divisible where
case (i)m = 1, 5, 9, ..., 97 correspondingly n = 3, 7, 11, ..., 99
case (ii) m = 2, 6, 10, ..., 98 correspondingly
n = 4, 8, 12, ..., 100
or the values of m and n can be reversed in each case mutually but then we get the same values.
Hence total distinct numbers so formed, divisible by 5, are 25 $\times$ 25 + 25 $\times$ 25 = 1250
Hence option (a) is correct.
Alternatively,
For Rem $(7^m+7^n)$/5 = 0 i.e. For Rem $\left(7^n\right(7^{m-n}+1\left)\right)$/5 = 0 , we need to have (m-n) of the form (4k - 2) where k lies between 1 and 25 including 1 & 25. Thus, there are a total of 25 values for k. Putting different values of k beginning from 1 till 25, we get the following arithmetic progression for the total number of cases divisible by 5.
Total number = 98 + 94 + 90 + ........ + 2 = (25/2)*100 = 1250