Math

Cindy had $1.08 using 9 coins.
She had as many pennies as dimes.
___
Danielle had 96 coins, using ten coins.
She had the same number of nickels as all the other coins put together.

__
The coin choices are.
Half Dollar, Quarters, Dimes, Nickels, and Pennies.
Help?

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  1. Cindy had $1.08 using 9 coins.
    She had as many pennies as dimes.

    There must be 4 pennies as 9 pennies would not allow for any other coins.

    Since there are supposed to be as many dimes as there are pennies, 4P + 4D = 44 cents.

    That leaves 65 cents to come from 1 coin which is impossible so there cannot be as many dimes as pennies.

    10D + 9P = $1.09 but 19 coins

    1Q + 8D + 4P = $1.09 but 13 coins

    2Q + 5D + 4P = $1.09 but 11 coins

    3Q + 3 D + 4P = $1.09 but 10 coins

    3Q + 2D + 2N + 4P = $1.09 but 11 coins

    Halves must be involved

    1H leaves 59 cents from 8 coins

    Must have 4 pennies

    1H + 4P leaves 55 cents from 4 coins

    1H + 2Q + 1N + 4P = $1.09 but 8 coins

    1H + 1 Q + 3D + 4P = $1.09 and 9 coins

    Hooray!!!

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  2. Danielle had 96 cents, using ten coins.
    She has the same number of nickels as all the other coins put together.

    Based on our prior experience, it is rather easy to zero in on
    1Q + 4D + 6N + 1P = 96 cents in 12 coins
    where 6N = 1Q + 4D + 1P coins

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  3. Danielle had 96 cents, using ten coins.
    She has the same number of nickels as all the other coins put together.

    Based on our prior experience, it is rather easy to zero in on
    1Q + 4D + 6N + 1P = 96 cents in 12 coins
    where 6N = 1Q + 4D + 1P coins

    OOPS - The hand was quicker than the eye. Lets try a more exact method.
    Assuming no halves are involved.

    1--25Q + 10CD + 5N + 1P = 96
    2--Q + D + N + P = 10
    3--N = Q + D + P
    4--Substituting (3) into (1) yields 30Q + 15D + 6P = 96
    5--Sustituting (3) into (2) yields Q + D + P = 5
    6--Multiplying (5) by 6 yields 6Q + 6D + 6P = 30
    7--Subtracting (6) from (4) yields 8Q + 3D = 22
    8--Dividing through by the lowest coefficient yields 2Q + 2Q/3 = 1 + 3/7
    9--(2Q - 1)/3 must be an integer as does (4Q - 2)/3
    10--Dividing by 3 again yields Q + Q/3 - 2/3
    11--(Q - 2)/3 is an integer k making Q = 3k + 2
    12--Substituting back into (7( yields D = 2 - 8k
    13--Clearly, k must be 0.
    14--For k = 0, Q = 2 and D = 2.
    15--From (5), P = 1 and hence, N = 4.
    16--Thus we have Q = 2, D = 2, N = 5 and P = 1 and 2Q's + 2D's + 1P = 5N as coins go.
    17--Also, 25(2) + 10(2) + 5(5) + 1(1) = 96 cents.

    At last.!!!

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  4. Cindy has 1 half dollar,1quater,3 dimes,1 nickel,and 3 penny's.

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