8 original problems at the level of the first olympiad stage: divisibility, remainders, divisor counting and equations in whole numbers. Each answer is a whole number. Try each one before opening the hints; the full solution explains why the method works and where the idea leads.
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By 3: one of n, n + 1, n + 2 is a multiple of 3, and the product misses out only n. So 3 divides it unless n ≡ 0 (mod 3): n ≡ 1 or 2 (mod 3).
By 4: exactly one factor is even, so that factor must be a multiple of 4: n + 1 ≡ 0 or n + 2 ≡ 0 (mod 4), that is n ≡ 3 or 2 (mod 4).
So 2 × 2 = 4 of every 12 consecutive values of n work. 500 = 41 × 12 + 8 gives 41 × 4 = 164 from the full blocks.
In the last 8 values (remainders 1 to 8 mod 12) only remainders 2 and 7 satisfy both conditions: 2 more.
Total: 164 + 2 = 166.
Why this works: Splitting 12 into the coprime parts 3 and 4 turns one condition into two independent ones, and the Chinese remainder idea lets you simply multiply the counts of good remainders.
Where it leads: Counting n with a polynomial condition modulo m by working prime power by prime power is the start of the theory of congruences and the Chinese remainder theorem.
Positive integers a and b with a < b have highest common factor 12 and lowest common multiple 720. How many such pairs (a, b) are there?
Hint
Write a = 12x and b = 12y. What must be true of x and y?
Second hint
x and y have no common factor and xy = 720 ÷ 12 = 60. Split the prime powers of 60 between x and y.
Full worked solution
Answer: 4
Write a = 12x and b = 12y. Then HCF(x, y) = 1 and LCM(a, b) = 12xy = 720, so xy = 60.
60 = 22 × 3 × 5. Because x and y share no prime, each whole prime power goes entirely to x or entirely to y.
Three prime powers, each with 2 choices: 23 = 8 ordered pairs (x, y).
x = y is impossible (60 is not a square), so exactly half have x < y: 4 pairs, giving (12, 720), (36, 240), (48, 180), (60, 144).
Why this works: HCF × LCM = ab always; dividing out the HCF leaves two coprime numbers, and coprime factor pairs are counted by sharing out whole prime powers.
Where it leads: The number of ways to write n as a product of two coprime factors is 2k, where k is the number of distinct primes of n: a fact used again in counting solutions of equations like 1/x + 1/y = 1/n.
p = 5: 25q ≤ 200, q ≤ 8, q ≠ 5: q = 2, 3, 7, so 3 numbers. p = 7: 49q ≤ 200, q = 2 or 3, so 2 numbers. p = 11: 121q > 200.
Total: 1 + 14 + 7 + 3 + 2 = 27.
Why this works: The divisor-count formula turns ‘exactly 6 divisors’ into a short list of shapes; after that it is careful casework on the size of the smallest prime.
Where it leads: Asking which n have a given number of divisors leads to highly composite numbers and to the study of how the divisor function grows.
CBSE Math Revision