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# Maths GCSE question? watch

1. Would someone give me steps to solve it without telling me the answer, I know it's simple algebra but I want to learn the quickest way.

Fernando chooses 3 different whole numbers between 1 and 40
The first number is a square number
The second number is 4 multiplied by the first number
The third is a prime number and less than the first number
The sum of all numbers is 50
What 3 numbers has he chosen
2. (Original post by Rakkidoesscience)
Would someone give me steps to solve it without telling me the answer, I know it's simple algebra but I want to learn the quickest way.

Fernando chooses 3 different whole numbers between 1 and 40
The first number is a square number
The second number is 4 multiplied by the first number
The third is a prime number and less than the first number
The sum of all numbers is 50
What 3 numbers has he chosen
Call the first number x. It must be a square number.

4 ≤ 4x ≤ 36. (Numbers need to be between 1 and 40).
So, 1 ≤ x ≤ 9. x must be a square.

I hope this helps.

The rest as you've said is easy algebra, but I don't want to give away the answer!
3. (Original post by Rakkidoesscience)
...
If you're stuck, writing them all out is also an option.

First number possibilities: 1, 4, 9, 16, 25, 36

Second number possibilities depending on the first choice: 4, 16, 36

Third number possibilities: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

Then just deduce the number by trial and error by going through the numbers in the first list.

For example;

Let's say you choose 4 from the first list. Then the second number must be 16. Then 4+16=20 thus to make 50 you need 30 but that number is neither a prime number nor is it less than the first number.
4. Well you know the first number is 4 or 9 because all the other squares wont work when multiplied by 4(16*4>50). So i guess the fastest way is just by inspection.
Call the first number x. It must be a square number.

4 ≤ 4x ≤ 36. (Numbers need to be between 1 and 40).
So, 1 ≤ x ≤ 9. x must be a square.

I hope this helps.

The rest as you've said is easy algebra, but I don't want to give away the answer!
Yep , that was the kind of algebra I was looking for I was trying to find a quicker way than trial and error

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