Do you want to be rich?

Let the there be $p$ pennies, $d$ dimes and $q$ quarters. So

$p+d+q=100 \\ p+10d+25q=500$

On subtracting the first equation from the second one, we get

$9d+24q=400$

$3|9q$ and $3|24q$ so $3$ must also divide $400$ if any integral values of $d$ and $q$ exist. But we see that $3\nmid 400$ . So no integral solutions of $d$ and $q$ exist.

Note

• $a|b$ means $a$ divides $b$ or in other words, $an=b$ for some integer $n$ . Examples: $3|6$ and $7|21$ .

• If $a|b$ and $a|c$ , then $a|(b+c)$

$x+10y+25(100-x-y)=500$

$2000=24x+15y=3(8x+5y)$

for x and y to be integers then 2000 must be divisible by 3 which is wrong therefore, no combination is possible.

1
2
3
4
5
6

#Python
for p in range(0,100,5):
    for d in range(0,100-p):
        q = 100-p-d
        if .01*p + .1*d + .25*q == 5.00:
            print "Success!", p, "pennies", d, "dimes", q, "quarters"

This has no output, so there are no successes. If you change the amount to $5.05, there are three such instances. By the way, the language is Python.

$x$ , $y$ , $z$ are pennies, dimes, and quarters respectively. Where $x$ , $y$ , and $z$ are integers.

Let's evaluate two information in the problem:

(1) The number of coins are 100. So, $x + y + z = 100$

(2) Total value of the coins are $5. So, $0.01x + 0.10y + 0.25z = 5$

We can also write both equations like this:

(1) $x + y + z = 100$

(2) $x + 10y + 25z = 500$

Subtract (2) with (1), we will get:

$9y + 24z = 400$

Now, divide both sides with $3$ and we get:

$3y + 8z = \frac{400}{3}$

If $y$ and $z$ are integers, it is impossible that $3y + 8y$ equals a fraction. Therefore we conclude that the system of equations has no solution.

Let x be number of coins of pennies

y be number of coins of dimes

z be number of coins of dimes

as

0.01x+0.10y+0.25z=5.00..............................(1)

x+y+z=100..................................................(2)

Multiply (1) by 100 on both sides

x+10y+25z=500................................................(3)

x+y+z=100..................................................(2)

take z on left side

x+10y=500-25z..................................(4)

x+y=100-z........................................(5)

Solve (4) and (5)

get x in terms of z only

x=(100-3z)/3

z can have values from 0 to 20 [from (3) we can find this range, put x=0, y=0, z=500/25=20]

100-3z can be never be a Multiple of 3 for any value of z in [0,20]

So the answer is NO