onose1: Impressive...

yea impressive (comments at FP is impressive)

adioolayi: Prof Einstein.. Read again..the Op said the number must end with 5!.....try 65.

prof Jega, same tin i said. D OP should av been more specific in his topic caption and not just say finding squares of numbers below 100. Incase u need to know. I av learnt from his method too and am not criticizing him, just informing.

Graviton2: Good job, OP. I'm following.

I lov that style too
17*11 = 1 (1+7) 7
= 187. This is just so cute

fxjones: hello fellow mathemacians,
just discover this while manipulating numbers, in class

how you quickly find the square of numbers without lifting a finger.

eg find the square of 15?

u can quickly do this without using calc or writing down

just say: 1*1+1 = 2
then find 5*5 = 25
lastly combine them together = 225

example 2

find the square of 25
again: 2*2+2 =6
then 5*5 again = 25
join them together = 625


example 3:

find the square of 75

again you say 7*7+7 = 56
and 5*5 = 25
finally join them together as we did before = 5625

this formula only work for numbers between 1-100 and the last number must end with 5

I will post the formula for other range: keep rocking!

Good job.

[quote author=vicdayo]
yea impressive (comments at FP is impressive)Sorri urs didn't make FP

nice

good to see something sensible being put on fp

An easier way would have been to multiply the right hand side by itself (i.e 5 x 5 = 25) then multiply the left number by the next higher number (e.g 2 x 3 = 6)

Therefore,

25 x 25 = 625
35 x 35 = 1225...etc

benbuks:

wow nice 1 bro......the spirit is 1 dsame tin av bin workin on too really learnt from this ....m working & thinking of finding roots of integers quicker & also multiplying /diving >3 digit numbers without calculator , our brain is a potential calculator just that sometyms we feel lazy to put it to work , ..

i"ll also share my discovery later ...

.

cool

Oga, you over complicated this matter. This has been discovered since the time of Pythagoras.

If you need to square a 2-digit number ending in 5,

Multiply the first digit by itself, add 1, and then put 25 on the end (It is ALWAYS 25).

That is all!

Am familiar wit dis rite frm my nursery school. Thank 2 ma headmaster who never stops teaching math even in his grave.

Chylo: Oga, you over complicated this matter. This has been discovered since the time of Pythagoras.

If you need to square a 2-digit number ending in 5,

Multiply the first digit by itself, add 1, and then put 25 on the end (It is ALWAYS 25).

That is all!

You are trying to say the exact same thing as the OP only that you got it wrong with the "add 1". You need to "add the first digit" which is the exact same thing the OP said, and it's not as complicated as you make it sound.

Nice one. Dad told me this trick when I was 6. I still remember that.

shollymata:

You are trying to say the exact same thing as the OP only that you got it wrong with the "add 1". You need to "add the first digit" which is the exact same thing the OP said, and it's not as complicated as you make it sound.

What he means to say is that it is an old discovery.

how cute so now u can show your children the same trick

ijbeats: Nice one. Dad told me this trick when I was 6. I still remember that.

#following

it doesn't work for 13

Zedric: Nyc 1 op. Biko, help me find square of 23.12 with ur method. I dey wait

winch

Wow! Educative.

Adesam01: Busted!

Your Example...
find the square of 25
again: 2*2+2 =6
then 5*5 again = 25
join them together = 625

My Problem!
find the square of 11
again: 1*1+1 =2
then 1*1 again = 1
join them together = 21 (Actual answer --> 11^2 = 121)


Same for 12 - 14

I conclude that to every rule or scientific formulation, there is a loop hole

op said number must end with 5 !!!

Well done OP . However you could have made it a bit simpler by multiplying the digit(s) before 5 by its next consecutive number .


Like this -


15 x 15 = (1 x 2 )25

which is 225


25 x 25 = (2 x 3 )25

which is 625


35 x 35 = (3 x 4)25

which is 1225


75 x 75 = (7 x 8 )25

which is 5625

115 x 115 = (11 x 12)25

which is 13225


and so on

In as much as I hate to rain on the OP's parade, I must state here that this isn't a new 'discovery' or his 'discovery' as he would like to think. This method has been around for ages and it originated from the indian punjab regions. It is a part of a series of fast methods for solving maths questions and its popularly known as VEDIC mathematics (Fast math). You can google it for more clarifications

truvine:

op said number must end with 5 !!!

My error!

olamikhail:



squares of numbers between 10 n 100 that end in 5.(the formulae doesnt work for 5;i stand to b corrected)

Or

squares of 2 digit numbers that end in 5..

Op take note 2.

All in all ,its a 9c post.kip it up.

is 5 between 10-100?

Great one Op.

People should read the disclaimer before faulting his discovery abegi.

Lemme share this; to get d square root of a number that has no perfect root, e.g 15. I know d nearest perfect roots are 9 n 16. Root of 9 is 3. Diff btw 9 n 16 is 7. Diff btw 9 n 15 is 6. Therefore, d root is
3. (6/7)

Nice bro

Shortcuts in multiplying / dividing any integer by 5 &
it exponential multiples
for '5'
the product of 5 & any integer , put a zero in front of
the number then divide by 2
e.g
5x1 =10/2 =5
5x2=20/2 =10
5x3=30/2=15
5x4=40/2=20
.
.
.& so on
thus
=> 5xn =n'0' /2
for '25'
to multiply 25 by any integer , put TWO zeros in
front of the number , then divide by 2
e.g
25 x5 =500/4 =125
25x8=800/4=200
25x 12=1200/4=300
& so on
thus
=> 25xn = n'00' /4
for '125'
put THREE zeros then divide by 8
e.g 125 x2=2000/8 =250
125x3=3000/8 = 375
125 x 4=4000/8=500
... & so on
hence , i can conclude that
5^(a) x n =( n 'a zeros' )/2^a ........for all a ,n € Z

Now the division
for '5'
to take the ratio of an integer with 5 , given the
integer is greater , multiply the integer by 2 ( i.e 2n)
then divide by 10 (i.e 2n/10)
e.g
10/5 =2x10/10 = 2
15/5 = (2x15)/10 =30/10 =3
20/5 =(2x20)/10 =40/10=4
.
.
.
& so on
=n/5 = 2n/10 ...(for n>5)
for
'25'
multiply by 4 then divide by 100
e.g
50/25 = (4x50)/100 =2
100/25 =(4x100)/100=4
150/25 =(4x150)/100 = 6
.
.
. & so on
=> n/25 =4n/100 ........(for n>25)
similarly , for '125'
*8 /1000
e.g
625/125 = (8x625)/1000 =5
1250/125 = (8x1250)/1000=10
2500/125= (8x2500)/1000 =20
.
.
.
& so on
thus => n/125 = 8n/1000
i could also conclude that
n/5^a = (2^a x n ) / 1'a zeros ' .......(for all a, n € Z)
that's it for now , more to come
feel free to make your contributions

to multiply by 6
6xn = n '0'/2 + n
e.g 6x1=10/2 + 1 = 5+1=6
6x2=20/2 =10 +2 =12
6x3=30/2 =15 +3 =18
6x4=40/2 =20 + 4 =24
& so on
try this in 5sec
6x666=?

LAWS in general, nay Mathematical "LAWS" are UNIVERSAL with no "iffs" and ""buts".

For example, Newton Laws of motion (Every 'body' continues in a state of rest or uniform motion unless and impressed force acts on it / Action and Reaction are equal and Opposite, e.t.c) have no "iffs" or "buts" and will ALWAYS apply to every object on this face of the erath which is where the law was written for.

Your so called "formula" is just an attempt to shorten a calculation, nice attempt though but you are very far from having "A LAW/FORMULA" because your "trick" is only valuable or valid ..for CERTAIN NUMBERS

I do applaud your effort though.....
You need to work more on it who knows, you might be Nigeria's Einstein!

Lastpage!

BTW: You need something that applies to ALL NUMBERS