Fun With Math: The Squaring Rule
In retrospect, if the 3 Rule is modified slightly, it can be applied to any number.
Check this out:
In the 3 Rule, we had:
n^ = (n-3)^ + ((n-3)x6) + 9
Now let's take a looke at those parts a little more closely.
Why 3, 6, and 9 in the equations? Because those are multiples of 3. You have 3x1 (3), 3x2 (6), and 3^ (9). Why didn't I say 3x3? Because it does not hold true with the other numbers (1, 2, 4, 5, 6, 7, 8, and 9)
In the first part, we have (n-3)
This is because we were using a number ending in "3". If it had been a number ending in "5", it would have been (n-5). This is because in the first step, you multiply the last digit by 1.
In the second part, we have ((n-3)x6)
Again, this is due to having a number ending in "3". If it had been a number ending in "4", it would have been ((n-4)x8). This is because to get the multiplier (8), you multiply the last digit by 2.
In the last part, we have 9
Once again, it is the result of having a number ending in "3". If it had been a number ending in 7, it would have been 49. This is because to get this last part, you square the last digit.
So, for numbers ending in "1", the formula will be:
n^ = (n-1)^ + ((n-1)x2) + 1
So, for numbers ending in "2", the formula will be:
n^ = (n-2)^ + ((n-2)x4) + 4
So, for numbers ending in "3", the formula will be:
n^ = (n-3)^ + ((n-3)x6) + 9
So, for numbers ending in "4", the formula will be:
n^ = (n-4)^ + ((n-4)x8) + 16
So, for numbers ending in "5", the formula will be:
n^ = (n-5)^ + ((n-5)x10) + 25
So, for numbers ending in "6", the formula will be:
n^ = (n-6)^ + ((n-6)x12) + 36
So, for numbers ending in "7", the formula will be:
n^ = (n-7)^ + ((n-7)x14) + 49
So, for numbers ending in "8", the formula will be:
n^ = (n-8)^ + ((n-8)x16) + 64
So, for numbers ending in "9", the formula will be:
n^ = (n-9)^ + ((n-9)x18) + 81
For example:
36 x 36 =
36^ = 30^ + (30 x 12) + 36
36^ = 900 + 360 + 36
36^ = 1296
21 x 21 =
21^ = 20^ + (20 x 2) + 1
21^ = 400 + 40 + 1
21^ = 441
69 x 69 =
69^ = 60^ + (60 x 18) + 81
69^ = 3600 + 1080 + 81
69^ = 4761
Ain't THAT neat?
Check this out:
In the 3 Rule, we had:
n^ = (n-3)^ + ((n-3)x6) + 9
Now let's take a looke at those parts a little more closely.
Why 3, 6, and 9 in the equations? Because those are multiples of 3. You have 3x1 (3), 3x2 (6), and 3^ (9). Why didn't I say 3x3? Because it does not hold true with the other numbers (1, 2, 4, 5, 6, 7, 8, and 9)
In the first part, we have (n-3)
This is because we were using a number ending in "3". If it had been a number ending in "5", it would have been (n-5). This is because in the first step, you multiply the last digit by 1.
In the second part, we have ((n-3)x6)
Again, this is due to having a number ending in "3". If it had been a number ending in "4", it would have been ((n-4)x8). This is because to get the multiplier (8), you multiply the last digit by 2.
In the last part, we have 9
Once again, it is the result of having a number ending in "3". If it had been a number ending in 7, it would have been 49. This is because to get this last part, you square the last digit.
So, for numbers ending in "1", the formula will be:
n^ = (n-1)^ + ((n-1)x2) + 1
So, for numbers ending in "2", the formula will be:
n^ = (n-2)^ + ((n-2)x4) + 4
So, for numbers ending in "3", the formula will be:
n^ = (n-3)^ + ((n-3)x6) + 9
So, for numbers ending in "4", the formula will be:
n^ = (n-4)^ + ((n-4)x8) + 16
So, for numbers ending in "5", the formula will be:
n^ = (n-5)^ + ((n-5)x10) + 25
So, for numbers ending in "6", the formula will be:
n^ = (n-6)^ + ((n-6)x12) + 36
So, for numbers ending in "7", the formula will be:
n^ = (n-7)^ + ((n-7)x14) + 49
So, for numbers ending in "8", the formula will be:
n^ = (n-8)^ + ((n-8)x16) + 64
So, for numbers ending in "9", the formula will be:
n^ = (n-9)^ + ((n-9)x18) + 81
For example:
36 x 36 =
36^ = 30^ + (30 x 12) + 36
36^ = 900 + 360 + 36
36^ = 1296
21 x 21 =
21^ = 20^ + (20 x 2) + 1
21^ = 400 + 40 + 1
21^ = 441
69 x 69 =
69^ = 60^ + (60 x 18) + 81
69^ = 3600 + 1080 + 81
69^ = 4761
Ain't THAT neat?