- #1
deep838
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Ok, I don't know if this method is already known or not, but I found this all by myself after some observations... so here it is...
Suppose we want to square a number, say 67.
What i have found is this:
1. First get 652 which is [6*7][5*5] = 4225
2. Forget the digit in the unit's place, ie, 5.
3. Take the remaining digits. In this case it's 422
4. This is my observation. For every 5, add 1 as many times as the original number is greater than the nearest 5 multiple. Since 65 is 5*13, we add 13 (67-65)=2 times.ie, we add 13*2=26 to the 'remaining number'. That is we do 422+26=448
5. Put the last digit of the square of the digit in the unit's place of the original number, after the sum obtained ie, put the last digit of 7*7 ie, 9, after 448, getting 4489 as the required square!
Another example:
Finding 822...
1. 802=6400
2. 80=5*16
3. 640+16*(82-80)=640+32=672
4. required square is 6724
The nearest multiple of 5 has to be taken. For 33, we must take 35 and then subtract instead of adding.
1. 352=1225
2. 35=5*7
3. 122 + 7*(33-35) = 122 - 14 = 108
4. required square is 1089
So what do you people think of this?
This is equally applicable for 3 digit numbers, though then the initial squaring of 5 multiples become another thing.
But we can still apply this.
2022 is:
1. 2002=40000
2. 200=5*40
3. 4000+40*2=4000+80=4080
4. required square is 40804.
Suppose we want to square a number, say 67.
What i have found is this:
1. First get 652 which is [6*7][5*5] = 4225
2. Forget the digit in the unit's place, ie, 5.
3. Take the remaining digits. In this case it's 422
4. This is my observation. For every 5, add 1 as many times as the original number is greater than the nearest 5 multiple. Since 65 is 5*13, we add 13 (67-65)=2 times.ie, we add 13*2=26 to the 'remaining number'. That is we do 422+26=448
5. Put the last digit of the square of the digit in the unit's place of the original number, after the sum obtained ie, put the last digit of 7*7 ie, 9, after 448, getting 4489 as the required square!
Another example:
Finding 822...
1. 802=6400
2. 80=5*16
3. 640+16*(82-80)=640+32=672
4. required square is 6724
The nearest multiple of 5 has to be taken. For 33, we must take 35 and then subtract instead of adding.
1. 352=1225
2. 35=5*7
3. 122 + 7*(33-35) = 122 - 14 = 108
4. required square is 1089
So what do you people think of this?
This is equally applicable for 3 digit numbers, though then the initial squaring of 5 multiples become another thing.
But we can still apply this.
2022 is:
1. 2002=40000
2. 200=5*40
3. 4000+40*2=4000+80=4080
4. required square is 40804.
Last edited: