Are mental math tricks worth learning, or just party pieces?

The ones worth learning are the ones with a reason. Every shortcut on this page is an algebraic identity in disguise - multiplying by 9 is multiplying by 10 and taking one lot back, because 9 is 10 minus 1. Learn the identity and you can rebuild the trick when you forget it, and you can tell when it applies. Learn the recipe alone and you have memorized something you cannot check.

Which mental math trick should I learn first?

Multiplying by 5, by 9 and by 11, and squaring numbers ending in 5. All four come up constantly, all four are one step, and all four rest on the same idea: replace an awkward number with a friendly one nearby and correct for the difference. Once that idea is familiar the rest of the list stops looking like a list.

Why does multiplying by 11 put the sum of the digits in the middle?

Because 11 is 10 plus 1. Multiplying 43 by 11 means adding 430 and 43, and when you line those up the tens column becomes 4 plus 3. The rule that says 'drop the digit sum in the middle' is that addition, done in advance. It is also why you carry when the sum reaches 10.

How does the digit-sum test for divisibility by 3 or 9 work?

Because 10 leaves a remainder of 1 when divided by 3 or by 9, and so does every power of 10. That means each digit contributes only itself to the remainder, no matter which column it sits in. So a number and the sum of its digits leave the same remainder - which is why adding the digits is a fair test.

36 Mental Math Tricks, and Why Each One Works

A mental-math trick with no reason attached is something you memorize and then forget. Every shortcut below is an algebraic identity wearing a disguise — and once you can see the identity, you can rebuild the trick from scratch and, more usefully, tell when it applies.

These are the 36 lessons in Math Challenge, listed in full. The tag marks the six that need no purchase; each of those links straight into its practice mode.

If you want the short version instead, the mental math tricks guide covers the dozen that come up most often, with more room for each.

Multiplication — 14 tricks

Nearly all of these are the same move: swap the awkward number for a friendly one nearby, then correct.

Multiply by 5

48 × 5 → half of 48 is 24 → 240

5 is half of 10, so ×5 is ×10 then halve — and 10 is the easiest number to multiply by in a base-10 system.

Worked example with figures → · Practice it →

Multiply by 9

48 × 9 → 480 − 48 = 432

9 is 10 − 1, so take ten lots and give one back. It is also why every answer in the 9 times table has digits that add to 9.

Worked example with figures → · Practice it →

Multiply by 11

43 × 11 → 4 (4+3) 3 → 473

11 is 10 + 1, so 43 × 11 is 430 + 43. Lined up, the tens column becomes 4 + 3 — the "sum in the middle" rule is that addition done in advance, which is also why you carry when the sum reaches 10.

Worked example with figures → · Practice it →

Multiply by 12

34 × 12 → 340 + 68 = 408

12 is 10 + 2: ten lots plus double.

Worked example with figures → · Practice it →

Multiply by 15

34 × 15 → 340 + 170 = 510

15 is 10 + 5, and 5 is half of 10 — so it is ×10, then add half of what you just made.

Worked example with figures → · Practice it →

Multiply by 25

32 × 25 → a quarter of 32 is 8 → 800

25 is a quarter of 100. Quarter the number, then multiply by 100.

Worked example with figures → · Practice it →

Rule of 99

43 × 99 → 4300 − 43 = 4257

99 is 100 − 1. The same shape as ×9, one place value up.

Worked example with figures → · Practice it →

Rule of 101

43 × 101 → 4343

101 is 100 + 1, so you get 4300 + 43. Because the number is under 100 the two copies never collide, and the answer is just the number written twice.

Worked example with figures → · Practice it →

Double and halve

14 × 45 → 7 × 90 = 630

Doubling one factor and halving the other multiplies by 2 and divides by 2, so the product cannot move. Repeat until one side is a number you like.

Worked example with figures → · Practice it →

Difference of squares

98 × 102 → 100² − 2² = 10000 − 4 = 9996

Two numbers the same distance either side of a friendly middle are (m − d) and (m + d), and that product is always m² − d².

Worked example with figures → · Practice it →

Cross-multiplication

23 × 12 → 2×1 | 2×2 + 3×1 | 3×2 → 2 | 7 | 6 → 276

It is the ordinary expansion of (20 + 3)(10 + 2), collected by place value instead of written out. One left-to-right pass, no intermediate rows.

Worked example with figures → · Practice it →

Just over 100

104 × 106 → 104 + 6 = 110 → 11000, plus 4 × 6 = 24 → 11024

(100 + a)(100 + b) = 10000 + 100(a + b) + ab. The first step handles the hundreds, the second the leftovers.

Worked example with figures → · Practice it →

Near-100 multiplication, below

97 × 94 → 97 − 6 = 91 → 9100, plus 3 × 6 = 18 → 9118

(100 − a)(100 − b) = 10000 − 100(a + b) + ab — the same identity with the sign flipped, which is why you subtract the other number's distance instead of adding it.

Worked example with figures → · Practice it →

Ends in 5, ten apart

35 × 45 → 3 × 5 = 15 → 1575

(10n + 5)(10n + 15) works out to 100 × n(n + 2) + 75, so it is "tens digit times two more than itself, then 75". Check it: 65 × 75 → 6 × 8 = 48 → 4875.

Worked example with figures → · Practice it →

Squaring — 5 tricks

Every one of these anchors on a round number and corrects with a small square.

Squaring numbers ending in 5

65 × 65 → 6 × 7 = 42 → 4225

A number ending in 5 is 10n + 5, and (10n + 5)² = 100n(n + 1) + 25. That is the rule exactly: n times the next number up, then 25 on the end.

Worked example with figures → · Practice it →

Squares near 100

96² → 96 − 4 = 92 → 9200, plus 4² = 16 → 9216

(100 − d)² = 100(100 − 2d) + d². Taking the distance off once gives the hundreds; the square of the distance is what is left.

Worked example with figures → · Practice it →

Squares in the 50s

54² → 25 + 4 = 29 → 2900, plus 4² = 16 → 2916

(50 + d)² = 2500 + 100d + d² = 100(25 + d) + d². The 25 is there because 50² is 2500.

Worked example with figures → · Practice it →

Squares in the 40s

48² → 25 − 2 = 23 → 2300, plus 2² = 4 → 2304

Same identity, approached from above: (50 − d)² = 100(25 − d) + d².

Worked example with figures → · Practice it →

Squares near 1000

996² → 996 − 4 = 992 → 992000, plus 4² = 16 → 992016

(1000 − d)² = 1000(1000 − 2d) + d². The near-100 trick, one place value up.

Worked example with figures → · Practice it →

Division — 3 tricks

Divide by 5

130 ÷ 5 → 260 → 26

Dividing by 5 is multiplying by 2 and dividing by 10, because 5 = 10 ÷ 2. Doubling is easy and dropping a zero is free.

Worked example with figures → · Practice it →

Divide by 25

800 ÷ 25 → 3200 → 32

25 is a quarter of 100, so ÷25 is ×4 then ÷100.

Worked example with figures → · Practice it →

Divide by 3

5712 → 5+7+1+2 = 15, divisible → 5712 ÷ 3 = 1904

Every power of 10 leaves remainder 1 when divided by 3, so each digit contributes only itself no matter which column it is in. That is why a number and its digit sum have the same remainder, and why the test is fair.

Worked example with figures → · Practice it →

Addition and subtraction — 3 tricks

Add reversed numbers

47 + 74 → 11 × (4 + 7) = 121

(10a + b) + (10b + a) = 11(a + b). The reversal is what makes the two place values swap and pair up.

Worked example with figures → · Practice it →

Subtract reversed numbers

82 − 28 → 9 × (8 − 2) = 54

(10a + b) − (10b + a) = 9(a − b). Every such difference is a multiple of 9 — that is not a coincidence, it is the identity.

Worked example with figures → · Practice it →

Subtract from 1000

1000 − 473 → 9−4, 9−7, 10−3 → 527

1000 is 999 + 1, and nothing borrows from 999. Take each digit from 9 and the last from 10, and the borrowing disappears entirely.

Worked example with figures → · Practice it →

Series and sequences — 2 tricks

Sum of consecutive odd numbers

1 + 3 + 5 + 7 + 9 = 25 = 5²

Each new odd number is exactly the L-shaped border that turns one square into the next. Add the first N and you have built an N × N square.

Worked example with figures → · Practice it →

Sum 1 to N (Gauss)

1 + 2 + … + 100 → 100 × 101 ÷ 2 = 5050

Pair the ends: 1 + 100, 2 + 99, 3 + 98 — fifty pairs, each making 101. The formula is that pairing written down.

Worked example with figures → · Practice it →

Infinite and telescoping sums — 2 tricks

Telescoping sums

1/(1×2) + 1/(2×3) + … + 1/(N×(N+1)) = 1 − 1/(N+1)

Each term is 1/n − 1/(n+1). Written that way, every middle piece is canceled by its neighbor and only the two ends survive.

Worked example with figures → · Practice it →

Zeno's paradox

1/2 + 1/4 + 1/8 + … = 1

Each step covers half of what is left, so the gap to 1 halves every time. Infinitely many steps, and the distance still runs out — which is what "converges" means.

Worked example with figures → · Practice it →

Number theory — 5 tricks

Last digit of a power

7¹, 7², 7³, 7⁴ end in 7, 9, 3, 1 — then it repeats

Only the units digit of one step feeds the units digit of the next, so the last digits must eventually cycle. Find the cycle, then reduce the exponent to its position.

Worked example with figures → · Practice it →

Last digit of a product

347 × 893 → 7 × 3 = 21 → ends in 1

Every other place value carries a factor of 10, and anything multiplied by 10 ends in 0. Only the units digits can reach the units column.

Worked example with figures → · Practice it →

Digital root (mod 9)

4573 → 4+5+7+3 = 19 → 1+9 = 10 → 1, and 4573 = 9 × 508 + 1

10 leaves remainder 1 when divided by 9, and so does every power of 10. Each digit therefore contributes only itself to the remainder, whatever column it sits in.

Worked example with figures → · Practice it →

Divisibility by 11

2728 → 2 − 7 + 2 − 8 = −11, so yes (2728 = 11 × 248)

10 leaves remainder −1 when divided by 11, so the powers of 10 alternate between +1 and −1. That is exactly the alternating sum.

Worked example with figures → · Practice it →

Large-exponent cycles

7¹⁰⁰ → the cycle is 4 long, and 100 divides by 4 → ends in 1

The same cycle as above, used on an exponent far too big to expand. You never compute the power — you compute where it lands in the loop.

Worked example with figures → · Practice it →

Fractions and continued fractions — 2 tricks

Flip the percent

8% of 50 → 50% of 8 = 4

Both sides are the same multiplication, 8 × 50 ÷ 100, read in a different order. There is more of this in the percentages guide.

Worked example with figures → · Practice it →

The golden ratio as a continued fraction

x = 1 + 1/(1 + 1/(1 + …)) → x = 1 + 1/x → x² = x + 1 → x ≈ 1.618

The fraction contains a perfect copy of itself, so you can name the whole thing x and substitute it back into itself. An infinite expression collapses into a quadratic.

Worked example with figures → · Practice it →

What to do with this list

Do not try to learn 36 shortcuts. People who are quick at mental arithmetic tend to have a handful they reach for without thinking, and everything else they work out.

Pick two or three that match sums you actually meet, and learn the reason rather than the recipe — the reason is what tells you the trick applies before you have finished reading the question. And if the times tables are not automatic yet, start there instead: every shortcut on this page assumes you are not spending working memory on 7 × 8.

Practice these in the game

Each of the 36 is an illustrated lesson in Math Challenge — the principle first, then practice on it with the difficulty adapting as you go. Six are free, and the rest come with the one-time unlock.

Your turn

Three to try — tap what you get.

65²

47 + 74

130 ÷ 5

Open the Magic Tricks →