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Math: Pattern of Squared Number Which Ends with 5

FINGERS
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 ...

25 as the Tail

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

Prefix — Yellow Colored Number

I put yellow color for the opening number sequence because it is interesting.

Let's see the 15 x 15 line.

If we isolate the first number, which is 1, and we do:

1 x (1 + 1) = 2

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

Another example 45 x 45 or 45²:

We take the first number, which is 4, and we do the same technique

4 x (4 + 1) = 20

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

Last example, the 5005 x 5005 or 5005².

500 x (500 + 1) = 250500

Which is also the first sequence of the result, 25050025.

Speed Squaring

  • From the examples above, we can generalize the pattern for the prefix (opening) number sequence as this:

    numberBefore5 x ( numberBefore5 + 1 )

  • The suffix (tail) will always be 25.

    We can attach it as the tail (suffix) of the result.

Usage

Let's find the result of 75².

  • The prefix sequence:

  • The suffix is 25.

Then the result of 75² is 5625.

Decimal Point

Let's use 0.75.

Calculate the result of 0.75²!

  • Take the 7. So the prefix number sequence:

  • The suffix is always 25.

  • Last step is to count how many digits after the decimal point.

    We have 0.75 (two digits) squared. Then we will have 4 digits behind the decimal point.

Result of 0.75² is 0.5625.

Thank you for visiting. I hope this bit is useful 👍

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  1. AnonymousJune 27, 2015 at 5:52 PM

    Hi Johan,

    That'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

    ReplyDelete

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