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Math Olympiad Help

I have to find all square numbers ending in 3 4s. The first number, that has a square ending in 3 4s is 38.
Here is the given solution to the problem:



Can someone explain to me how this person got to that solution of N = 500k +/- 38?
Reply 1
Original post by WassupLadz
I have to find all square numbers ending in 3 4s. The first number, that has a square ending in 3 4s is 38.
Here is the given solution to the problem:



Can someone explain to me how this person got to that solution of N = 500k +/- 38?

If it's just the last part before it,
N +/- 38 is divisible by 500
So N = 500k +/- 38
Where k is the quotient and +/-38 the remainder (if we allow the remainder to be negative)
Reply 2
I don't get how and why N +/- 38 is divisible by 500
Reply 3
Original post by WassupLadz
I don't get how and why N +/- 38 is divisible by 500

Which parts do you understand up to?
Reply 4
Original post by mqb2766
Which parts do you understand up to?

I don't understand anything that he did. If you can explain it all that would be great.
Reply 5
Original post by WassupLadz
I don't understand anything that he did. If you can explain it all that would be great.


You must understand some parts? Not sure I can give a totally different take on the model solution. Also the question is missing and it looks like the first part of the model solution isn't attached.

At least break it up into chunks which you're happy/unhappy with and be clear about which parts you're not clear about.
(edited 3 years ago)
Reply 6
Original post by mqb2766
You must understand some parts? Not sure I can give a totally different take on the model solution. Also the question is missing and it looks like the first part of the model solution isn't attached.

At least break it up into chunks which you're happy/unhappy with and be clear about which parts you're not clear about.

Ok so I understand the question.
I understand 38 is the first number that has a square ending in 444.
I understand how he got to n -38, n + 38
Now I don't understand how he got greatest common factor as 78.
I understand the prime factorisation of 1000
Why did he use the prime factorisation?
I also don't understand why he said N is 500k +/- 38, where did he get 500 from? He used 3 5s which I understand but why not 3 2s?why only 2 2s?
Reply 7
Original post by WassupLadz
Ok so I understand the question.
I understand 38 is the first number that has a square ending in 444.
I understand how he got to n -38, n + 38
Now I don't understand how he got greatest common factor as 78.
I understand the prime factorisation of 1000
Why did he use the prime factorisation?
I also don't understand why he said N is 500k +/- 38, where did he get 500 from? He used 3 5s which I understand but why not 3 2s?why only 2 2s?

Ok, is there a part before the "alternatively".
Edit - Also is it 76 or.78 (typo?)
(edited 3 years ago)
Reply 8
Original post by WassupLadz
Ok so I understand the question.
I understand 38 is the first number that has a square ending in 444.
I understand how he got to n -38, n + 38
Now I don't understand how he got greatest common factor as 78.
I understand the prime factorisation of 1000
Why did he use the prime factorisation?
I also don't understand why he said N is 500k +/- 38, where did he get 500 from? He used 3 5s which I understand but why not 3 2s?why only 2 2s?

The
HCF(N-38,76)
Is syraightforward. The HCF must divide both N+38 and N-38, so it must divide their difference. This is used repeatedly in Euclids algorithm for calculating the HCF. So the HCF must divide 76.

Is that ok?
Reply 9
Original post by mqb2766
The
HCF(N-38,76)
Is syraightforward. The HCF must divide both N+38 and N-38, so it must divide their difference. This is used repeatedly in Euclids algorithm for calculating the HCF. So the HCF must divide 76.

Is that ok?

No, why is that true? why can't it just be 38 as the HCF?
Original post by WassupLadz
No, why is that true? why can't it just be 38 as the HCF?

The HCF of a and b must divide into both a and b.
So it will also divide into a-b. If you've gone over Euclid's HCF method, this is a central part.

As a-b=76=2^2*19 in this case, that's the maximum the HCF can be and the answer then goes into showing the HCF is 2*19.

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