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**Proofing****Member**- Registered: 2011-02-10
- Posts: 2

Now i realize i could use the division algorithm and get to the correct answer, but i was looking for a different way. However this is my first proofing class was wondering if this is valid or not.

5|a if a has a 0 or a 5 in the one's position. So 5|n^2+2 if n^2 has a 3 or an 8 in the ones position. When multiplying 2 numbers the one's position of the product is only affected by the one's position of the multiplicand and the multiplier. This is shown by abc*def=f(abc)+10e(abc)+100d(abc) where a and d are single digit integers in the hundred's place of the number of the multiplicand and multiplier, b and e are single digit integers in the ten's place of the multiplicand and multiplier, and c and f are single digit integers in the one's position of the multiplicand and multiplier. From this result we see that only f(abc) will have an impact on the one's position of the product because any integer multiplied by a factor of ten will have a 0 in the one's position. Also f*(abc)= f*c+f*(10b)+f*(100a) so only f*c will affect the one's position of abc*def. Any integer squared will have the same value for the number in the one's position. We can see that 1^2=1 2^2=4 3^2=9 4^16 5^2=25 6^2=36 7^2=49 8^2=64 9^2=81 0^2=0. Since all of the squares of single digit integers do not end in 3 or 8 in the one's position 5|n^2 +2 is never true.

Was my idea on how to approach it differently. This is my 3rd proof to date.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi Proofing;

Or you could have tested just 5 cases:

Statements 1 - 5 can all be proved easily with algebra and since they exhaust all the cases we are done.

Welcome to the forum.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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