Math amc 10 prob 25 2018

Combining $\gcd(a,b) = 24$ and $\gcd(b,c) = 36$ , we get that $72\mid b$ ; similarly, combining $\gcd(b,c) = 36$ and $\gcd(c,d) = 54$ tells us that $108\mid c$ .

Let $a=24\alpha$ and $d=54\delta$ . Note that $\color{#3D99F6} 3\nmid \alpha$ , else $72\mid a$ which would mean that $\gcd(a,b) \ge 72$ ; by the same reasoning, $\color{#3D99F6} 2\nmid \delta$ , else $108\mid d$ which would mean that $\gcd(c,d) \ge 108$ . This has two consequences:

• First, as $\gcd(24,54)=6$ , then if $\gcd(\alpha,\delta)=\color{#20A900}k$ , it must be true that $70<6k<100$ .
• Second, $2\nmid k$ and $3\nmid k$ .

These two facts in turn imply that $k$ is itself prime, for if $k$ were a combination of two or more primes, each $>3$ , then $k\ge 25$ and consequently $6k>100$ .

Finally, the only prime $k$ such that $70<6k<100$ , and therefore the requisite prime divisor of $a$ , is $\boxed{13}$