Pounds & Shillings Swap: Find The Math!

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Multiplying Pounds and Shillings: Swapping the Numbers

Hey guys! Ever stumbled upon a cool math trick that just makes you go, "Whoa!"? Well, get ready for one! We're diving into a fascinating puzzle involving good old British currency – pounds (£) and shillings (s). The question is: can we discover sums of money for which multiplying them by a whole number simply swaps the pounds and shillings places? Let's investigate this mathematical quirk and discover its hidden depths.

The Curious Case of £6 13s

So, the prompt mentions an interesting example: £6 13s. When you double this amount, you get £13 6s. Notice anything special? The pounds and shillings have simply swapped places! This immediately sparks a question: Are there any other sums of money that exhibit this peculiar property? And if so, how can we find them?

This is where the fun begins! To solve this, we need to convert everything into a single unit. Since there are 20 shillings in a pound, we can express any amount in terms of shillings. Let's say we have x pounds and y shillings. The total amount in shillings would be 20x + y. When we multiply this by an integer, let's call it n, we want the result to be y pounds and x shillings, or 20y + x shillings. Therefore, we need to find integer solutions to the equation:

n(20x + y) = 20y + x

Rearranging this equation, we get:

20nx + ny = 20y + x

20nx - x = 20y - ny

x(20n - 1) = y(20 - n)

This gives us the ratio:

x / y = (20 - n) / (20n - 1)

Now, remember that x and y represent pounds and shillings, respectively. This means x must be a non-negative integer, and y must be an integer between 0 and 19 (inclusive), since you can't have 20 or more shillings. Also, n has to be an integer greater than 1, since multiplying by 1 wouldn't swap anything, and if n=20, the numerator becomes zero and x will also be zero.

Finding the Solutions

Okay, time to put on our detective hats and hunt for solutions. We know that x and y have to be whole numbers and y must be between 0 and 19. So, let's test different values of n and see what we get. We already know that doubling £6 13s works. So, n=2 is one of our multipliers. Let's start by listing out all the integer values for n greater than 1 and less than 20.

  • n = 2: x / y = (20 - 2) / (40 - 1) = 18 / 39 = 6 / 13. This gives us x = 6 and y = 13, which is the example we started with! £6 13s doubled is £13 6s.
  • n = 3: x / y = (20 - 3) / (60 - 1) = 17 / 59. Since 17 and 59 are prime numbers, the fraction is already in its simplest form. To find integer values for x and y, we can try multiplying the numerator and denominator by a common factor. However, since 59 > 19, there is no such number that satisfies the requirements.
  • n = 4: x / y = (20 - 4) / (80 - 1) = 16 / 79. Same as the case with n = 3, since 79 > 19, there is no integer value that satisfies the requirements.
  • n = 5: x / y = (20 - 5) / (100 - 1) = 15 / 99 = 5 / 33. Same as the previous case, since 33 > 19, there is no integer value that satisfies the requirements.
  • n = 6: x / y = (20 - 6) / (120 - 1) = 14 / 119 = 2 / 17. This gives us x = 2 and y = 17! So, £2 17s multiplied by 6 is £17 2s.
  • n = 7: x / y = (20 - 7) / (140 - 1) = 13 / 139. Once again, since 139 > 19, there is no integer value that satisfies the requirements.
  • n = 8: x / y = (20 - 8) / (160 - 1) = 12 / 159 = 4 / 53. Since 53 > 19, there is no integer value that satisfies the requirements.
  • n = 9: x / y = (20 - 9) / (180 - 1) = 11 / 179. Since 179 > 19, there is no integer value that satisfies the requirements.
  • n = 10: x / y = (20 - 10) / (200 - 1) = 10 / 199. Since 199 > 19, there is no integer value that satisfies the requirements.
  • n = 11: x / y = (20 - 11) / (220 - 1) = 9 / 219 = 3 / 73. Since 73 > 19, there is no integer value that satisfies the requirements.
  • n = 12: x / y = (20 - 12) / (240 - 1) = 8 / 239. Since 239 > 19, there is no integer value that satisfies the requirements.
  • n = 13: x / y = (20 - 13) / (260 - 1) = 7 / 259. Since 259 > 19, there is no integer value that satisfies the requirements.
  • n = 14: x / y = (20 - 14) / (280 - 1) = 6 / 279 = 2 / 93. Since 93 > 19, there is no integer value that satisfies the requirements.
  • n = 15: x / y = (20 - 15) / (300 - 1) = 5 / 299. Since 299 > 19, there is no integer value that satisfies the requirements.
  • n = 16: x / y = (20 - 16) / (320 - 1) = 4 / 319. Since 319 > 19, there is no integer value that satisfies the requirements.
  • n = 17: x / y = (20 - 17) / (340 - 1) = 3 / 339 = 1 / 113. Since 113 > 19, there is no integer value that satisfies the requirements.
  • n = 18: x / y = (20 - 18) / (360 - 1) = 2 / 359. Since 359 > 19, there is no integer value that satisfies the requirements.
  • n = 19: x / y = (20 - 19) / (380 - 1) = 1 / 379. Since 379 > 19, there is no integer value that satisfies the requirements.

So far, we've found two solutions: £6 13s and £2 17s.

Generalizing the Solution

We can generalize the solution by expressing x as k(20-n) and y as k(20n-1), where k is any constant integer.

x = k(20-n) y = k(20n-1)

However, remember that y must be between 0 and 19. So, we can say that:

0 <= k(20n - 1) <= 19

Let's analyze our current solutions to understand better.

For £6 13s, n = 2, x = 6, y = 13. So, 6 = k(20-2) => 6 = 18k => k = 1/3 13 = k(20*2-1) => 13 = 39k => k = 1/3

For £2 17s, n = 6, x = 2, y = 17. So, 2 = k(20-6) => 2 = 14k => k = 1/7 17 = k(20*6-1) => 17 = 119k => k = 1/7

From the above analysis, we can modify the equations to remove the constant k.

x = (20-n)/d y = (20n-1)/d

where d is the greatest common divisor of (20-n) and (20n-1) such that y is between 0 and 19.

Let's Summarize the Solutions

After our investigation, here are two sets of amounts that work with the specified conditions:

  • £6 13s: Double it, and you get £13 6s.
  • £2 17s: Multiply it by 6, and you get £17 2s.

It's important to remember the constraint that the number of shillings must always be less than 20. Otherwise, we need to convert any additional shillings into pounds, which will change the problem entirely.

Further Exploration

This exploration demonstrates how number theory concepts can pop up in unexpected places, even in something as seemingly mundane as money! You can continue exploring by trying different currencies or creating your own variations of this problem.

Isn't math just awesome? Keep exploring, keep questioning, and keep having fun with numbers!