Olympiad number questions reward whoever can avoid the arithmetic. A divisibility test, a factor count or the HCF × LCM rule will usually get you there in one line.
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Learn these and you will never do a trial division again.
➗ The tests
| Divisor | Test | Example |
|---|---|---|
| 2 | last digit is even | 4 386 ✅ |
| 3 | digit sum divisible by 3 | 1233 → 9 ✅ |
| 4 | last two digits divisible by 4 | 13 16 ✅ |
| 5 | ends in 0 or 5 | 2465 ✅ |
| 6 | passes both the 2 and the 3 test | 4386 ✅ |
| 8 | last three digits divisible by 8 | 12 136 ✅ |
| 9 | digit sum divisible by 9 | 5346 → 18 ✅ |
| 11 | difference of alternate digit sums is 0 or 11 | 918082 ✅ |
The 11 test in practice. For 918082, add alternate digits: 9 + 8 + 8 = 25 and 1 + 0 + 2 = 3. The difference is 22, a multiple of 11, so the number is divisible by 11.
There is no useful test for 7 at this level, and none for 6 or 12 on their own — you test their factors instead. For 12, check 3 and 4.
Prime factorisation answers more questions than any other single tool.
🌳 What factorisation gives you
How many factors. Write the number as pᵃ × qᵇ and multiply (a + 1)(b + 1). So 36 = 2² × 3² has 3 × 3 = 9 factors.
HCF: take each common prime to its lowest power.
LCM: take every prime to its highest power.
HCF × LCM = the product of the two numbers. This turns most HCF–LCM questions into one division.
The rule in action. Two numbers have HCF 6 and LCM 36, and one of them is 12. The other is (6 × 36) ÷ 12 = 18. No factor trees needed.
Word-problem signal. "Largest that divides…", "greatest length of rope…" means HCF. "Smallest that is divisible by…", "bells ring together again…" means LCM.
Digit puzzles are solved by testing the divisibility rule, not by guessing.
Find the missing digit. If 3 4 ? 2 is divisible by 9, the digit sum 3 + 4 + ? + 2 = 9 + ? must be a multiple of 9, so ? = 0 or 9.
The reversing trick. Any two-digit number minus its reverse is always a multiple of 9; the sum of a two-digit number and its reverse is always a multiple of 11. Test it: 73 − 37 = 36 and 73 + 37 = 110.
Unit digits of powers repeat in cycles. The units digit of 2ⁿ runs 2, 4, 8, 6, 2, 4, 8, 6 … So 2¹⁰⁰ ends in 6, because 100 is a multiple of 4.
12 Olympiad questions on divisibility and factors. 🍀
