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Problem #38 · September 3, 2026

Greatest common divisor EKUB and least common multiple EKUK

For the equation EKUK(n;12)=60EKUK(n; 12) = 60, how many natural number values of nn satisfy it?
  1. A2
  2. B3
  3. C4
  4. D5
  5. E6

WORKED EXAMPLE

Solution

60=22⋅3⋅560 = 2^2 \cdot 3 \cdot 5, 12=22⋅312 = 2^2 \cdot 3. For EKUK(n;12)=60EKUK(n; 12) = 60 to hold, nn must be a divisor of 60 and must include the prime factor 5 (otherwise, the least common multiple will not contain the factor 5). Thus, n=2a⋅3b⋅5n = 2^a \cdot 3^b \cdot 5, where a∈{0;1;2}a \in \{0; 1; 2\} and b∈{0;1}b \in \{0; 1\}, giving a total of 3⋅2=63 \cdot 2 = 6 numbers: 5, 10, 15, 20, 30, 60.
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