Question 562260: the digit in the ones place is the sum of the digits in a dozen
please help
Answer by issacodegard(60)   (Show Source):
You can put this solution on YOUR website! Hi, i think the problem is to find a positive integer x such that the ones place of x is the sum of the digits of 12x. Denote this as ones(x)=dsum(12x).
You can check that x=9 works, but i'll try to give some explanation for the problem.
We know that the ones place of x can be 0,1,2,3,4,5,6,7,8, or 9
If ones(x)=0, then dsum(12x)=0, which can't be true
If ones(x)=1, then dsum(12x)>=ones(12x)=ones(12)*ones(x)=2*1=2, so it can't equal 1
If ones(x)=2, then dsum(12x)>=ones(12x)=2*2=4, so it can't equal 2
If ones(x)=3, then dsum(12x)>=ones(12x)=2*3=6, so it can't equal 3
If ones(x)=4, then dsum(12x)>=ones(12x)=2*4=8, so it can't equal 4
This at least eliminates half of the possible x's, since it shows that ones(x) must be 5,6,7,8, or 9.
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