The number \(a\) is divisible by \(b\) if the remainder after dividing \(a \div b\) is zero. For example, 12 is divisible by 3 because \(12 \div 3 = 4\) (with no remainder).
Remember these divisibility tests — they save time in olympiads:
Divisibility, prime numbers, digits, and their secrets. Numbers hide beautiful patterns — let’s uncover them.
The number \(a\) is divisible by \(b\) if the remainder after dividing \(a \div b\) is zero. For example, 12 is divisible by 3 because \(12 \div 3 = 4\) (with no remainder).
Remember these divisibility tests — they save time in olympiads:
| Divisible by | Rule | Example |
|---|---|---|
| 2 | The last digit is even (0, 2, 4, 6, 8) | 374 → 4 is even ✓ |
| 3 | The sum of the digits is divisible by 3 | 528 → 5+2+8 = 15 ✓ |
| 4 | The last two digits form a number divisible by 4 | 1 316 → 16 ÷ 4 = 4 ✓ |
| 5 | The last digit is 0 or 5 | 2 735 → 5 ✓ |
| 6 | Divisible by both 2 and 3 | 354 → even + (3+5+4=12) ✓ |
| 9 | The sum of the digits is divisible by 9 | 729 → 7+2+9 = 18 ✓ |
| 10 | The last digit of 0 | 1 530 → 0 ✓ |
| 11 | The difference of the alternating digit sums is divisible by 11 | 2 574 → (2+7)−(5+4) = 0 ✓ |
A prime number is a number greater than 1 that is divisible only by 1 and itself.
The first primes are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47…
Prime numbers from 1 to 50 (highlighted in blue):
How many prime numbers are there between 30 and 50?
Check each one: 31 ✓, 32 ✗, 33 ✗, 34 ✗, 35 ✗, 36 ✗, 37 ✓, 38 ✗, 39 ✗, 40 ✗, 41 ✓, 42 ✗, 43 ✓, 44 ✗, 45 ✗, 46 ✗, 47 ✓, 48 ✗, 49 ✗.
Answer: 5 (they are 31, 37, 41, 43, 47).
GCD (greatest common divisor) — the largest number that divides both numbers.
LCM (least common multiple) — the smallest number divisible by both numbers.
Let us find GCD(18, 24).
Divisors of 18: 1, 2, 3, 6, 9, 18
Divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Common divisors: 1, 2, 3, 6The largest one → GCD = 6.
Let us find LCM(12, 15).
Multiples of 12: 12, 24, 36, 48, 60, 72…
Multiples of 15: 15, 30, 45, 60, 75…
The first common one → LCM = 60.
A two-digit number with digits \(a\) (tens) and \(b\) (ones) can be written as:
The sum of the digits of a two-digit number is 12. If the digits are swapped, the new number is 36 less than the original. What is the number?
Let the number be \(\overline{ab}\). Then:
\(a + b = 12\)
\(\overline{ab} - \overline{ba} = 36\), so \(9(a-b) = 36\), hence \(a - b = 4\).
From the system: \(a + b = 12\) and \(a - b = 4\) → \(a = 8\), \(b = 4\).
Answer: 84.
This is a favourite CMS topic! The last digit of powers . repeats in cycles. You only need to find the cycle.
| Base | \(n=1\) | \(n=2\) | \(n=3\) | \(n=4\) | Cycle |
|---|---|---|---|---|---|
| 2 | 2 | 4 | 8 | 6 | 2, 4, 8, 6 (length 4) |
| 3 | 3 | 9 | 7 | 1 | 3, 9, 7, 1 (length 4) |
| 4 | 4 | 6 | 4 | 6 | 4, 6 (length 2) |
| 7 | 7 | 9 | 3 | 1 | 7, 9, 3, 1 (length 4) |
| 8 | 8 | 4 | 2 | 6 | 8, 4, 2, 6 (length 4) |
| 9 | 9 | 1 | 9 | 1 | 9, 1 (length 2) |
The last digit of \(7^{100}\) equals:
Cycle for 7: 7, 9, 3, 1 (length 4).
\(100 \div 4 = 25\) (remainder 0).
Remainder 0 → the last position in the cycle → 1.
When \(17 \div 5 = 3\) with remainder \(2\), we write: \(17 = 5 \times 3 + 2\).
A remainder is what is “left over.” It is always smaller than the divisor.
What is the remainder when \(176543 + 92841\) is divided by 4?
For divisibility by 4, it is enough to look at the last two digits:
\(176543\): the last two digits are 43. \(43 \div 4 = 10\) with remainder 3.
\(92841\): the last two digits are 41. \(41 \div 4 = 10\) with remainder 1.
Remainder of a sum: \(3 + 1 = 4\). \(4 \div 4 = 1\) with remainder 0.