1x1=1, so numbers ending in 1 will result in a number also ending in 1 when multiplied by themselves (or another number ending in one). Add the 4 from 84, and you know the end result of (84+(21^10)) will end in 5, meaning it will be divisible by 5
2020-11-18 23:10
Well because 21 ends in 1, any power of it will also end in 1
and since you're adding 84
which makes the last digit of the number always 5
it's divisible by 5 because it ends in 5
very logical, unless you need mathematical proof, then I can't help you
2020-11-18 23:10
No matter how many times you multiply 21 by itself, the result will always be something with a 1 at the end. This is a proof by intuition, but it's not too difficult to see why; 21*21 can be written as 20*20 + 20*1 + 1*20 + 1*1. In general, if you multiply two numbers with a 1 at the end, the resulting number will also have a 1 at the end. So 21^10 is equal to some number with a 1 at the end, add 84 to that and you got yourself a number with a 5 at the end. Not exactly Q.E.D. but you get the jist.
EDIT: Okay so just to get a little more detail; Imagine that you have two arbitrary numbers both with a 1 at the end. Let A = a_1 + 10*a_2 + 100*a_3 + ... + 10^(n - 1)b_n, and analogously B = b_1 + 10*b_2 + ... + 10^(m -1)b_m, with a_1, ..., a_n, b_1, ..., b_m being the digits of the respective number. If you multiply these two numbers you get 10^(n - 1)a_n*(10^(m - 1)b_m + ... + 1) + 10^(n - 2)a_(n - 1)*(10^(m - 1)b_m + ... + 1) + ........... + 1*(10^(m - 1)b_m + ... + 10b_2 + 1). As you can kind of see, the only resulting summand with a factor of 10^0 is 1*1 = 1. Now this was really ugly, but I hope it's a bit clearer as to why it has to be this way.
2020-11-18 23:25