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Posted by Denis Borris on 21/9/2009, 1:50 pm, in reply to "Re: Piggy back"
Yes; limitless solutions, BUT Ryan's nickels = 30 is unique.
Make 'em dollars instead of cents ($1, $5, $10, $25).
Let Ryan's number of 1's, 5's, 10's, 25's = a, b, c, d.
Let Alyssa's number of 1's, 10's = e, f (5's are given, 25's are 4 more than Ryan).
Since they have same amount of money:
a + 5b + 10c + 25d = e + 115 + 10f + 25d + 100 ; simplify:
a - e = 10f - 5b - 10c + 215 [1]
Since Ryan has 32 more bills than Alyssa:
a + b + c + d = e + 23 + f + d+4 + 32 ; simplify:
a - e = f - b - c + 59 [2]
[1],[2]: 10f - 5b - 10c + 215 = f - b - c + 59 ; simplify:
c - f = 4(39 - b) / 9
Since c - f is an integer, then 4(39 - b) / 9 must be an integer;
numerator is even (multiplication by 4), so must be 18 or 36 or 54 or 72....(divisible by 9);
can't be 18 or 54 (result not integers), 36 makes b=30, 72 makes b=21 (too low,since b > 23).
SO b = 30 = Ryan's $5's
A little substituting back in shows that Ryan has 4 more $10's and 25 more $1's than Alyssa.
All solutions will reflect that.
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