Kinda awkward title.

It's the pattern in decimal.

Shall we?

Let's take a look at this multiplication table

5

x

5

=

25

15

x

15

=

225

25

x

25

=

625

35

x

35

=

1225

45

x

45

=

2025

55

x

55

=

3025

65

x

65

=

4225

75

x

75

=

5625

85

x

85

=

7225

95

x

95

=

9025

105

x

105

=

11025

115

x

115

=

13225

...

1005

x

1005

=

1010025

...

5005

x

5005

=

25050025

...

et cetera

The 25

As you can see above, any number which ends with 5 if __squared__will have 25 as the "tail".

The yellow colored numero

I put yellow color for the "opening" number sequence because it's interesting.Let's see the

**line.**

`15 x 15`

If we isolate the first number, which is 1, and we do this1 x (1 + 1) =2

It yields 2, the opening of 15² result,225.

Another example, the

**or**

`45 x 45`

**.**

`45²`

We take the first number, which is 4, and we do the same technique4 x (4 + 1) =20

20 is the starting sequence (before the ending, 25) of the result,2025.

Last example, the

**or**

`5005 x 5005`

**.**

`5005²`

500 x (500 + 1) =, which is also the first sequence of the result,25050025050025.

In conclusion

For speed "squaring" a number which ends with 5, we can do this:
- The first number sequence:
the_first_number_before_5 x (the_first_number_before_5 + 1)
- The last (or the tail) will always be
**25**. We can just attach it to the first number sequence.

Usage

Let's find the result of 75².
- The first number sequence
- The tail is
**25**.

Number with decimal point?

Let's use the 75 sequence, but it's 0.75 now.Calculate the result of 0.75² !

- Take the 7. So the first number sequence
- The tail is always
**25**. - Last step is to count how many digits after the decimal point.

We have 0.75 (two digits) squared. Then we'll have 4 digits behind the thingy.

It sort of happens too in other number systems. But then, my brain hurts. So, thanks for visiting.

Hi Johan,

ReplyDeleteThat's some very keen thinking and a neat trick you have there! There's also another pattern: each number that we're squaring is an odd multiple of 5 (that is, 5*1, 5*3, 5*5,...), which we can write as (5*(2n-1))^2. Therefore, the prefactor must depend on which n we choose. If we pair up the first few n's with the prefactors, we find 1:0, 2:2, 3:6, 4:12, 5:20, etc. The relationship is that each n is multiplied by (n-1) (ex: 1*(1-1) = 0, 2*(2-1) = 2, 3*(3-1) = 6). Of course, the prefactors are multiples of 100, so we multiply by 100. Altogether we find that (5*(2n-1))^2 = 100*n*(n-1) + 25, which we could simplify to (2n-1)^2 = 4*n*(n-1) + 1.

Best,

Another Math Wizard

Wow that's really structured. I dig it!

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