### Math trick and its proof: square a number ending in 5

I will be hosting the blog carnival Math Teachers at Play next week. (You can send in submissions here.)

One submission I got about various multiplication tricks or shortcuts got me inspired to write a proof of the particular trick.

You could definitely use this in algebra class. First explain the shortcut or trick itself. Then ask students to prove it, or to explain WHY it works, using algebra.

You could also explain this to younger students as an additional "neat trick" and let them explore and play with it.

THE "TRICK"

If a number ends in 5, then its square can be calculated using this "trick" (I like to call it a shortcut because there's nothing magic about it):

Let's say we have 75 × 75 => Go 7 × 8 = 56. Then tag 25 (or 5 × 5) into that. You get 5625.

Let's say we have 35 × 35 => Go 3 × 4 = 12. Then tag 25 into that. You get 1225.

Let's say we have 115 × 115 => Go 11 × 12 = 132. Then tag 25 into that. You get 13,225.

Let's say we have 245 × 245 => Go 24 × 25 = 600. Then tag 25 into that. You get 60,025.

So, you simply take the digit or digits in front of the 5 and consider those as a number in itself. Multiply that by the next number. Then "tag" 25 to the answer you got in the previous step.

PROOF

Any whole number that ends in five is of the form A + 5, and A is a multiple of 10. Since A is a multiple of 10, we can write A = 10b, where b is now some whole number. So, our number is of the form 10b + 5. Now, let's square it.

(10b + 5)(10b + 5) and we use the distributive property to multiply this out.

(10b + 5)(10b + 5) = 100b

Now, notice those 100b's there. We can gather that as a common factor for the first two terms:

= 100 b (b + 1) + 25

This is now essentially in the form that the trick is using.

The trick says to take b, or the number formed by the digits in front of the 5. That corresponds exactly with our b! (For example, in number 645, b is 64. Our number 645 is 10b + 5, or 640 + 5.)

So we take b, multiply it by (b + 1) which is the next number, and also by 100, and lastly add 25.

Now, b × (b+1) is the part of the trick where you multiply the digits in front of the 5 by the next number. To "tag" 25 to those digits means you add 25 only after having multiplied the number by 100 so that it would end in "00". Once it ends in "00" you can add 25 (or any two-digit number) and it is the same as "tagging" 25 to the digits without the "00".

I hope this is clear enough.

By the way, I do not feel all students must learn this shortcut for finding the square of numbers ending in 5. It is a nice addition to one's mathematical knowledge, but not any necessity. However, it is useful as an algebra problem, and of course has been used as such over the course of centuries, I'm sure.

One submission I got about various multiplication tricks or shortcuts got me inspired to write a proof of the particular trick.

You could definitely use this in algebra class. First explain the shortcut or trick itself. Then ask students to prove it, or to explain WHY it works, using algebra.

You could also explain this to younger students as an additional "neat trick" and let them explore and play with it.

THE "TRICK"

If a number ends in 5, then its square can be calculated using this "trick" (I like to call it a shortcut because there's nothing magic about it):

Let's say we have 75 × 75 => Go 7 × 8 = 56. Then tag 25 (or 5 × 5) into that. You get 5625.

Let's say we have 35 × 35 => Go 3 × 4 = 12. Then tag 25 into that. You get 1225.

Let's say we have 115 × 115 => Go 11 × 12 = 132. Then tag 25 into that. You get 13,225.

Let's say we have 245 × 245 => Go 24 × 25 = 600. Then tag 25 into that. You get 60,025.

So, you simply take the digit or digits in front of the 5 and consider those as a number in itself. Multiply that by the next number. Then "tag" 25 to the answer you got in the previous step.

PROOF

Any whole number that ends in five is of the form A + 5, and A is a multiple of 10. Since A is a multiple of 10, we can write A = 10b, where b is now some whole number. So, our number is of the form 10b + 5. Now, let's square it.

(10b + 5)(10b + 5) and we use the distributive property to multiply this out.

(10b + 5)(10b + 5) = 100b

^{2}+ 50b + 50b + 25 = 100b^{2}+ 100b + 25Now, notice those 100b's there. We can gather that as a common factor for the first two terms:

= 100 b (b + 1) + 25

This is now essentially in the form that the trick is using.

The trick says to take b, or the number formed by the digits in front of the 5. That corresponds exactly with our b! (For example, in number 645, b is 64. Our number 645 is 10b + 5, or 640 + 5.)

So we take b, multiply it by (b + 1) which is the next number, and also by 100, and lastly add 25.

Now, b × (b+1) is the part of the trick where you multiply the digits in front of the 5 by the next number. To "tag" 25 to those digits means you add 25 only after having multiplied the number by 100 so that it would end in "00". Once it ends in "00" you can add 25 (or any two-digit number) and it is the same as "tagging" 25 to the digits without the "00".

I hope this is clear enough.

By the way, I do not feel all students must learn this shortcut for finding the square of numbers ending in 5. It is a nice addition to one's mathematical knowledge, but not any necessity. However, it is useful as an algebra problem, and of course has been used as such over the course of centuries, I'm sure.