Maths Hacks – Learn Fast Techniques to Solve Maths Problems Like a Pro
Explore these fun math tricks to simplify calculations, check divisibility, and improve mental math skills! Each trick below includes the rule, a plain-English explanation of why it actually works, several worked examples, and a quick question so you can try it yourself.
Divisibility by 2
The Rule: If a number ends in an even digit — 0, 2, 4, 6, or 8 — it is divisible by 2.
Why It Works: Think of any number as a stack of place-value chunks: hundreds, tens, and ones. Every chunk except the very last one is a multiple of 10 (like 10, 20, 100, 240 …), and multiples of 10 are always even. So the only chunk that can possibly make the whole number odd is the ones digit — the single digit you see at the very end. That is why checking just that one digit is always enough.
Formal Proof (for advanced learners)
Let N be any whole number, and write it as N = 10a + b, where b is N’s units digit (0–9) and a is the number formed by all the remaining digits. Since 10a is always even for any whole number a, N is even exactly when b is even. In modular notation, N ≡ b (mod 2), so checking whether b ∈ {0, 2, 4, 6, 8} fully decides divisibility by 2.
Examples
- 28 ends in 8, so it’s divisible by 2.
- 46 ends in 6, so it’s divisible by 2.
- 133 ends in 3, so it is not divisible by 2.
Divisibility by 3
The Rule: If the sum of all the digits of a number is divisible by 3, then the number itself is divisible by 3.
Why It Works: Every place value in our number system — 1, 10, 100, 1000 — is just one more than a multiple of 3 (1, 9+1, 99+1, 999+1 …). So when a digit sits in the hundreds place, say, almost all of its value is already a multiple of 3, and only a tiny leftover of “1 per digit” escapes. Adding up the digits is really just collecting all those small leftovers — and if they add up to a multiple of 3, the whole number must be too.
Formal Proof (for advanced learners)
Write N = ∑ a_i·10^i, where a_i are the digits of N. Since 10 ≡ 1 (mod 3), it follows that 10^i ≡ 1 (mod 3) for every power i. Therefore N ≡ ∑ a_i·1 = ∑ a_i (mod 3): N and its digit sum always leave the same remainder when divided by 3, so one is divisible by 3 exactly when the other is.
Examples
- 123 → 1 + 2 + 3 = 6, divisible by 3.
- 453 → 4 + 5 + 3 = 12, divisible by 3.
- 217 → 2 + 1 + 7 = 10, not divisible by 3.
Divisibility by 4
The Rule: If the last two digits of a number form a number divisible by 4, then the entire number is divisible by 4.
Why It Works: Any number can be split into “full hundreds” plus “whatever is left in the last two digits.” A hundred happens to be 4 × 25, so every full hundred is already a clean multiple of 4 and never affects the test. That leaves only the number formed by the last two digits to actually check.
Formal Proof (for advanced learners)
Write N = 100a + b, where b is the two-digit number (0–99) formed by N’s last two digits and a is everything before that. Since 100 = 4·25, we have 100a ≡ 0 (mod 4), so N ≡ b (mod 4). Hence N is divisible by 4 exactly when its last two digits, read as a number, are divisible by 4.
Examples
- 132 → last two digits 32, divisible by 4.
- 3116 → last two digits 16, divisible by 4.
- 750 → last two digits 50, not divisible by 4.
Divisibility by 5
The Rule: If a number ends in 0 or 5, it is divisible by 5.
Why It Works: This works for the same reason as the rule for 2. Every digit except the last one sits in a place value that is a multiple of 10, and every multiple of 10 is automatically a multiple of 5 as well. So once again, only the very last digit can possibly change the answer.
Formal Proof (for advanced learners)
Let N = 10a + b, where b is the units digit. Since 10a is divisible by 5 for any whole number a, N ≡ b (mod 5). So N is divisible by 5 exactly when b is 0 or 5.
Examples
- 35 ends in 5, divisible by 5.
- 120 ends in 0, divisible by 5.
- 47 ends in 7, not divisible by 5.
Divisibility by 6
The Rule: A number is divisible by 6 if it is divisible by both 2 and 3.
Why It Works: 6 is built from two smaller building blocks, 2 and 3, which share no common factors of their own. If a number passes both the “is it even” test and the “digit sum” test, it already contains a hidden factor of 2 and a hidden factor of 3 — and multiplying those two together guarantees a factor of 6 sitting inside the number.
Formal Proof (for advanced learners)
6 = 2·3, and gcd(2, 3) = 1 (they share no common factors). If 2 | N and 3 | N, then because 2 and 3 are coprime, their product also divides N, i.e. 6 | N. So N is divisible by 6 if and only if it is divisible by both 2 and 3.
Examples
- 54 is even, and 5 + 4 = 9 (divisible by 3). So 54 is divisible by 6.
- 36 is even, and 3 + 6 = 9 (divisible by 3). So 36 is divisible by 6.
- 27 is odd, so it fails the first test and is not divisible by 6.
Divisibility by 7
The Rule: Double the last digit, subtract it from the rest of the number, and if the result is divisible by 7, so is the original number.
Why It Works: This one is genuinely less obvious than the others — and that is exactly why it feels clever rather than routine. Splitting off the last digit, doubling it, and subtracting is a way of quietly “cancelling out” the effect that the number 10 has, in a way that happens to line up perfectly with 7. It is not something you can see at a glance; it only falls out once you follow the arithmetic all the way through. But it always works, every single time.
Formal Proof (for advanced learners)
Let N = 10a + b, where b is the units digit and a is the rest of the number. Claim: 7 | N ⇔ 7 | (a − 2b). Multiply (a − 2b) by 10: 10a − 20b. Since −20 ≡ 1 (mod 7) (because −21 = −3·7), we get −20b ≡ b (mod 7), so 10(a − 2b) ≡ 10a + b = N (mod 7). Because gcd(10, 7) = 1, multiplying by 10 never changes whether a number is ≡ 0 (mod 7), so N ≡ 0 (mod 7) ⇔ a − 2b ≡ 0 (mod 7).
Examples
- 203 → double 3 to get 6, then 20 − 6 = 14, which is divisible by 7.
- 168 → double 8 to get 16, then 16 − 16 = 0, which counts as divisible by 7.
- 92 → double 2 to get 4, then 9 − 4 = 5, not divisible by 7.
Divisibility by 8
The Rule: If the last three digits of a number are divisible by 8, the whole number is divisible by 8.
Why It Works: This is the same idea as the rule for 4, just carried one step further. A thousand happens to equal 8 × 125, so every full thousand hiding inside a number is already a clean multiple of 8 and can be safely ignored. That leaves only the last three digits to decide the answer.
Formal Proof (for advanced learners)
Write N = 1000a + b, where b is the three-digit number formed by N’s last three digits (0–999). Since 1000 = 8·125, we have 1000a ≡ 0 (mod 8), so N ≡ b (mod 8). Hence N is divisible by 8 exactly when its last three digits are.
Examples
- 1000 → last three digits 000, divisible by 8.
- 3128 → last three digits 128 (128 ÷ 8 = 16), divisible by 8.
- 1234 → last three digits 234, not divisible by 8.
Divisibility by 9
The Rule: If the sum of all the digits of a number is divisible by 9, then the number itself is divisible by 9.
Why It Works: This works for exactly the same reason the rule for 3 does — just with 9 instead. Every place value (1, 10, 100 …) is also one more than a multiple of 9, so once again, adding up the digits collects all those tiny leftovers, and the digit sum always carries the same remainder as the full number.
Formal Proof (for advanced learners)
Since 10 ≡ 1 (mod 9), it follows that 10^i ≡ 1 (mod 9) for every power i. So N = ∑ a_i·10^i ≡ ∑ a_i (mod 9), meaning N and its digit sum always share the same remainder when divided by 9, and so are either both divisible by 9 or neither is.
Examples
- 729 → 7 + 2 + 9 = 18, divisible by 9.
- 981 → 9 + 8 + 1 = 18, divisible by 9.
- 514 → 5 + 1 + 4 = 10, not divisible by 9.
Divisibility by 10
The Rule: If a number ends in 0, it is divisible by 10.
Why It Works: Ten is the very foundation our whole counting system is built on. Any number that is not a clean multiple of 10 will always show exactly that leftover amount as its last digit — which is why checking for a zero at the end really is the entire test.
Formal Proof (for advanced learners)
Write N = 10a + b, where b is the units digit. Since 10a ≡ 0 (mod 10) for any whole number a, N ≡ b (mod 10). So N is divisible by 10 exactly when b = 0.
Examples
- 130 ends in 0, divisible by 10.
- 500 ends in 0, divisible by 10.
- 125 ends in 5, not divisible by 10.
Divisibility by 11
The Rule: Add and subtract the digits alternately, from right to left. If the result is 0 or a multiple of 11, the number is divisible by 11.
Why It Works: Ten sits just one small step away from eleven, one less than it. That tiny one-step gap means that every time you move one place value to the left — from ones to tens, tens to hundreds, and so on — the effect on the “leftover after dividing by 11” flips sign instead of staying the same. Adding and subtracting the digits alternately is really just tracking that flip, one place value at a time.
Formal Proof (for advanced learners)
Since 10 ≡ −1 (mod 11), it follows that 10^i ≡ (−1)^i (mod 11). Therefore N = ∑ a_i·10^i ≡ ∑ a_i·(−1)^i (mod 11), which is precisely the alternating sum of the digits — adding digits in even positions and subtracting digits in odd positions (or vice versa, depending on how you index from the right).
Examples
- 121 → 1 − 2 + 1 = 0, divisible by 11.
- 2728 → 2 − 7 + 2 − 8 = −11, divisible by 11.
- 3729 → 3 − 7 + 2 − 9 = −11, divisible by 11.
Multiplying by 5
The Rule: To multiply by 5 quickly, divide the number by 2 and then multiply by 10.
Why It Works: 5 is exactly half of 10. Multiplying by 10 just means sliding every digit one place to the left — something your brain can do instantly — so if you halve the number first and then tack on that shift, you have finished the whole multiplication in two very easy moves instead of one hard one.
Formal Proof (for advanced learners)
5 = 10 ÷ 2, so for any n: n × 5 = n × (10 ÷ 2) = (n ÷ 2) × 10 = (n × 10) ÷ 2. Both orders give the same result, by the associativity and commutativity of multiplication and division.
Examples
- 36 × 5 → 36 ÷ 2 = 18, then ×10 = 180.
- 84 × 5 → 84 ÷ 2 = 42, then ×10 = 420.
- 123 × 5 → 123 ÷ 2 = 61.5, then ×10 = 615.
Multiplying by 9
The Rule: To multiply by 9, multiply by 10 and subtract the original number.
Why It Works: 9 is just one small step below 10. Multiplying by 10 is instant — you simply add a zero — and since “9 groups of something” is the same as “10 groups, minus 1 group,” subtracting the original number once quietly corrects for that small overshoot from rounding up to 10.
Formal Proof (for advanced learners)
n × 9 = n × (10 − 1) = n×10 − n×1 = 10n − n, a direct application of the distributive law.
Examples
- 8 × 9 → 8 × 10 = 80, then 80 − 8 = 72.
- 23 × 9 → 230 − 23 = 207.
- 150 × 9 → 1500 − 150 = 1350.
Multiplying by 11
The Rule: To multiply a two-digit number by 11, add the digits together and place the sum in between them. If the sum is 10 or more, carry the extra 1 into the first digit.
Why It Works: Any two-digit number can be thought of as “ten lots of the first digit, plus the second digit.” Multiplying that whole thing by 11 is the same as multiplying by 10 and then adding one more full copy of the original number — and adding that extra copy happens to slot the digit sum neatly into the middle, between the two original digits.
Formal Proof (for advanced learners)
Let the two-digit number be N = 10a + b (digits a and b). Then N×11 = N×10 + N = (100a + 10b) + (10a + b) = 100a + 10(a+b) + b. When a+b < 10, this is literally the three-digit number with digits a, (a+b), b. When a+b ≥ 10, the tens column overflows into the hundreds column — which is exactly the “carry the 1” step.
Examples
- 23 × 11 → 2 + 3 = 5, placed between: 253.
- 45 × 11 → 4 + 5 = 9, placed between: 495.
- 57 × 11 → 5 + 7 = 12, so carry the 1: (5+1) 2 7 = 627.
Multiplying by 15
The Rule: To multiply by 15, multiply the number by 10 and add half of that result.
Why It Works: 15 is simply 10 plus half of 10. So once you have multiplied by 10 — an instant step — all that is left is to add half of that very same answer, which is usually easy to work out in your head.
Formal Proof (for advanced learners)
15 = 10 + 5 = 10 + (10 ÷ 2), so n×15 = n×10 + n×(10÷2) = 10n + (10n)÷2, by the distributive law.
Examples
- 40 × 15 → 400 + 200 = 600.
- 12 × 15 → 120 + 60 = 180.
- 50 × 15 → 500 + 250 = 750.
Multiplying by 25
The Rule: To multiply by 25, divide by 4 and then multiply by 100.
Why It Works: 25 is exactly a quarter of 100. Dividing by 4 first keeps the numbers small and friendly to work with, and multiplying by 100 at the very end is just adding two zeros — no hard multiplication required at any point.
Formal Proof (for advanced learners)
25 = 100 ÷ 4, so n×25 = n×(100÷4) = (n÷4)×100, again by associativity of multiplication and division.
Examples
- 16 × 25 → 16 ÷ 4 = 4, then ×100 = 400.
- 48 × 25 → 48 ÷ 4 = 12, then ×100 = 1200.
- 37 × 25 → 37 ÷ 4 = 9.25, then ×100 = 925.
Squaring Numbers Ending in 5
The Rule: Square any two-digit number ending in 5 by multiplying the first digit by itself plus one, then appending 25.
Why It Works: Every two-digit number ending in 5 is really “some leading digit, then a 5” — like “4, then 5” for 45. It turns out that when you square any such number, the answer always ends in exactly 25, and the digits in front are simply that leading digit multiplied by one more than itself. It looks like a magic shortcut, but it drops straight out of expanding the square with ordinary algebra.
Formal Proof (for advanced learners)
A number ending in 5 can be written as 10a + 5, where a represents the leading digit(s). Expanding the square: (10a + 5)² = 100a² + 2·(10a)·5 + 25 = 100a² + 100a + 25 = 100·a(a+1) + 25. This is precisely “a times (a + 1), followed by the digits 25.”
Examples
- 25² → 2 × 3 = 6, append 25: 625.
- 45² → 4 × 5 = 20, append 25: 2025.
- 105² → 10 × 11 = 110, append 25: 11025.
Finding 10% of a Number
The Rule: To find 10% of a number, simply move the decimal point one place to the left.
Why It Works: “Percent” literally means “per hundred,” so 10% means 10 parts out of every 100 — which simplifies to exactly one-tenth. Dividing by 10 in our number system is the same as sliding every digit one place to the right, which is exactly why moving the decimal point does the job.
Formal Proof (for advanced learners)
10% = 10÷100 = 1÷10. Dividing any number n by 10 shifts its decimal representation one place value to the right, i.e. n÷10 is n with the decimal point moved one place to the left.
Examples
- 10% of 250 = 25.
- 10% of 84 = 8.4.
- 10% of 1500 = 150.
Adding 9 Quickly
The Rule: To add 9 to a number, add 10 and then subtract 1.
Why It Works: Adding exactly 9 is a little fiddly, but adding 10 is instant — you just raise the tens digit by one. Since 9 is only 1 less than 10, adding 10 and then stepping back down by 1 lands you on precisely the right answer, with far less mental effort than adding 9 directly.
Formal Proof (for advanced learners)
For any n: n + 9 = n + 10 − 1, a direct rearrangement of the addition — adding 10 and subtracting 1 always gives the same result as adding 9.
Examples
- 56 + 9 → 56 + 10 = 66, then 66 − 1 = 65.
- 28 + 9 → 38 − 1 = 37.
- 145 + 9 → 155 − 1 = 154.
Subtracting from 1000
The Rule: To subtract a number from 1000, subtract each digit from 9, except the last digit, which you subtract from 10.
Why It Works: 999 is an unusually forgiving number to subtract from, because every single one of its digits is a 9 — so subtracting any digit from 9 never requires borrowing from the next column. Subtracting from 1000 is really just subtracting from 999 and then adding back the 1 you “borrowed” to make 999 in the first place. That extra +1 is exactly what happens when the very last digit is subtracted from 10 instead of from 9.
Formal Proof (for advanced learners)
For a three-digit number n with digits d₂d₁d₀: 1000 − n = (999 − n) + 1. Since every digit of 999 is 9, computing 999 − n digit-by-digit never requires borrowing (each digit of n is at most 9), giving digits (9−d₂), (9−d₁), (9−d₀). Adding the final +1 to the units column is equivalent to computing (10−d₀) for the last digit instead of (9−d₀) — exactly the stated rule.
Examples
- 1000 − 356 → 9−3=6, 9−5=4, 10−6=4: 644.
- 1000 − 472 → 9−4=5, 9−7=2, 10−2=8: 528.
- 1000 − 89 → treat as 089: 9−0=9, 9−8=1, 10−9=1: 911.
Finding 20% of a Number
The Rule: To find 20% of a number, find 10% first and then double it.
Why It Works: 20% is simply double 10%, and 10% happens to be the easiest percentage of all to find. So instead of tackling 20% head-on, this trick quietly turns it into two small, easy steps: find the easy one first, then double it.
Formal Proof (for advanced learners)
20% = 20÷100 = 2×(10÷100) = 2×10%. So computing 20% of n is algebraically identical to computing 10% of n and then multiplying that result by 2.
Examples
- 20% of 150 → 10% is 15, doubled is 30.
- 20% of 90 → 10% is 9, doubled is 18.
- 20% of 250 → 10% is 25, doubled is 50.