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03 Aug 2020, 18:54
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
23% (01:59) correct
77%
(02:01)
wrong
based on 112
sessions
History
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Find the number of ordered quadruples of positive integers (a,b,c,d) satisfying \(a^3b-ab^3 = cd\), in which c and d are prime numbers.
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
04 Aug 2020, 09:21
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nick1816
\(a^3b-ab^3 = cd\)
i.e. \(ab(a^2-b^2) = cd\)
i.e. \(ab(a-b)(a+b) = cd\)
i.e. a-b must be 1 for the condition to be true
i.e. a=2 and b=1 is the only case that works
a+b = 2+1 = 3
a-b = 2-1 = 2
a*b = 2*1 = 2
i.e. i.e. \(ab(a-b)(a+b) = 2*1*(1)*(3) = cd\)
now, c and d may be (2,3) or (3,2)
Answer: Option C
04 Aug 2020, 20:27
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nick1816
\(a^3b-ab^3 = cd\)
\(ab*(a^2-b^2) = cd\)
\(ab*(a-b)(a+b) = cd\)
We want c*d to be two primes multiplied, so the left side should be something like 2*3, 3*5 etc. However on the left side, we seem to have 4 factors. To "eliminate" the factors, we have to set some of them as 1. First, note that a > b since we have (a - b) as a "factor". b = 1 will eliminate the "factor" of b. Next, we can let \(a - b = 1\) so that \(a - b\) is not a real factor either. With that, we confirm b = 1 and a = 2. Check that we are left with 2 * 3 on the left side which is only two prime factors, and other numbers would create too many factors.
Finally, we confirm c*d = 6 but we can have either c = 2, d = 3 or c = 3, d = 2, which is two different pairings in an ordered set. Thus the answer is 2.
Ans: C
