which amount up to £1 cannot be paid exactly with fewer than 6 coins?
is this a trick question and can all the amounts be paid?! I am really struggling with this one
Is this using current British coinage, or pre-decimalisation? There is a quite famous problem by Dudeney about this, but it uses old coinage - farthings and suchlike. The book "Amusements in Mathematics by Henry Ernest Dudeney" is available ibn Project Gutenberg, if you want to look it up.
Still, it's a nice example of proving by exhaustion, using current coinage.
We have: 1p, 2p, 5p, 10p, 20p and 50p
Let's consider the values up to 10, in terms of numbers of coins needed
1 -> 1
2 -> 1
3 -> 2
4 -> 2
5 -> 1
6 -> 2
7 -> 2
8 -> 3
9 -> 3
OK, so second digit can be satisfied with at most 3 coins.
But first digit is same case as second digit, multiplied by 10, since for every unit, there exists an equivalent 10 times bigger. Therefore we have proved that 10p, 20p, ... 90p can be given with three coins.
Therefore every integer 1 through 99 nclusive can be paid with no more than 6 coins.
But do check out Dudeney - there are some fun puzzles there! :-)
thankyou so much for your help I am vey grateful, I felt you could do every value but wasn't sure if I was missing something.
No, it is not a trick question. The amount that cannot be paid exactly with fewer than 6 coins is £.85. Let me break it down for you step-by-step:
Step 1: Start with the highest denomination coin, which is the £0.50 coin.
Step 2: Subtract £0.50 from £1, which leaves you with £.50.
Step 3: Now, let's try to use the second highest denomination coin, which is the £0.20 coin.
Step 4: Subtract £0.20 from £.50, which leaves you with £.30.
Step 5: Repeat this process, using the £0.10 coin.
Step 6: Subtract £0.10 from £.30, which leaves you with £.20.
Step 7: Again, use the £0.05 coin.
Step 8: Subtract £0.05 from £.20, which leaves you with £.15.
Step 9: Finally, try to use the £0.01 coin.
Step 10: Subtract £0.01 from £.15, leaving you with £.14.
As you can see, this process requires a total of 6 coins - a £0.50, a £0.20, a £0.10, a £0.05, a £0.01, and a £0.01, which totals £1.02. Therefore, the amount of £.85 cannot be paid exactly with fewer than 6 coins.
No, this is not a trick question. There are indeed certain amounts that cannot be paid exactly with fewer than 6 coins. To solve this problem, we need to consider the possible combinations of coins that can be used to make up a pound (£1).
In the UK, the coins currently in circulation include the following denominations: £2, £1, 50p, 20p, 10p, 5p, 2p, and 1p. In order to find the amount that cannot be paid with fewer than 6 coins, we need to exhaustively check all possible combinations of these coins.
To simplify our analysis, we can start by considering the largest possible number of £2 coins that can be used to make up a pound. Since £1 = 50p × 2, the highest number of £2 coins we can use is 0.
Now, let's analyze the remaining denominations one by one, from highest to lowest, to determine the possible combinations and find the amount that cannot be paid with fewer than 6 coins:
- £1 coin: Since we have already used the maximum number of £2 coins, we cannot use a £1 coin in this case.
- 50p coin: We can have at most 2 of these coins. If we use 2 x 50p coins, the remaining amount we need to make up is 0p.
- 20p coin: We can have at most 5 of these coins. If we use 5 x 20p coins, the remaining amount we need to make up is 0p.
- 10p coin: We can have at most 10 of these coins. If we use 10 x 10p coins, the remaining amount we need to make up is 0p.
- 5p coin: We can have at most 20 of these coins. If we use 20 x 5p coins, the remaining amount we need to make up is 0p.
- 2p coin: We can have at most 50 of these coins. If we use 50 x 2p coins, the remaining amount we need to make up is 0p.
- 1p coin: We can have at most 100 of these coins. If we use 100 x 1p coins, the remaining amount we need to make up is 0p.
Based on this analysis, we see that we can make up any amount less than or equal to £1 using a combination of fewer than 6 coins. Therefore, there is no amount that cannot be paid exactly with fewer than 6 coins.
I hope this explanation helps!