Lesson: Squaring Numbers Ending in 5

Comment on Squaring Numbers Ending in 5

Does this work if the base is negative?
gmat-admin's picture

Yes. Keep in mind that x² = (-x)².

For example, 3² = 9, and (-3)² = 9.

Likewise, 35² = (-35)² = 1225, and 65² = (-65)² = 4225

Sir, How can we Find Last 2 Digit and Last unit of a number whenever it is multiplied , Divided, Subtracted or Added in to any other number?
gmat-admin's picture

The last (units) digit is pretty straightforward, but in most cases, the last 2 digits is outside the scope of the GMAT.

Does this technique is also useful for squares higher than 2 (i.e. 25^3)
gmat-admin's picture

Good question.

The answer is no; the technique can't be extended to powers other than 2.

I think this technique can only be used for numbers ending in 5? I tried with 6, for example- 76 and it didn't work.
gmat-admin's picture

That's correct; the technique works on for numbers ending in 5 (thus the title :-)

gmat-admin's picture

Indeed :-)

Yulia's picture

Hi Brent,

I have a question regarding this problem https://gmatclub.com/forum/6-226675.html

Automatically I looked at cyclicity of numbers in which 4 has cyclicity of 2 and 6 has cyclicity of 1, meaning that 6⁴ - 4⁴ will end up with unit digits 6-6= 0 so the result must be ending with zero. However, we have several answer choices ending with zero. So now since the exponent is 4 which means that 6⁴ has 4 digits and 4⁴ has 4 digits then the answer is E.

Is my approach is valid?
gmat-admin's picture

Question link: https://gmatclub.com/forum/6-226675.html

That's a perfectly valid approach. Nice work!!

Hi Brent,

Are there more techniques like this , which can help us to save time on gmat?

gmat-admin's picture

There are tons of time-saving techniques sprinkled throughout the course.

May I ask why?

Why does the technique work?

Once I know it, it should remind me
gmat-admin's picture

Here's why the technique works:

First recognize that any integer with units digit 5 can be expressed in the form 10k + 5
Some examples: 35 = (10)(3) + 5, 85 = (10)(8) + 5, 115 = (10)(11) + 5

So, squaring an integer with units digit 5 is the same as squaring 10k + 5.

Now let's see what happens when we square 10k + 5.
(10k + 5)² = 100k² + 100k + 25
= (100)(k)(k + 1) + 25

As we can see, (k)(k + 1) represents the product of the value that comes before units digit 5 and 1 greater than that value (just like in the video)
We then multiply that product by 100, and add 25 to the end (just like in the video).

Does that help?

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