Question

Suppose that the least common multiple of the first $25$ positive integers is equal to $26A7114B4C0$. Find $100 \times A + 10 \times B + C$.

Answer 1

Start by removing 1 and the primes from the set {1, 2, 3, …, 25}. The LCM among these numbers will be their product.

{1, 2, 3, 5, 7, 11, 13, 17, 19, 23}   ==>   product = 223,092,870

Factorize the remaining numbers in the set:

{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25}   ==>

{2², 2×3, 2³, 3², 2×5, 2²×3, 2×7, 3×5, 2⁴, 2×3², 2²×5, 3×7, 2×11, 2³×3, 5²}

From each number above, remove any factor already accounted for in the product of primes:

{2, 1, 2², 3, 1, 2, 1, 1, 2³, 3, 2, 1, 1, 2², 5}

The LCM among these factors is 2³×3×5 = 120.

Then the LCM of the numbers in {1, 2, 3, …, 25} is

223,092,870 × 120 = 26,771,144,400

so that A = 7, B = 4, and C = 0. Then

100A + 10B + C = 740