Help me with this math riddle...it is driving me CRAZY!?

Question

I don't really have a mathematical mind and I'm trying to solve this crazy riddle. You have 100 coins and no nickles and the coins add up to $5.00. What coins do you have?

Answer 1

1)
75 pennies = $0.75

15 dimes = $1.50

9 quarters = $2.25

1 halfdollar = $0.50

2)

90 pennies = $0.90

6 dimes = $0.60

1 half dollar coin = $0.50

3 dollar coins = $3.00

100 coins = $5.00

3)

75 pennies = $0.75

20 dimes = $2.00

1 quarter = $0.25

4 half dollars = $2.00

That's 100 coins having a value of $5.00. There's more ways out there.

Answer 2

Wow, lots of people not getting this question.

Let p, d, and q be the number of pennies, dimes and quarters. For now, let's skip the case of half-dollars and silver dollars.

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

9d + 24q = 400.

But the left hand side is divisible by 3, and the right hand side is not, so you can't do it with just pennies, dimes and quarters.

So let h be the number of half-dollars.

Then:
p + d + q + h = 100
p + 10d + 25q + 50h = 500

The difference here gives us:

(*) 9d + 24q + 49h = 400

You need 400-49h to be divisible by 3. That means that h must be 1, 4, or 7. (You can do this by trial and error.)

Case:

h=1:

9d + 24q = 400 - 49 = 351

Divide by 3, we get:

3d + 8q = 117

Since 117 is divisible by 3, that means q must be a multiple of 3, say q=3q'. Dividing by 3 again, we get:

d + 8q' = 39

This gives us (d,q') = (7,4), (15,3), (23,2), (31,1), (39,0).

This in turn gives us:

7 dimes, 12 quarters, 1 half-dollar, 80 pennies
15 dimes, 9 quarters, 1 half-dollar, 75 pennies
23 dimes, 6 quarters, 1 half-dollar, 70 pennies
31 dimes, 3 quarters, 1 half-dollar, 65 pennies
39 dimes, 0 quarters, 1 half-dollar, 60 pennies

h=4:
9d + 24q = 400 - 49*4 = 204

Dividing by 3, we get:

3d + 8q = 68

68 is divisible by 4, so d must be divisible by 4, say d=4d'. Dividing by 4 we get:

3d' + 2q = 17

You can solve this by trial, giving: (d',q) = (5,1), (3,4), (1,7)

Giving solutions:

20 dimes, 1 quarter, 4 half-dollars, 75 pennies
12 dimes, 4 quarters, 4 half-dollars, 80 pennies
4 dimes, 7 quarters, 4 half-dollars, 85 pennies

h=7:

9d + 24q = 400-49*7 = 57

divide by 3:

3d + 8q = 19

Trial and error shows only one solution, (d,q) = (1,2).

That yields:

1 dime, 2 quarters, 7 half-dollars, and 90 pennies.

So, that's all solutions without using dollar coins. Dollar coins adds another level.

Answer 3

well r the coins all the same? ( like , r they all nickles or pennies)
if not they could be arranged in a couple of ways:
1$ : 50 pennies/5 dimes
1$ : 10 dimes...etc.

Answer 4

First let p=# of pennies, n=# of nickles, d=# of dimes, q = # of quarters.
Now, make some equations based on the sentences.
P+N+D+Q = 100
.01p+.05N+.1d+.25q = 5.00

n=0
This means that you can simplify the first equation to
p+d+q=100 and .01 p + .1d + .25q = 5.00

I would multiply the second equation by 100 to get rid of all the decimals and get
1p+10d+25q=500

Beyond that, you would need to get rid of one more variable to solve or add in another line comparing two of the 3 coins so that we could get rid of another variable.

Answer 5

i'm using a selection A...B. the consequence of the multiplication subject is going to be B...A. i exploit ... by way of fact i don't comprehend how many digits i visit prefer. A must be a million or 2 by way of fact that's the comparable sort of digits. B is then the two 4 or 8. by using a tiny little bit of trial and mistake, we discover a must be 2 and B must be 4. (4*8 = 32 (2!), and four*2 = 8). We additionally observe that for this to paintings, the digit next to A in A...B must be a million or 2 (we don't prefer to hold some thing over). permit's call this digit C. by using trial and mistake, it could't be a three digit integer, so we attempt a 4 digit integer. permit the subsequent integer be D. immediately, we bypass over the two cases: 21D8 and 22D8. in case you write them out, you ought to hold over a three (from 4*8=32). 4D + 3 can in no way bring about 2, so C must be a million and 4D + 3 ought to end in a million. Subtract the three (observe 4D + 3 would not equivalent a million, it in effortless terms leads to a million) to discover that 4D leads to eight. the only 2 posibilities for D are then 2 and seven. attempt them out: 2178 works! i could discover a speedier / extra algebraic way and positioned up it here (by using enhancing my positioned up). save checking back! besides, from here, we are able to branch in many diverse guidelines: permit's attempt for a 5 digit selection! we could have 4*2ABC8 = 8CBA2 B is going to be the comparable in the two cases; A has the comparable regulations as before ( it ought to equivalent a million or 2). in basic terms like before, we discover that A must be a million and C must be 7 with trial and mistake. we've: 4*21B78 = 87B12. in case you write it out, you will see that 4*a million + x (permit x be what we grant over from 4*B) = 7, so x = 3. for this reason, B must be 7, 8, or 9. yet submit to in concepts that we even have 4B + 3 leads to B. the only case that works that's 21978. If this sounds extremely confusing, attempt doing the extremely multiplication out by way of fact then each little thing will replace into clearer.

Answer 6

There are many answers, but the first one I found was:

75 pennies = 0.75

15 dimes = 1.50

9 quarters = 2.25

1 halfdollar = 0.50

That's 100 coins totally $5.00 in value.

Answer 7

Dimes, Pennies, Quarters, 50 cent pieces, and Susan B. Anthony Dollars..... those are what coins you may have -- but I just don't know how many!!!

Answer 8

42 dimes, 55 pennies, and 3 quarters I think

Answer 9

half dollar coins

Answer 10

pennies and quarters