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Some of you geeks probably have a slide-rule in your head — there I go dating myself again (and we all know that dating yourself isn’t sexy), I should have said “math processor” or “calculator”. Anyway, I am personally much better than average at math, but I still can’t do some of the bigger problems in my head.

Via Mercola, I found this list of 10 Easy Arithmetic Tricks to save you the trouble of getting out a calculator for solving some math problems.

The “11 Times Trick” is one I hadn’t thought of before, but it makes perfect sense. To multiply any two-digit number by 11, you add the digits together and put the result between them, adding any excess digit to the first digit. It makes sense because multiplying it out long-hand you get (for any two digit number *m**n*):

* m n*X 1 1

*———-*

m n

m n

*m n*

————

*m*(

*m*+

*n*)

*n*

Some of the tricks are a bit obvious, like the multiply/divide by 5 and how to compute a 15% tip (a lot of places are starting to expect 20% now anyway). But all the tricks are good exercises for understanding how numbers work.

Here are a few more tricks that I learned somewhere along the way for determining if a decimal integer is a multiple of:

- Duh. Of course it is.
- Is the last digit even? Another easy one.
- Add up the digits in the number. If the result is a multiple of 3, then so is the number. If the sum of the digits is too big for you to know whether it’s a multiple of 3 or not, recurse. Add up
*its*digits and see if*that’s*a multiple of 3. - Take the number formed from the last two digits. If that’s a multiple of 4, so is the number. Why? Because 100 is a multiple of 4.
- Is the last digit a 0 or a 5? Too easy.
- Is the number a multiple of 2
*and*3? - Double the last digit and subtract it from the number formed by the remaining digits. If that result is a multiple of 7, so is the original number. If you can’t tell, recurse. For instance, take 357. Double 7 to get 14, subtract that from 35 and you get 21. Since 21 is a multiple of 7, so is 357.
- Take the number formed from the last three digits. If that’s a multiple of 8, so is the number — because 1000 is a multiple of 8.
- Add up the digits in the number. If the result is a multiple of 9, then so is the number. Again, you can recurse if you’re not sure. It’s no accident that this rule for 9 is the same as the rule for 3.
- Does it end with a 0? Now we’re back in elementary school.
- Add up all the odd digits to get one number, then add up all the even digits to get a second number. If the difference between them is a multiple of 11 (zero included), then so is the number. If you think about it, this is really the “11 Times Trick” reversed. Let’s take 26719 as an example. 2 + 7 + 9 = 18, 6 + 1 = 7, 18 – 7 = 11, so 26719 is a multiple of 11.
- Is the number a multiple of 3
*and*4?

Looking at this list, the relationship between 3, 6, and 9 is obvious — as is the relationship between 4 and 8. The unique characteristics of 7 and 11 are intriguing, don’t you think?

What other arithmetic tricks do you know?

A quick way to calculate the square of a number that ends with a 5.

Example: 65 x 65

1. Add 1 to the first 6, equals 7.

2. Multiply this 7 with the second 6, equals 42.

3. The last 2 digits will always be 25.

4. So the answer is 4225.

Thanks, Awangku. That’s actually #2 in the post linked to above.

Ah yess… my mistake. Stupidly, I did not click the link to the 10 tricks. Okay, I’m embarrassed now. Hahahahaha…

Not a problem. Thanks for reading and commenting!

A quick way to calculate the square of a number that ends with a 5.

Example: 65 x 65

1. Add 1 to the first 6, equals 7.

2. Multiply this 7 with the second 6, equals 42.

3. The last 2 digits will always be 25.

4. So the answer is 4225.

Thanks, Awangku. That's actually #2 in the post linked to above.

Ah yess… my mistake. Stupidly, I did not click the link to the 10 tricks. Okay, I'm embarrassed now. Hahahahaha…

Not a problem. Thanks for reading and commenting!

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Pingback: Neat math tricks: Integer divisibility | An Obvious Hack

Hi Geeks,

I have the tests for divisibility by all primes below 50 up on my website.

Go look at http://www.savory.de/maths1.htm

Hi Geeks,

I have the tests for divisibility by all primes below 50 up on my website.

Go look at http://www.savory.de/maths1.htm

And I can also show you how to take cube roots in your head too. See

http://home.egge.net/~savory//blog_feb_08.htm#200…

Good stuff, Stu! Thanks.

And I can also show you how to take cube roots in your head too. See

http://home.egge.net/~savory//blog_feb_08.htm#200…

Good stuff, Stu! Thanks.

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Most of them know it b4:)try t includ someth new..thankyou

Anu

Are you the same anu who did a course in AMES training centre back in 2000 in Auckland?.IF you are then reply back – naren_2010@hotmail.com

Most of them know it b4:)try t includ someth new..thankyou

35

x 67

245

210

2345

35

x 67

245

210

2345

Sorry, it turned out different than I typed in. It uses a variant of the first one, but it's actually able to do it with more that just x11.

35 x 7 = 245, and 35 x 60 = 2100, and 2100 + 245 = 2345. Mayb a big number isn't practical for an example, but 25 x 15, or smaller number with lower digits.

Sorry, it turned out different than I typed in. It uses a variant of the first one, but it’s actually able to do it with more that just x11.

35 x 7 = 245, and 35 x 60 = 2100, and 2100 + 245 = 2345. Mayb a big number isn’t practical for an example, but 25 x 15, or smaller number with lower digits.

Actually, there really are only three divisibility tricks. Interestingly, they work in any base, not just base 10.

For a given base B:

1) n B – 1: Take the last digit, multiply by n, and add it to remaining digits. Repeat until you can tell whether the number is divisible by n B – 1

2) B^n: If the last n digits are divisible by B^n, so is the number

3) n B + 1: Take the last digit, multiply by n, and subtract it from the remaining digits. Repeat until you can tell

In each case, if the relevant number isn't prime, the rule also applies to all of its factors. For base 10:

2, 4, 5, 8 are all cases of rule 2. n=1 for 2 and 5, 2 for 4, 3 for 8

3 and 9 are cases of rule 1 with n=1

7 and 11 (and 3) are cases of rule 3. For 11, n=1. For 7, n=2 (2 * 10 + 1 = 21, factors are 3 and 7)

That's fascinating, Ari.

Actually, there really are only three divisibility tricks. Interestingly, they work in any base, not just base 10.

For a given base B:

1) n B – 1: Take the last digit, multiply by n, and add it to remaining digits. Repeat until you can tell whether the number is divisible by n B – 1

2) B^n: If the last n digits are divisible by B^n, so is the number

3) n B + 1: Take the last digit, multiply by n, and subtract it from the remaining digits. Repeat until you can tell

In each case, if the relevant number isn’t prime, the rule also applies to all of its factors. For base 10:

2, 4, 5, 8 are all cases of rule 2. n=1 for 2 and 5, 2 for 4, 3 for 8

3 and 9 are cases of rule 1 with n=1

7 and 11 (and 3) are cases of rule 3. For 11, n=1. For 7, n=2 (2 * 10 + 1 = 21, factors are 3 and 7)

That’s fascinating, Ari.