# Thread: Help with Proof: A cents in B coins --> B dollars in A coins?

1. ## Help with Proof: A cents in B coins --> B dollars in A coins?

Hi, I need to solve this proof put I cannot seem to figure it out. I'd really appreciate some help!

If a dollar is 100 cents and coins come in 1, 2, 5, 10, 20, 50 and 100 cents. Suppose that one can make A cents using exactly B coins. Prove that it is possible to make B dollars using exactly A coins

2. I need to solve this proof put I cannot seem to figure it out. I'd really appreciate some help!
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3. Originally Posted by jsimon Hi, I need to solve this proof put I cannot seem to figure it out. I'd really appreciate some help!

If a dollar is 100 cents and coins come in 1, 2, 5, 10, 20, 50 and 100 cents. Suppose that one can make A cents using exactly B coins. Prove that it is possible to make B dollars using exactly A coins
If we can make A cents using B coins, that means there are non-negative integers
$\displaystyle x_1, x_2, \dots ,x_7$
such that
$\displaystyle x_1 + 2 x_2 + 5 x_3 + 10 x_4 + 20 x_5 + 50 x_6 + 100 x_7 = A$
where
$\displaystyle x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = B$

Multiplying the previous equation by 100,
$\displaystyle 100 x_1 + 100 x_2 + 100 x_3 + 100 x_4 + 100 x_5 + 100 x_6 + 100 x_7 = 100 B$
which we can write as
$\displaystyle 100 x_1 + 50(2 x_2) + 20(5 x_3) + 10(10 x_4) + 5(20 x_5) + 2(50 x_6) + 1(100 x_7) = 100B$

Does that give you any ideas?

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