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.

]]>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

]]>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

]]>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

]]>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.

]]>(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.

]]>Try this 35 * 35 =

5*5 = 25

3*4 = 12

answer 1225

How does this work every time?

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