# Small change

itisi
Posers and Puzzles 04 Mar '07 19:23
1. 04 Mar '07 19:23
"Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.

In how many ways can you make up 20 pence using 20p, 10p, 5p, 2p and 1p coins?"

On a paper entitled "Advanced Problems in Core Mathematics" I acquired recently. I can think of no ways of solving it other than listing all the possibilities. Any ideas?
2. 04 Mar '07 20:31
Start with the largest denomination and work your way down, starting with the most of that denomination and work down to none of that denomination.

1) With 1 10p we have 0p left, so one combination with a single 10p. (*).

2) With 2*5p we have 0p left, so one combination with 2*5p (*).

3) With 1*5p we have 5p left, so find number of ways of making 5p with denominations less than 5p:

3A) With 2*2p we have 1p left, so find number of ways of making 1p with denominations less than 2p.

3Aa) With 1p we have 0p left, so one combination with 1*5p, 2*2p, 1*1p (*)

3B) With 1*2p we have 3p left, so find number of ways of making 3p with denominations less than 2p.

3Ba) With 3*1p we have 0p left, so one combination with 1*5p, 1*2p, 3*1p (*)

3Bb) With 2*1p, we have 1p left, clearly no way to make this up as no denominations smaller than 1p.

3Bc) With 1*1p, we have 2p left, clearly no way to make this up as no denominations smaller than 1p.

3C) With no 2ps, we have 5p left, so find number of ways of making 5p with denominations less than 2p.

3Ca) With 5*1p we have 0p left, so one combination with 1*5p, 5*1p (*)

3Cb) Clearly no solutions with less than 5*1p (see 3Bb and 3Bc).

4) With no 5ps we have 10p left, so find number of ways of making 10p with denominations less than 5p.

4A) With 5*2p we have 0p left, so one combination with 5*2p (*)

4B) With 4*2p we have 2p left, so one combination with 4*2p, 2*1p (*)

4C) With 3*2p we have 4p left, so one combination with 3*2p, 4*1p (*)

4D) With 2*2p we have 6p left, so one combination with 2*2p, 6*1p (*)

4E) With 1*2p we have 8p left, so one combination with 1*2p, 8*1p (*)

4F) With no 2ps we have 10p left, so one combination with 10*1p (*)
3. TheMaster37
Kupikupopo!
04 Mar '07 20:531 edit
As for the second question; use the results of the first.

1) Using one 20p piece, we're done.

2) Using two 10p pieces, we're done.

3) Using one 10p, we have 11 possibilities to make the rest. Done.

4) Using no 10p or 20p pieces, we have 11*11=121 possibilities to make the rest. Done.

In total 1 + 1 + 11 + 121 = 134 possibilities.
4. 04 Mar '07 21:39
Originally posted by TheMaster37
As for the second question; use the results of the first.

1) Using one 20p piece, we're done.

2) Using two 10p pieces, we're done.

3) Using one 10p, we have 11 possibilities to make the rest. Done.

4) Using no 10p or 20p pieces, we have 11*11=121 possibilities to make the rest. Done.

In total 1 + 1 + 11 + 121 = 134 possibilities.
I don't think that works unfortunately, there will be a lot of repititions if you simply use 11*11.
5. TheMaster37
Kupikupopo!
04 Mar '07 23:02
I don't think that works unfortunately, there will be a lot of repititions if you simply use 11*11.
True, I was a bit hasty. I found another error as well. How's this for a second attempt:

As for the second question; use the results of the first.

1) Using one 20p piece, we're done.

2) Using two 10p pieces, we're done.

3) Using one 10p, we have 10 possibilities to make the rest. Done.

4) Using no 10p or 20p pieces, we have 10*10=100 possibilities to make the rest. Ten of those are symmetric, the others have every possibility doubled. 10 + 100/2 = 60. Done.

In total 1 + 1 + 10 + 60 = 72 possibilities.
6. 05 Mar '07 02:07
I think the answer is… 41

With a 20p 1 way
With 2x10p 1
With 1x10p 10

With no 20 or 10
And with 4x5p 1
With 3x5p 3
With 2x5p 6
With only 1x5p 8

With no 20 or 10 or 5 11
__
41
7. 05 Mar '07 03:45
Originally posted by luskin
I think the answer is… 41

With a 20p 1 way
With 2x10p 1
With 1x10p 10

With no 20 or 10
And with 4x5p 1
With 3x5p 3
With 2x5p 6
With only 1x5p 8

With no 20 or 10 or 5 11
__
41
Just trying to make it easier to read

I think the answer is… 41

With a 20p ........................1 way
With 2x10p .......................1
With 1x10p .....................10

With no 20 or 10
And with 4x5p................... 1
With 3x5p ........................3
With 2x5p ........................6
With only 1x5p .................8

With no 20 or 10 or 5 ......11
.....................................__
.............................total..41
8. 05 Mar '07 06:35
Originally posted by luskin
I think the answer is… 41

With a 20p 1 way
With 2x10p 1
With 1x10p 10

With no 20 or 10
And with 4x5p 1
With 3x5p 3
With 2x5p 6
With only 1x5p 8

With no 20 or 10 or 5 11
__
41
well done Arthur
9. TheMaster37
Kupikupopo!
06 Mar '07 00:03
Drats.
10. 07 Mar '07 06:56
Thanks everybody. My first answer for the second part was 67. 11 ways of making 10 hence (11c2) ways of making 20 considering it as two tens and 1 with just the 20p = 67. But again this suffers from repetition!

It's a good job the rest of the questions on the paper are algebra and trig because I can do those.