Quick maths tricks

[aanraag]

To multiply any number with 11, just add the first and last digit and the answer should be placed between the two digits which are being multiplied by 11.

eg 11 x 11= 121 (1+1=2 which is placed betwwen 11)
12 x 11= 132 (1+2=3 which is placed between 1 and 2)
13 x 11= 143 and so on.

[aanraag]

If a number is divisible by 3 then its also divisible by 9.
to chech if a number id divisible by 3, add all the digits and the anwer should be divisible by 3.
eg 153 add 1+5+3= 9 which can be divided by 3

[daveg]

To multiply any number with 11, just add the first and last digit and the answer should be placed between the two digits which are being multiplied by 11.

eg 11 x 11= 121 (1+1=2 which is placed betwwen 11)
12 x 11= 132 (1+2=3 which is placed between 1 and 2)
13 x 11= 143 and so on.

19 x 11 = 1109?

[daveg]

If a number is divisible by 3 then its also divisible by 9.

6 is divisible by 3. 6 is not divisible by 9.

[yoyo123]

exactly, a number which is divisible by 18,12, 9, 6 is also divisible by 3, it doesn't work the other way round

[aliportico]

To multiply any number with 11, just add the first and last digit and the answer should be placed between the two digits which are being multiplied by 11.

eg 11 x 11= 121 (1+1=2 which is placed betwwen 11)
12 x 11= 132 (1+2=3 which is placed between 1 and 2)
13 x 11= 143 and so on.

19 x 11 = 1109?

Kind of You have to carry - so 1+9 is 10 - 0 in the middle, carry the 1, and add that to the 1_9 to get 209

Works for 2 digit numbers.

Divisible by 4 - take the last two digits, halve that, if the result is even then the number is divisible by 4.

Divisible by 6 if even and divisible by 3.

By 8 - take the last 3 digits, halve and halve again, if the result is even then divisible by 8.

Divisible by 9 if the sum of its digits is divisible by 9.

[daveg]

19 x 11 = 1109?

Kind of You have to carry - so 1+9 is 10 - 0 in the middle, carry the 1, and add that to the 1_9 to get 209

Exactly. But the problem with a lot of these allegedly "quicker" methods is that they're applied blindly and then, if they're also quoted in a "works some of the time" mistaken version, errors ensue. Fast ways to multiply by 11 seem barely worthwhile, as at worst you just need to perform one addition (if 37x11 = 370+37 = 407 why mess around mis-remembering some other method?)

Divisible by 9 if the sum of its digits is divisible by 9.

The proofs for that (or the equivalent test for divisibility by 3) are worth looking at. There's several ways to prove it. The test works because 10 is congruent to 1 modulo 3 or modulo 9, and a decent GCSE student should be able to follow most versions of the proof.

[WP]

Divisibility by 11: sum the even digits and sum the odd digits. When the difference between the sums is divisible by 11, so is the original number. Example, 62458 is divisible by 11, because (6+4+8)-(2+5) = 11.

Squaring a number ending in 5: multiply the front bit by itself plus 1, and append 25. Example: 85^2 = 7225.

As with the other tricks, proving these is instructive.

[aliportico]

Squaring a number ending in 5: multiply the front bit by itself plus 1, and append 25. Example: 85^2 = 7225.

This is useful for squaring numbers ending in "and a half" e.g. 5.5 * 5.5 = 5*6 + 0.25