SOLUTION: From a pile of 100 pennies, 100 nickles , and 100 dimes . select 21 coins which have a total of exactly $1.00 . how many of each of the three types should be selected?

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Question 320851: From a pile of 100 pennies, 100 nickles , and 100 dimes . select 21 coins which have a total of exactly $1.00 . how many of each of the three types should be selected?
Found 2 solutions by Edwin McCravy, JBarnum:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

system%28p%2Bn%2Bd=21%2Cp%2B5n%2B10d=100%29

Eliminate n by multiplying the first equation through by -5

system%28-5p-5n-5d=-105%2Cp%2B5n%2B10d=100%29

Adding these term by term:

-4p%2B5d=-5

The smallest absolute value of a coefficient is 4,
so write all integers in terms of their nearest multiple
of 4

-4p%2B%284%2B1%29d=-%284%2B1%29
-4p%2B4d%2Bd=-4-1

Divide through by 4

-p%2Bd%2Bd%2F4=-1-1%2F4

Isolate fractional terms:

d%2F4%2B1%2F4=p-d-1

The right side is an integer, so let that integer be A, 
set both sides equal to integer A:

system%28d%2F4%2B1%2F4=A%2Cp-d-1=A%29

Clear the first equation of fractions:

system%28d%2B1=4A%2Cp-d-1=A%29

Solve the first equation for d

d=4A-1

Substitute that for d in p-d-1=A

p-d-1=A
p-%284A-1%29-1=A
p-4A%2B1-1=A
p-4A=A
p=5A

Subsitutute p=5A and d=4A-1 in p%2Bn%2Bd=21

p%2Bn%2Bd=21
5A%2Bn%2B4A-1=21
9A%2Bn-1=21
n=22-9A

So now we have the numbers of coins in terms of integer A

system%28p=5A%2C+n=22-9A%2C+d=4A-1%29

10 dimes makes a dollar so there can't be as many as 10 dimes, since
we have to have 21 coins, so

0%3C=d%3C10

0%3C=4A-1%3C10

Add 1 to all three sides:

1%3C=4A%3C11

Divide all three sides by 4

1%2F4%3C=A%3C11%2F4
1%2F4%3C=A%3C2%263%2F4

Since A is an integer then A is either 1 or 2, for
they are the only integers between 1%2F4 and 2%263%2F4
So there will be two solutions,

If A = 1, then 

system%28p=5A=5%281%29=5%2C+n=22-9A=22-9%281%29=22-9=13%2C+d=4A-1=4%281%29-1=4-1=3%29

So that's one solution: 5 pennies, 13 nickels and 3 dimes.

If A = 2, then 

system%28p=5A=5%282%29=10%2C+n=22-9A=22-9%282%29=22-18=4%2C+d=4A-1=4%282%29-1=8-1=7%29

So that's the other solution: 10 pennies, 4 nickels and 7 dimes.

Edwin

Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
p+n+d=21
.01p+.05n+.1d=100
if there are any pennies they have to be in 5,10,15,or 20
if there are 20 penies then only room for 1 more coin which cant add to 1.00
15 pennies then there are 6 coins left but even at 6 dimes thats only 0.75
10 pennies makes 11 coins left n+d=11 and .05n+.1d=.90 is a possibility
5 pennies makes 16 coins left n+d=16 and .05n+.1d=.95 is a 2nd possibility
according to the problem it seems there has to be a number for each coin


.05n%2B.1d=.90 multiply by 100
n%2Bd=11 multiply by 5
.
.
5n%2B10d=90
5n%2B5d=55
5d=35
d=7


p%2Bn%2Bd=21
p=10
n=7
17%2Bd=21
d=4
check:
highlight%2810%2B7%2B4=21%29
10 pennies, 7 nickels, 4 dimes
this wasnt that hard...lol that other tutor confused even me.


other solution:
n%2Bd=16 multiply by 10
.05n%2B.1d=.95 multiply by 100
.
.
10n%2B10d=160
5n%2B10d=95
5n=65
n=13
.
13%2Bd=16
d=3
.
highlight%285%2B13%2B3=21%29
5 pennies, 13 nickels, 3 dimes