# Lesson: Squaring Numbers Ending in 5

## Comment on Squaring Numbers Ending in 5

### Does this work if the base is

Does this work if the base is negative? ### Yes. Keep in mind that x² = (

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

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? ### The last (units) digit is

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

Does this technique is also useful for squares higher than 2 (i.e. 25^3) ### Good question.

Good question.

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

### I think this technique can

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. ### That's correct; the technique

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

### Majestic!

Majestic! Indeed :-)

### Amazing!

Amazing! ### Hi Brent,

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

### Hi Brent,

Hi Brent,

Are there more techniques like this , which can help us to save time on gmat? ### There are tons of time-saving

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

Why does the technique work?

Once I know it, it should remind me ### Here's why the technique

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?