nahid078 wrote:In how many different ways can pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents) be combined for a total of $1.10 (110 cents), if at least one of each type of coin must be included?
(A) 36
(B) 51
(C) 70
(D) 87
(E) 101
Since the sum of 110 is a multiple of 5 -- and the amounts yielded by the quarters, dimes and nickels will each be a multiple of 5 -- the number of 1-cent pennies must also be a multiple of 5.
Thus, to use at least 1 penny, we must use at least 5 pennies.
Amount yielded by 1 quarter, 1 dime, 1 nickel and 5 pennies = 25 + 10 + 5 + 5*1 = 45 cents.
Remaining amount to be yielded = 110-45 = 65 cents.
Since the remaining amount is a multiple of 5 -- and the amounts yielded by the quarters, dimes and nickels will each be a multiple of 5 -- the remaining number of 1-cent pennies must also be a multiple of 5.
Let 5x = the total number of remaining pennies.
Since the remaining amount to be yielded by the 4 types of coins is 65 cents, we get:
25q + 10d + 5n + 5x = 65
5q + 2d + n + x = 13.
Note the following:
If n+x=
1, there are
2 options for n and x:
n=0, x=1
n=1, x=0.
If n+x=
2, there are
3 options for n and x:
n=0, x=2
n=1, x=1
n=2, x=0.
Implication of the values in red:
For every value of n+x, the number of options for n and x = n+x+1.
Case 1: q=2
Substituting q=2 into 5q + 2d + n + x = 13, we get:
5*2 + 2d + n + x = 13
2d + n + x = 3.
If d=1, then n+x=1, with the result that the number of options for n and x = 1+1 = 2.
If d=0, then n+x=3, with the result that the number of options for n and x = 3+1 = 4.
Total options for Case 1 = 2+4 = 6.
Case 2: q=1
Substituting q=1 into 5q + 2d + n + x = 13, we get:
5*1 + 2d + n + x = 13
2d + n + x = 8.
If d=4, then n+x=0, with the result that the number of options for n and x = 0+1 = 1.
If d=3, then n+x=2, with the result that the number of options for n and x = 2+1 = 3.
If d=2, then n+x=4, with the result that the number of options for n and x = 4+1 = 5.
If d=1, then n+x=6, with the result that the number of options for n and x = 6+1 = 7.
If d=0, then n+x=8, with the result that the number of options for n and x = 8+1 = 9.
Total options for Case 2 = 1+3+5+7+9 = 25.
Case 3: q=0
Substituting q=0 into 5q + 2d + n + x = 13, we get:
5*0 + 2d + n + x = 13
2d + n + x = 13.
If d=6, then n+x=1, with the result that the number of options for n and x = 1+1 = 2.
If d=5, then n+x=3, with the result that the number of options for n and x = 3+1 = 4.
If d=4, then n+x=5, with the result that the number of options for n and x = 5+1 = 6.
If d=3, then n+x=7, with the result that the number of options for n and x = 7+1 = 8.
If d=2, then n+x=9, with the result that the number of options for n and x = 9+1 = 10.
If d=1, then n+x=11, with the result that the number of options for n and x = 11+1 = 12.
If d=0, then n+x=13, with the result that the number of options for n and x = 13+1 = 14.
Total options for Case 3 = 2+4+6+8+10+12+14 = 56.
Total options = Case 1 options + Case 2 options + Case 3 options = 6 + 25 + 56 = 87.
The correct answer is
D.
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