Lcm = 22 + gcd

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

Correct answer:

a =  5

Step-by-step explanation:

$$\begin{gathered} a_1=8;b_1=6 \\ {a}_2=23;b_2=1 \\ {a}_3=24;b_3=2 \\ {a}_4=33;b_4=11 \\ {a}_5=44;b_5=22 \\ a=5 \\ 8=2^3 \\ 6=2\cdot 3 \\ \text{GCD}(8,6)=2 \\ {d}_1=GCD(8,6)=2 \\ 8=2^3 \\ 6=2\cdot 3 \\ LCM(8,6)=2^3\cdot 3=24 \\ {n}_1=LCM(8,6)=24 \\ {d}_1+22=n_1 \\ 23\mathrm{.}\mathrm{.}\mathrm{.}\text{ prime number } \\ 1\mathrm{.}\mathrm{.}\mathrm{.}\text{ prime number } \\ \text{GCD}(23,1)=1 \\ {d}_2=GCD(23,1)=1 \\ 23\mathrm{.}\mathrm{.}\mathrm{.}\text{ prime number } \\ 1\mathrm{.}\mathrm{.}\mathrm{.}\text{ prime number } \\ LCM(23,1)=23 \\ {n}_2=LCM(23,1)=23 \\ {d}_2+22=n_2 \\ 24=2^3\cdot 3 \\ 2\mathrm{.}\mathrm{.}\mathrm{.}\text{ prime number } \\ \text{GCD}(24,2)=2 \\ {d}_3=GCD(24,2)=2 \\ 24=2^3\cdot 3 \\ 2\mathrm{.}\mathrm{.}\mathrm{.}\text{ prime number } \\ LCM(24,2)=2^3\cdot 3=24 \\ {n}_3=LCM(24,2)=24 \\ {d}_3+22=n_3 \\ 33=3\cdot 11 \\ 11\mathrm{.}\mathrm{.}\mathrm{.}\text{ prime number } \\ \text{GCD}(33,11)=11 \\ {d}_4=GCD(33,11)=11 \\ 33=3\cdot 11 \\ 11\mathrm{.}\mathrm{.}\mathrm{.}\text{ prime number } \\ LCM(33,11)=3\cdot 11=33 \\ {n}_4=LCM(33,11)=33 \\ {d}_4+22=n_4 \\ 44=2^2\cdot 11 \\ 22=2\cdot 11 \\ \text{GCD}(44,22)=2\cdot 11=22 \\ {d}_5=GCD(44,22)=22 \\ 44=2^2\cdot 11 \\ 22=2\cdot 11 \\ LCM(44,22)=2^2\cdot 11=44 \\ {n}_5=LCM(44,22)=44 \\ {d}_5+22=n_5 \end{gathered}$$

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