Question 751645
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The only way to solve this is to find a logical way to list them.


Two quarters is 50 cents, so it is only possible to have either one quarter or zero quarters.


Start with 1 quarter.  That leaves you with 18 cents to make up.  So, given that you have 1 quarter, the most dimes you can have is 1, leaving you 8 cents to make up, so you can have, at most, 1 nickel, and then 3 pennies.


Now take away the nickel and add 5 pennies, so 1 Q, 1 D, 0 N, 8 P.


Take away the dime, so:


1 Q, 0 D, 3 N, 3 P, or
1 Q, 0 D, 2 N, 8 P, or
1 Q, 0 D, 1 N, 13 P, or
1 Q, 0 D, 0 N, 18 P


Take away the quarter, and now you can have up to 4 dimes:


0 Q, 4 D, 0 N, 3 P


And so on until you get down to 0 Q, 0 D, 0 N, 43 P.  You can follow the pattern and then count the possibilities for yourself.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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