Help with a math problem please.

[Aintnosunshinewhenimgone]

Dd has a problem in her math book and we are struggling to answer it- there is no answer in the book. The question is: can you make a system of 3 coins that would allow you to make every number from 1-50 if you only had one of each coin?? Ypu can make up your own coins e.g. 3p 15p etc but can only use each one once to make each number. I'm stuck and stubborn so someone please help!! Thank you

How can you make 1 from 3 coins?

@TheOrigRights

How can you make 1 from 3 coins?

By using just a 1p coin?

I'm not entirely sure I understand the question though. Can you only have 3 coin values?

What age is your dd?

Do they mean you can add up your three chosen coins in different combinations to make the numbers 1-50?

No. You have to have 1p because you need odd amounts. You need 2p because how else do you make 2p if you can't use 2 x 1p and you have to have 5p because you can't 5p without using using 2 x 2ps.

But then you cannot make 4p without using 2 x 2p coins so impossible.

I don't think this is possible as I understand the question.

To make the low numbers you would need multiple low value coins. To make the high numbers you would need a high value coin.

The only thing I can think of is if 'coin' is a red herring and as well as addition and subtraction you are meant to use multiplication and division. So just find three numbers that can make every combo 1-50 Countdown style. Even then, not sure if that works.

Or try and make 50p. You need largish denominations. Which rules out making small amounts.

This sounds ridiculous.

As a primary school teacher, many children have the misconception that there are 3 pence coins / 4 pound coins etc, and tasks like this are reinforcing it.

I also don't understand what you're being asked to do.

Can you post a picture of the actual homework?

If she can make up her own coins then 1/3 p, or something. How old is she?

Ignore me! People usually do!

If you can only use each number once for each number I don’t think it’s possible.
1-1
2-2
3-2+1
4-4
5-4+1
6-4+2
7-4+2+1

But then you couldn’t go further without having a double of some of those coins.

Let her have double sided coins with a different value on each side

Is the 3 in the OP a typo?

You're missing the 's' on the end.

Problem solved.

The question says that you can use made up coins like 3p

One of the coins needs to be 1p to make 1p
2p can only be made with a 2p
3p can't be made with a 2p+1p (you need them to make 1p and 2p) so say 3p.

The biggest number you can make with a 1p,2p and 3p is 6p

Scratch that 3p can be made with 2p + 1p
4p =1p+3p or 4p

1,2,3 biggest number you can make is 6
1,2,4 biggest number you can make is 7

No. You can't. Unless you're allowed to use multiple coins of the same value (in which case you'd only need a 1p coin).

Unless this is a 'change' problem? So for example to pay 4p you could give a 5p and get 1p change? Not actually worked thought about that though - still don't think it's possible.

Or is this a question asking for up to 3 different numbers to make each number?

So 1p= 1p
7p=2p+5p
34p= 20p+ 10p+ 4p
49p = 25p + 20p +4p

Etc

Can they be different coins each time?
Or the same 3? Which would be impossible.

As it stands, the answer to your post is NO.

Is it possible that you have paraphrased and lost the original meaning OP?

If you had only 3 types of coin but were able to use a number of each it might be doable.

Could you possibly post a photo of the question as it is presented in the maths book?

If you have to use 3 coins 1p could be made from fractions of a penny or negative amounts

1p = 0.5p + 0.2p+ 0.3p

But I'm assuming your dd is old enough to have done decimals

What a random question.

If it helps, the minimum number of coins you need is 6. Think about binary numbers.
1p,2p,4p,8p,16p and 32p
Then you can make numbers up to 63p

I think what they're after is a convincing argument as to why the answer is "no, this is impossible."

As others have said, you need a 1p to make 1p, and a 2p (because you're not allowed to use the same coin twice, hence you can't do 1+1). Then 3 is easy 2+1 or you could opt for a 3p coin. 4 you need either a 4p coin (cos you're not allowed 2+2) or if you chose to have a 3p coin you could have 3+1.

By this point you have to have minted either:
1p, 2p, 3p, in which case the largest number you can get to is 6
Or you've minted
1p 2p 4p, in which case you're stuck at 7.

Neither option gets you anywhere near 50.

Any child who gets that under their own steam at primary is going to end up doing maths at university!

Oh darn!! Sorry 3 should be 6 coins in the OP!! Oops. So she needs to develop a system of 6 coins to make all numbers from 1-50 using each coin only once per number e.g. 1p + 5p + 10p = 16p cant use 1p and 1p again

Looks impossible. Have you copied the wording exactly? (Apologies if that is patronising but it is the sort of mistake I might make.)