Question 891345: In the following equation, A, B, C, and D are all positive integers such that A has 4 factors, B has 6 factors, and D has 8 factors. What is the smallest possible sum of A, B, C, and D, if (A×B)/C = D ?
Found 2 solutions by richard1234, Edwin McCravy:
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Rearrange to AB = CD
Since D has 8 factors, D can equal p^7 or (p^3)(q^1) or pqr where p, q, r are primes.
Similarly, A is of the form p^3 or pq, and B is of the form p^5 or (p^2)q.
Greedily let D = (2^3)(3) = 24, A = 2*3 = 6, B = (2^2)(3) = 12, so C = 3. A+B+C+D = 6+12+3+24 = 45.
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
positive integers of the form pq, and only those of the form pq,
have 4 factors, 1,p,q, and pq where p and q are different primes.
So let A = pq
Positive integers of the form rs², and only those of the form rs²,
have 6 factors, 1,r,s,s²,rs and rs² where r and s are different primes.
So let B = ra²
Positive integers of the form tuv, and only those of the form tuv,
have 8 factors, 1,t,u,v,tu,tv,uv and tuv where t, u and v are
3 different primes.
So let D = tuv
Since D = AB/C, C = AB/D = pqrs²/(tuv)
So we have
A = pq we can assume p < q
B = ra²
C = pqrs²/(tuv)
D = tuv we can assume t < u < v
We will try to use only the smallest three primes, {2,3,5} if possible.
If so, then A and B can be made smaller than D which is the product
of three different primes.
We make D as small as possible by choosing t=2, u=3, v=5, so D=30.
We make B as small as possible by choosing r=3, s=2, so B=3×2² = 12.
We would like to make A as small as possible, which would be 6 with p=2 and q=3.
But that won't do, because since v=5, C = pqrs²/(tuv), then pqrs² must
be divisible by 5, and since neither r nor s are 5, the larger of p,q
must be 5, so q=5, and we can choose p=2. So A = pq = 2×5 = 10, and
C = pqrs²/(tuv) = 2×5×3×2²/(2×3×5) = 4
So A+B+C+D = 10+12+4+30 = 56, the smallest possible sum.
Edwin
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