Lcm = 22 + gcd

The least common multiple of two numbers is 22 more than their greatest common divisor. Find these numbers.

Final Answer:

a = 5

a_{1} = 8; b_{1} = 6 a_{2} = 23; b_{2} = 1 a_{3} = 24; b_{3} = 2 a_{4} = 33; b_{4} = 11 a_{5} = 44; b_{5} = 22 a = 5 8 = 2^{3} 6 = 2 \cdot 3 GCD (8, 6) = 2 d_{1} = G C D (8, 6) = 2 8 = 2^{3} 6 = 2 \cdot 3 L C M (8, 6) = 2^{3} \cdot 3 = 24 n_{1} = L C M (8, 6) = 24 d_{1} + 22 = n_{1} 23 . . . prime number 1 . . . prime number GCD (23, 1) = 1 d_{2} = G C D (23, 1) = 1 23 . . . prime number 1 . . . prime number L C M (23, 1) = 23 n_{2} = L C M (23, 1) = 23 d_{2} + 22 = n_{2} 24 = 2^{3} \cdot 3 2 . . . prime number GCD (24, 2) = 2 d_{3} = G C D (24, 2) = 2 24 = 2^{3} \cdot 3 2 . . . prime number L C M (24, 2) = 2^{3} \cdot 3 = 24 n_{3} = L C M (24, 2) = 24 d_{3} + 22 = n_{3} 33 = 3 \cdot 11 11 . . . prime number GCD (33, 11) = 11 d_{4} = G C D (33, 11) = 11 33 = 3 \cdot 11 11 . . . prime number L C M (33, 11) = 3 \cdot 11 = 33 n_{4} = L C M (33, 11) = 33 d_{4} + 22 = n_{4} 44 = 2^{2} \cdot 11 22 = 2 \cdot 11 GCD (44, 22) = 2 \cdot 11 = 22 d_{5} = G C D (44, 22) = 22 44 = 2^{2} \cdot 11 22 = 2 \cdot 11 L C M (44, 22) = 2^{2} \cdot 11 = 44 n_{5} = L C M (44, 22) = 44 d_{5} + 22 = n_{5}

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