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**justme****Member**- Registered: 2006-01-29
- Posts: 1

When you want to square a number ending in a 5, you take the tens column and add one and multiply it to the original tens digit and place it in front of 25... not a good description.

Try this 35 * 35 =

5*5 = 25

3*4 = 12

answer 1225

How does this work every time?

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

this is what i came up with:

(10a + 5)^2 and a<10

100a^2 + 100a + 25 as you see you will always have to add 25.

100a^2 + 100a = 100a(a + 1) so you multiply tens by tens plus one, by putting it in front of 25 you "multiply" by 100.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

100a(a + 1) so you multiply tens by tens plus one, by putting it in front of 25 you "multiply" by 100.

If that wasn't clear, what you are doing is getting them number a(a+1) like you said (3*4), and then shifting that over two digits to make room for the 25 on the right.

Nicely done kempos.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**RauLiTo****Member**- Registered: 2006-01-11
- Posts: 142

to square a 2 digit number ending with 5 you have to follow these steps :

1- Multiply the first digit by the next consecutive number.

2- The product is the first two digits: XX _ _.

3- The last part of the answer is always 25: _ _ 2 5.

lets try '' 35 '' as you said :

1- the number is 35 ... 3 * 4 = 12 X X

2- The last part of the answer is always 25

3- 35 * 35 = 1225 that is

try it with any number you want

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

Raulito this we already know the question was about proving it.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I didn't. Never heard this before.

A logarithm is just a misspelled algorithm.

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**RauLiTo****Member**- Registered: 2006-01-11
- Posts: 142

kempos wrote:

Raulito this we already know the question was about proving it.

he said that it doesnt work with 35 ! so i showed him ... sorry maybe i didn't get your question explain more man

*Last edited by RauLiTo (2006-01-29 20:30:41)*

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

he gave 35*35 as an example. he said it works with 35 but asked why :-)

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**RauLiTo****Member**- Registered: 2006-01-11
- Posts: 142

kempos wrote:

he gave 35*35 as an example. he said it works with 35 but asked why :-)

i got the right question now (( sorry for bothering man but i am new here )) ... i think i agree with your solution

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

In general any two digit square equals;

100f² + 20fs + s²; where f = 1st digit and s = 2nd digit

A number ending in 5 is just a special case where the first two terms above equals;

100f (f + 1)

How about if f = s?

It's square is simply 121f²

I'm sure that anyone who feels like playing around with relationships can find a lot of "tricks" like those above.

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