School curriculum· 25 min· 21 questions
Learn first: Natural numbers and operations
Divisibility is the foundation of the arithmetic of natural numbers. Divisibility tests tell you whether a number is divisible by another without doing the division. Primes and prime factorisation reveal the “make-up” of any number. The HCF and LCM are needed to simplify fractions, find common denominators and solve “when will they meet again” problems. The topic links directly to number theory, fractions and integer solutions of equations.
A number a is divisible by b (a ⋮ b) if a = b·k for some integer k. Then b is a divisor (factor) of a, and a is a multiple of b.
Basic properties:
A prime is a natural number with exactly two divisors (1 and itself): 2, 3, 5, 7, 11, 13, …. A composite number has more than two divisors. 1 is neither prime nor composite. 2 is the only even prime.
Fundamental theorem of arithmetic: every composite number factorises into primes in exactly one way
To test whether n is prime it is enough to try the primes not exceeding √n: if n = a·b with both factors greater than √n, the product would exceed n.
By 2 — last digit even; by 5 — last digit 0 or 5; by 10 — ends in 0
By 4 — the number formed by the last two digits is divisible by 4; by 25 — ends in 00, 25, 50, 75
By 8 — the number formed by the last three digits is divisible by 8
By 3 — digit sum divisible by 3; by 9 — digit sum divisible by 9
By 11 — the difference between the sums of digits in odd and even positions is divisible by 11
Why do these work? 100 is divisible by 4 and 25, so everything from the hundreds upwards is always divisible by 4 — only the last two digits matter. Since 1000 ⋮ 8, three digits suffice for 8. Because 10 = 9 + 1, 100 = 99 + 1 and so on, a number differs from its digit sum by a multiple of 9 — the reason for the tests for 3 and 9.
Composite divisors: to be divisible by 6 a number must be divisible by 2 and 3; by 12 — by 3 and 4; by 15 — by 3 . The factors must be : testing as “2 and 6” is wrong ( is divisible by both 2 and 6 but not by 12).
Is 918 082 divisible by 11?
The four-digit number 34*6 is divisible by 12. Which digit replaces the star?
The largest number that divides each of the given numbers (also called the GCD). Methods:
Why does Euclid’s algorithm work? If a = bq + r, any common divisor of a and b also divides r = a − bq, and conversely. So the pairs (a, b) and (b, r) have the same common divisors.
If HCF(a, b) = 1, the numbers are coprime: e.g. 8 and 15.
Find HCF(84, 126) in two ways.
The smallest natural number divisible by each of the numbers. Take the primes to their highest powers.
HCF(a, b)·LCM(a, b) = a·b
LCM(84, 126) = 2²·3²·7 = 252. Check: 42·252 = 10 584 = 84·126.
For coprime numbers LCM(a, b) = a·b. When adding fractions the common denominator is exactly the LCM — that keeps the numbers smallest.
One bus leaves every 12 minutes, another every 18 minutes. They left together at 8:00. When will they next leave together?
If n = p₁^a₁ · p₂^a₂ · … · pₖ^aₖ, every divisor of n has the form p₁^b₁ · … · pₖ^bₖ with 0 ≤ bᵢ ≤ aᵢ. Each exponent has aᵢ + 1 choices.
Number of divisors τ(n) = (a₁ + 1)(a₂ + 1)…(aₖ + 1)
For instance 360 = 2³·3²·5 has 4·3·2 = 24 divisors. Exactly the perfect squares have an odd number of divisors (100 has 9).
Division with remainder: a = bq + r, 0 ≤ r < b. The remainder is always smaller than the divisor.
“When divided by 2, 3, 4, 5, 6 the remainder is always 1” means number − 1 is divisible by all of them: number − 1 = LCM(2, 3, 4, 5, 6)·k = 60k. The smallest is 61.
If “the remainder is always one less than the divisor” (by 2 — 1, by 3 — 2, by 4 — 3), then number + 1 is divisible by all of them.
The three-digit number 5a7 is divisible by 9. Find a.