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July 5, 2015

July 5, 2015

Posted by **Bob** on Wednesday, May 21, 2008 at 7:56pm.

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?

- Math -
**tchrwill**, Thursday, May 22, 2008 at 11:59amCindy 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!!!

- Math -
**tchrwill**, Thursday, May 22, 2008 at 12:39pmDanielle 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

- Math -
**tchrwill**, Thursday, May 22, 2008 at 2:12pmDanielle 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.!!!

- Math -
**Aitana**, Tuesday, May 21, 2013 at 11:20pmCindy has 1 half dollar,1quater,3 dimes,1 nickel,and 3 penny's.