sana.noor wrote:If the integers m and n are chosen at random from integers from integers 1 to 100 with replacement, then the probability that a no. of the form 7^m + 7^n is divisible by 5 equals?

(1)1/2
(2)1/4
(3)1/8
(4)1/16
(5)None of these

I dont have its answer

this is a pretty cool problem.

when you think about divisibility by 5, that's really just an issue of the last digit (the units digit). remember, if the units digit is 0 or 5, then the number is a multiple of 5.

so, what's the pattern of units digits in powers of 7?
7^1 = 7
7^2 = ends in 9 (it's just 49, but keeping any other digits is a waste of time)
7^3 = ends in 3 (since 9 x 7 = 63)
7^4 = ends in 1 (since 7 x 3 = 21)
and we're back to 7 again with 7^5.

so, the units digits repeat: 7, 9, 3, 1, etc.

here we're given the powers 7^1 through 7^100. conveniently, that's 25 sets of the above four digits, so you just have a 1/4 chance of getting each digit.

let's say the powers you pick are x and y.
how can you get a multiple of 5?
* clearly you can't get a number ending in "5", since you're adding two odd numbers together.
* you can have x ending in "7" and y ending in "3", or vice versa.
* you can have x ending in "9" and y ending in "1", or vice versa.

so, there are 4 x 4 = 16 possibilities. 4 of them are winners. so, 1/4.