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Topic: Infinite System of Equations (Read 497 times) |
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ThudnBlunder
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Infinite System of Equations
« on: Jan 20th, 2009, 7:38pm » |
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Solve the following infinite system of equations, where N is an integer and A,B,X,Y are distinct, one-digit integers:
A*N = XY
AB*N = XXY
ABB*N = XXXY
ABBB*N = XXXXY
etc.
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balakrishnan
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Re: Infinite System of Equations
« Reply #1 on: Jan 21st, 2009, 1:05am » |
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The solution set
{A,B,X,Y,N} can be
{3,1,7,5,25}
{8,3,6,4,8}
and
{8,3,9,6,12}
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| « Last Edit: Jan 21st, 2009, 9:21am by balakrishnan » |
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codpro880
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Re: Infinite System of Equations
« Reply #2 on: Mar 3rd, 2009, 6:04pm » |
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They all seem to work for the first few equations....but how did you come up with those answers and how can you prove they'll always work?
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Eigenray
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Re: Infinite System of Equations
« Reply #3 on: Mar 4th, 2009, 12:33am » |
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The first equation tells us obviously A*N = 10X + Y. Now suppose
AB...B*N = XX...XY. Then
AB...BB*N = [10*AB...B + B]*N
= 10*(AB...B*N) + B*N
= XX...XY0 + B*N
So to get from one step to the next, it is necessary and sufficient that Y0 + B*N = XY. Since also XY = A*N, we get N*(A-B) = 10*Y. So for each value of Y we can just look at the divisors of 10Y and see which pairs work for N, A-B.
I actually did it differently: divide the n-th line by 10n-1. Then
A.B...B * N = XX.X...Y,
and taking the limit gives (A + B/9)*N = X*100/9. Combined with A*N=XY this is equivalent to Y0+B*N=XY, but if you do it this way I guess it's less obvious that any solution will work.
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| « Last Edit: Mar 4th, 2009, 12:35am by Eigenray » |
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