Mental Math "tricks" you learned in elementary school

[Godhumor]

Call this a sister thread to the mnemonic one floating around now.

Just curious as to any little shortcuts your teachers might have taught you as a youngster that you still find yourself using. For me, there are a few (I think they all came from the same teacher, too.):

Check if a number is divisible by 3:
Add the digits of the number together. Is that sum divisible by 3? If so, then the original number is as well. For example, 123 = 1+2+3 = 6. 6/3 is 2, so 123 is divisible by 3.

Check if a number is divisible by 9:
Same method as the one listed above, but divide the sum by 9 instead of 3. For example, 954 = 9+5+4 =18. 18/9 = 2. 954 is divisible by 9.

Squaring any 2 digit number ending in 5:
Multiple the tens digit by the tens digit plus 1. Stick the answer on the front of 25. For exampled 25*25 = 2*(2+1) = 625. 55*55 = 5*(5+1) = 3025

Multiplying a 2 digit number by 11:
Add the digits of the number together. Put the sum of that between the two digits to get your answer. For example, 54*11 = 594 (5(5+4)4). If the sum is greater than nine, don't forget to carry. 99*11 = 1089 (9(9+9)9 = 9(18)9, so carry the 1. (9+1)89 = 1089.)

Any others?

[Shell Beau]

1. I remember in first grade learning "greater than, less than" and

the teacher said this > was an alligator mouth and it wanted to eat the biggest number. Silly!

[Godhumor]

2. Oh yeah, I remember that one, too. And the "Left - Right" check

Make an L with your index finger and thumb on both hands. Whichever one makes the correct L is your left hand.

[pokerfan]

5. There's a right hand rule for handling three dimensional vectors

When our electromagnetics prof was teaching us this trick, my buddy couldn't get it to work right. Turned out he was using his left hand. :rofl:

Everyone had a good laugh at that.

[WolverineDG]

9. The arrow always points to the smaller number

not as cute as the alligator mouth, but it works. :)

dg

[Kali]

3. multiplying by 9

take the easy "times ten" answer and subtract the number
ex: 7x9 >>> think: 7x10 is 70, minus 7 is 63

[pokerfan]

17. use your fingers to multiply by nine

I know using fingers is frowned upon but it illustrates a useful property of the number 9.

Assign the integers 1 through 10 to your fingers, left to right. Then to multiply any number by nine just fold down the finger corresponding to that number. You will be left with two groups of numbers. The number of fingers in each of those groups represent the answer and of course, they will always add up to nine.

[pokerfan]

4. a couple

I suppose everyone knows these....

A) Dividing by 5 is the same as multiplying by 2 and moving the decimal point:
612/5 => 612*2=1224 => 122.4
It's easier for me to double a number in my head than do long division.

B) Summing a series of numbers: take them in pairs. For example to sum all the integers from 1 to 100, pair 1 with 100, 2 with 99, 3 with 98, etc. Each pair sums to 101 and there are fifty such pairs (half of 100). Then multiply 101 by 50 to get 5050.

C) Rule of 72: Years To Double = 72/Interest Rate

D) Decimal exponents of 2:
2^10 ~ 1K
2^20 ~ 1M
2^30 ~ 1B

[struggle4progress]

7. There's a generalization of B that has been known for thousands of years:

to sum an arithmetic series, multiply the number of terms by the average of the first and last terms

"Arithmetic series" refers to a list of numbers with constant successive difference: for example, 124, 124 + 7 = 131, 131 + 7 = 138, 138 + 7 = 145, ...

As an example, what is the sum of the numbers from 34 to 78? The successive difference for the numbers 34, 35, 36, ... , 78 is always 1; there are 78 - 34 + 1 = 45 of these numbers; the average of the first and last is (34 + 78)/2 = 56, so the sum is 34 + 35 + 36 + ... + 78 = 45*56 = 2000 + 240 + 250 + 340 = 2490 + 30 = 2520

As another example, what is the sum of the odd numbers from 87 to 311? The successive difference for the numbers 87, 89, 91, ... , 311 is always 2; 87 is the 44th odd number and 311 is the 156th, so there are 156 - 44 + 1 = 113 of these numbers; the average of the first and last is (87 + 311)/2 = 199, so the sum is 87 + 89 + 91 + ... + 311 = 199*113 = 200*113 - 113 = 22600 - 113 = 22487

Another example: what is the sum of the numbers 117 + 7*n for n = 1, 2, ..., 10? The average of the first and last of these ten numbers is (117 + 7 + 117 + 70)/2 = 117 + 38.5 = 155.5, so the sum is 124 + 131 + 138 + ... + 187 = 155.5*10 = 1555

The rule has some interesting consequences, also known from ancient times:

For example, what is the sum of the first n odd numbers? The kth odd number is 2*k - 1. So the average of the first and the nth is (1 + 2*n - 1)/2 = n. The sum of the first n odd numbers is therefore n*n, a perfect square

One can use the rule to find the difference of squares. For m > n, what is the difference between the mth and nth squares? These squares are (respectively) the sum of the first m odd numbers and the sum of the first n odd numbers, so the difference is the sum of the odd numbers from the (n + 1)st to the mth. But this sum is easily found: it has m - (n + 1) + 1 = m - n terms; the first term is 2*(n + 1 - 1) + 1 = 2n + 1, the last is 2(m - 1) + 1 = 2m - 1, and their average is (2n + 1 + 2m - 1) = m + n. Thus the difference between the mth and nth squares is (m - n)*(m + n)

As a result, one can use the rule to find Pythagorean triples. For example, pick an odd square p*p and factor it any way you like: say, p*p = r*s where r > s. Next find m and n such that m + n = r and m - n = s: obviously, m = (r + s)/2 and n = (r - s)/2. Since p is odd, both r and s are odd, so m and n are both whole numbers. Moreover, p*p = r*s = (m + n)*(m - n) is the difference between the mth square and the nth square: in other words, m*m = n*n + p*p. As an illustration, take p = 5, r = 25, s = 1; then m = 13 and n = 12; so 5*5 + 12*12 = 13*13. As another illustration, take p = 21, r = 49, s = 9; then m = 29 and n = 20; so 21*21 + 20*20 = 29*29. Another illustration: p = 21, r = 63, s = 7 gives m = 35, n = 28 so 21*21 + 28*28 = 35*35. It's actually possible to find the general rule for constructing Pythagorean triples by examining thse ideas more carefully

<edit: typos>

[crimsonblue]

6. I learned the trick on the calculator

that made it say BOOBIES.

[conscious evolution]

11. 707!!!

[conscious evolution]

12. My sister was taught

how to do a abacus type calculator with her fingers.She still uses it and can do simple equations as fast as someone using regular calculator.

A trick I remember being taught was for checking multiples of 9.Add the answer numbers together and it almost always equls 9.
9x9=81 8+1=9
9x127= 1143 1+1+4+3=9
Sometimes the answer adds up to more than 9.In that case add the numbers from that answer together and it will the add up to 9.
9x3267=29403 2+9+4+0+3=18 1+8=9
I have no idea if this is a law of mathematics that always holds true.But it always seems to work for me.

[applegrove]

8. To find out your average in your report card take a base number say 75.

If your first mark is 80 add +5. If your next number is 78 add +3 so you are not at +8. If your next mark is 65 then minus - 10 so you are at -2. If this is the last of your marks then -2 from your base number of 75 and the average of all your marks is 75 - 2 = 73. So 73 is the average mark.

[Fire Walk With Me]

10. It is far easier to add, even large numbers, by going left to right

and not right to left, as we are taught.

134+
213

Add the hundreds first, then the tens, or of course, the remainder if it's simple enough.

[Dr. Strange]

13. Check if a number is divisible by 11:

Similar to the one for 9 but alternate between addition and subtraction.

To test if 759154 is divisible by 11, look at 7 - 5 + 9 - 1 + 5 - 4 = 11; since 11 is divisible by 11, so is 759154.

Incidentally, on the divisibility by 9: the sum will have the same remainder as the original number does after dividing by 9. So, for example, 5123 is not divisible by 9, since 5 + 1 + 2 + 3 = 11 is not divisible by 9. But, since 11 has a remainder of 2 after dividing by 9, so will 5123.

This is because the universe works on a base 10 number system--don't listen to BlooinBloo and Rabrrrrrrrrrrr. :party:

[struggle4progress]

16. Check whether a number is divisible by 7, 11, or 13:

(1) beginning at the left, first divide the number into 3-digit pieces
(2) add the pieces, alternating signs
(3) if the result is negative, add a convenient multiple of 1001 to make it positive, before going to step (4): for example, if the negative result has 3 or fewer digits, you can add 1001; if the negative result has four digits, you can add 10010; if the negative result has five digits, you can add 100100; if the negative result has six digits, you can add 1001000; & so on
(4) now that you have a positive result:
(4a) if it has more than three digits, you can (if you like) go back to step (1) & restart with the new (smaller) number; otherwise continue with step (4b)
(4b) divide your result by 7, 11, & 13; you will get the same remainders as the original number would have produced

Example. Is 1,234,567,890 divisible by 7, 11, or 13? Examine 890 - 567 + 234 - 1 = 556; since 556 leaves remainders of 3, 6, & 10 when divided by 7, 11, & 13, conclude 1,234,567,890 also leaves remainders of 3, 6, & 10 when divided by 7, 11, & 13, so 1,234,567,890 is not divisible by any of 7, 11, or 13

Example. Is 112,294 divisible by 7, 11, or 13? Examine 294 - 112 = 182; since 182 leaves remainders of 0, 6, & 0 when divided by 7, 11, & 13, conclude 112,294 is divisible by 7 & 11 but leaves a remainder of 6 when divided by 13

Example. Is 11,395,059 divisible by 7, 11, or 13? Examine 59 - 395 + 11 = -325; since this is negative & has three digits, add 1001 to get 1001 - 325 = 676; since 676 leaves remainders of 4, 5, & 0 when divided by 7, 11, & 13, conclude 11,395,059 is divisible by 13 but leaves remainders of 4 & 5 when divided by 7 & 11

[graywarrior]

14. Sweet. I have to take the census test next week

:rofl:

[pokerfan]

15. big list of multiplication tricks

4: double twice

5: Multiply by 10 and divide by 2

6: Sometimes multiplying by 3 and then 2 is easy

8: double three times

9: Multiply by 10 and subtract the original number

12: Multiply by 10 and add twice the original number

13: Multiply by 3 and add 10 times original number

14: Multiply by 7 and then multiply by 2

15: Multiply by 10 and add 5 times the original number, as above

16: You can double four times, if you want to. Or you can multiply by 8 and then by 2

17: Multiply by 7 and add 10 times original number

18: Multiply by 20 and subtract twice the original number

19: Multiply by 20 and subtract the original number

24: Multiply by 8 and then multiply by 3

27: Multiply by 30 and subtract 3 times the original number

45: Multiply by 50 and subtract 5 times the original number

90: Multiply by 9 (as above) and put a zero on the right

98: Multiply by 100 and subtract twice the original number

99: Multiply by 100 and subtract the original number