Every numbers multiple of 5 ends in 0 zero.Cite your evidence to this statement
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Mathematics
Is there a mathematical proof that shows all multiples of 55 either end with a 00 or 55?
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I know that all multiples of 55 end up with a 00 or 55 as the last digit. But there are an infinite amount of numbers. Is there a way to formally prove that this is true for all numbers using variables?
sequences-and-series
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edited Dec 30, 2015 at 1:58
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Consider the number 5n5n where nn is an integer. when nn is even, 5n5n is a multiple of 1010 so it ends with digit 00. Otherwise 5n=10k+55n=10k+5 for some kk, so it ends with digit 55. –
user160738
Dec 30, 2015 at 1:37
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@user160738 Is it obvious all multiples of 10 end with digit 0? ;) –
Corellian
Dec 30, 2015 at 2:05
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Let nn be a multiple of 55, say n=5mn=5m for some integer mm. If mm is even, there is an integer kk such that m=2km=2k, and then n=10kn=10k. If, on the other hand, mm is odd, there is an integer kk such that m=2k+1m=2k+1, and in that case n=10k+5n=10k+5. To complete the argument, we need only show that every multiple of 1010 ends in 00.
Suppose that nn is a multiple of 1010, and suppose that when written in ordinary decimal notation, it is drdr−1…d0drdr−1…d0, where the dkdk are the digits. Then
n=10rdr+10r−1dr−1+…+10d1+d0=10(10r−1dr+10r−2dr−1+…+10d2+d1)+d0,
n=10rdr+10r−1dr−1+…+10d1+d0=10(10r−1dr+10r−2dr−1+…+10d2+d1)+d0,
where the quantity in parentheses is an integer. Thus,
d0=n−10(10r−1dr+10r−2dr−1+…+10d2+d1),(1)
(1)d0=n−10(10r−1dr+10r−2dr−1+…+10d2+d1),
and if nn is a multiple of 1010, the righthand side of (1)(1) is a multiple of 1010. Thus, d0d0 is a multiple of 1010. But 0≤d0≤90≤d0≤9, so d0=0d0=0. This proves that every multiple of 1010 ends in 00 and hence that every multiple of 55 ends in 00 or 55.