Algebraic Extensions of GCD and LCM

GCD(12, 18) = 6.

That means x shares exactly a factor of 7 with 7, so x is a multiple of 7.

21 = 3 × 7, so x must be 7.

4 = 2², 28 = 2² × 7, so x must supply the 7 → x = 7.

To keep only 3 as the GCD, 3x must be odd multiple of 3 (so x odd). From choices, x = 5 works (15 & 12 share only 3).

x must share exactly 6 as greatest common factor. x = 18 works (12 & 18 share 6; larger common factors don’t appear).

8 = 2³; 40 = 2³ × 5, so x must bring in a 5 but not extra primes → x = 10.

14 = 2 × 7. To have GCD 7, x must be a multiple of 7 but not even (otherwise GCD could be 14). x = 21 fits.

Both are multiples of 2a, so the greatest common divisor is 2a.

6 = 2 × 3; 18 = 2 × 3². We need an extra 3 → x = 9.

We need 2 as the only common factor with 6 (=2×3), so x must not be a multiple of 3. The smallest positive integer is x = 1.

20 has factors 2² × 5. To keep the GCD at 5 (not 10), x must be odd and not introduce factor 2. From choices, x = 3 works.

x must be a multiple of 10 but not of 20 (or else GCD would be 20). x = 30 works.

9 = 3²; 36 = 2² × 3², so x needs a 2² factor → x = 12.

x must be a multiple of 9 but not share a factor 2 with 18 (otherwise GCD could be 18). x = 27 works.

5 and 20 give LCM 20 (note: 4 also works in general, but among choices, 20 is the one that works).

x must be a multiple of 4 but not of 8. x = 12 fits (gcd(8,12)=4).

For two positive integers a, b: LCM(a, b) × GCD(a, b) = a × b.

2 and 7 combine to 14 without extra primes.

We want the smallest x so that 3 and x reach 12. 3 and 4 give LCM 12.