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Mario has $773500$ gold coins to purchase a number of stars and comets. Each star costs $299$ gold coins, and each comet costs $208$ gold coins. If the number of stars that Mario buys is at least twice the number of comets, how many ways can Mario spend his gold coins? List one of the ways in which Mario could buy the stars and comets. Note: Mario needs to spend all of his gold coins.

My work: Used EEA to find $\gcd(299, 208) = 13$ and $(299 \cdot 7) - (208 \cdot 10) = 13$.

Did some more work to get the two inequalities for the LDE solution, $n > -25911.2$, $n < -25869.6$.

I don't know where to go from the simplified inequalities.

Jake
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1 Answers1

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299x + 208y = 773500

299x/208 + 208y/208 = 773500/208 (Simplifying)

91x/208 - 156/208 = 3718 -x -y

7x/16 - 3/4 = k(constant)

*7 => 49x/16 - 21/4 = 7k

x/16 - 1/4 = 7k - 3x + 5

x/16 - 1/4 = c (constant)

=> x = 16*(c + 1/4)

= 16c + 4 c x y

1 20 3690

2 36 3667

3 52 3644

4 68 3621

..... ..... .....

119 1908 976

120 1924 953

We can easily see that the values of x and y are forming an AP.

For c=120, the value of x goes over the double of y for the first time, so this is the first solution.

Similarly continuing, you can find out the rest of the solutions by counting the terms of y, till it is positive.

Om Prakash
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  • Check the fifth line. It should be $\dfrac{49x}{16}-\dfrac{21}{4}=7k$. –  Oct 22 '14 at 06:28
  • k was a constant there and no matter how many times you multiply a constant value with an integral value it remains constant only. So only for simplification purpose later I replaced it with another constant c. You can do that. – Om Prakash Oct 22 '14 at 06:32