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United States SLAM_THE_FART 
How do you show that (84+(21^10)) is divisible by 5?
2020-11-18 23:06
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wait ill do it for u
2020-11-18 23:07
1 reply
alright, 21 to the power of 10 is always going to be something ending with 1, it just comes that way round if you multiply it enough. so adding 84 will round it out so that it ends with 5 therefore dividable by 5. im not the best at explaining it and im not sure how i can lay it out for you like #3 but heres the answer
2020-11-18 23:14
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
3 replies
nice
2020-11-18 23:14
ty kind sir
2020-11-18 23:16
1 reply
If you also need to show why a number is divisible by 5 this is a really good proof proofsfromthebook.com/2014/04/09/divisib..
2020-11-18 23:27
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
#5
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Brazil Beckenbauer
idk im retarded kkkk
2020-11-18 23:13
#6
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Switzerland NotJuan
you dont
2020-11-18 23:14
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