2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 22 Hard

Let a,b,c and d be positive integers such that \gcd(a,b)=24, \gcd(b,c)=36, \gcd(c,d)=54, and 70<\gcd(d,a)<100. Which of the following must be a divisor of a? (2018 AMC 10A Problem, Question#22)

  • A.

    5

  • B.

    7

  • C.

    11

  • D.

    13

  • E.

    17

Answer:D

The gcd information tells us that 24 divides a, both 24 and 36 divide b, both 36 and 54 divide c, and 54 divides d. Note that we have the prime factorizations:24=2^3\cdot 3,36=2^2\cdot 3^2,54=2\cdot 3^3.

Hence we have a=2^3\cdot 3\cdot w,b=2^3\cdot 3^2\cdot x,c=2^2\cdot 3^3\cdot y,d=2\cdot 3^3\cdot z for some positive integers w,x,y,z. Now if 3 divdes w, then \gcd(a,b) would be at least 2^3\cdot 3^2 which is too large, hence 3 does not divide w. Similarly, if 2 divides z, then \gcd(c,d) would be at least 2^2\cdot 3^2 which is too large, so 2 does not divide z. Therefore, \gcd(a,d)=2\cdot 3\cdot \gcd(w,z) where neither 2 nor 3 divide \gcd(w,z). In other words, \gcd(w,z) is divisible only by primes that are at least 5. The only possible value of \gcd(a,d) between 70 and 100 and which fits this criterion is 78=2\cdot 3\cdot 13, so the answer is \boxed{\rm (D)~13}.

We can say that a and b 'have' 2^3\cdot 3, that b and c have 2^2\cdot 3^2, and that c and d have 3^3\cdot 2. Combining 1 and 2 yields b has (at a minimum) 2^3\cdot 3^2, and thus a has 2^3\cdot 3 (and no more powers of 3 because otherwise \gcd(a,b) would be different). In addition, c has 3^3\cdot 2^2, and thus d has 3^3\cdot 2 (similar to a, we see that d cannot have any other powers of 2). We now assume the simplest scenario, where a=2^3\cdot 3 and d=3^3\cdot 2. According to this base case, we have \gcd(a,d)=2\cdot 3=6. We want an extra factor between the two such that this number is between 70 and 100, and this new factor cannot be divisible by 2 or 3. Checking through, we see that 6\cdot 13 is the only one that works. Therefore the answer is \boxed{\rm (D)~13}.

Elaborating on to what Solution 1 stated, we are not able to add any extra factor of 2 or 3 to \gcd(a,d) because doing so would later the \gcd of (a,b) and (c,d). This is why: The \gcd(a,d) is 2^3\cdot 3 and the \gcd of (c,d) is 2\cdot 3^3. However, the \gcd of (b,c)=2^2\cdot 3^2 (meaning both are divisible by 36). Therefore, a is only divisible by 3^1 (and no higher power of 3), while d is divisible by only 2^1 (and no higher power of 2).

Thus, the \gcd of (a,d) can be expressed in the form 2\cdot 3\cdot k for which k is a number not divisible by 2 or 3. The only answer choice that satisfies this (and the other condition) is \boxed{\rm (D)~13}.

First off, note that 24, 36, and 54 are all of the form 2^x\times 3^y. The prime factorizations are 2^3\times 3^1, 2^2\times 3^2 and 2^1\times 3^3, respectively. Now, let a_2 and a_3 be the number of times 2 and 3 go into a, respectively. Define b_2, b_3, c_2, and c_3 similiarly. Now, translate the lcms into the following: \min(a_2,b_2)=3\min(a_3,b_3)=1\min(b_2,c_2)=2\min(b_3,c_3)=2\min(a_2,c_2)=1, \min(a_3,c_3)=3.

Notice that gcd(a,b,c,d)=gcd\left(gcd\left(a,b\right),gcd\left(b,c\right),gcd\left(c,d\right)\right)=gcd(24,36,54)=6, so gcd(d,a) must be a multiple of 6. The only answer choice that gives a value between 70 and 100 when multiplied by 6 is \boxed{\rm (D)~13}.