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# A few tough maths Qs (not from A-lvl syllabus) watch

1. Heya I am trying to crack my way through a UKMT past paper (maths question of varying levels; get very tough towards the end!), and was hoping to get some help for those I don't understand. Thnx a lot!

1) How many pairs of positive integers (x, y) are solutions to the equation
1/x + 2/y = 3/19

2) Given an unlimited supply of 50p, £1 and £2 coins, in how many different ways can we make a sum of £100?

Thnx again!
2. (Original post by Wagamuffin)
Heya I am trying to crack my way through a UKMT past paper (maths question of varying levels; get very tough towards the end!), and was hoping to get some help for those I don't understand. Thnx a lot!

1) How many pairs of positive integers (x, y) are solutions to the equation
1/x + 2/y = 3/19

2) Given an unlimited supply of 50p, £1 and £2 coins, in how many different ways can we make a sum of £100?

Thnx again!
How far have you got with them? Let us know what you've done, so we know how much to explain.
3. I can only find one solution: (7,133).
4. (Original post by Ralfskini)
I can only find one solution: (7,133).
Keep looking then
6. (Original post by Wagamuffin)
Heya I am trying to crack my way through a UKMT past paper (maths question of varying levels; get very tough towards the end!), and was hoping to get some help for those I don't understand. Thnx a lot!

1) How many pairs of positive integers (x, y) are solutions to the equation
1/x + 2/y = 3/19

2) Given an unlimited supply of 50p, £1 and £2 coins, in how many different ways can we make a sum of £100?

Thnx again!

I reckon there to be 2,601 ways.
7. 1/x+2/y=3/19

can be rearranged to:

3xy-38x-19y=0
(3x-19)(y-38/3)-(19*38)/3=0
3(3x-19)(y-38/3)=19*38
(3x-19)(3y-38)=722

722=1*722=2*361=19*38 and these are the only ways of expressing 722 as a product of two numbers, therefore the only solutions are the values of x and y for which either (3x-19) or (3y-38) is equal to one of these numbers.

This gives the solutions (x,y):

(7,133)
(19,19)
(247,13)
8. Thnx Ralfskinni! I managed to factorise it by completing square and all; but I had trouble determining the factors of 722 without a calculator (they aren't allowed) do you have a fast method of determining this or maybe just checking whether 361 is prime within 60 or so seconds Thnx !!!!

As for the coin question, I have no idea how to go about it without any kind of calculator either... I was also searching for some general method. 1st Question sorted now thnx guys!
9. (Original post by Wagamuffin)
Thnx Ralfskinni! I managed to factorise it by completing square and all; but I had trouble determining the factors of 722 without a calculator (they aren't allowed) do you have a fast method of determining this or maybe just checking whether 361 is prime within 60 or so seconds Thnx !!!!

As for the coin question, I have no idea how to go about it without any kind of calculator either... I was also searching for some general method. 1st Question sorted now thnx guys!
361 = 19^2
10. To answer number 2, first note there are 51 possibilities for the number of two pounds coins. Say the number of two pound coins is x, then the number of possibilities of 1 pound coins is 101-2x (it can 0 or any number between 1 and 100-2x), after this the number of 50s is fixed, as the total is fixed.

So sum over 0 to 50: 101 - 2x = 101.51 - 50.51 = 51^2 = 2601.
11. (Original post by Wagamuffin)
Thnx Ralfskinni! I managed to factorise it by completing square and all; but I had trouble determining the factors of 722 without a calculator (they aren't allowed) do you have a fast method of determining this or maybe just checking whether 361 is prime within 60 or so seconds Thnx !!!!
For reasonably small numbers ( < 2500) the easiest way to check if they're prime is to check, if the number is x, whether any primes less than or equal to root(x) divide x, if they don't, then none greater can either, so x must be prime. This only gives you a few primes (those < 50) to check.
12. Aha I didn't remember we could discard the 3rd variable thnx I hope I will be able to apply that method in future!

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