Find the number of ordered quadruples of positive integers (a,b,c,d) s

Question

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

GMATinsight · Answer

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

TestPrepUnlimited · Answer

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

Anuraagsaboo1 · Answer

Just one question - how do we ensure no other combination exists?

sambitspm · Answer

Kindly see the attachment.

JUST MAKE SURE THAT YOU CONSIDER COMBINATIONS in sets.

a>b needs to be as c and d are prime.
c and d can be anything.

2,1,2,3
2,1,3,2

Fdambro294 · Answer

1st) Factor out the Equation and Simplify the Expression

(a^3)*b - a*(b^3) = c * d ------ where c and d are 2 Prime Bases

Looking at the Answer Choices, there are not many possibilities we can have. Notice that the question asks for "Ordered Solutions" for the quadruple (a, b, c , d)

In this case, the order of the Solutions Matters. If we get a result that is (c * d) then we must necessarily have 2 Ways to write that Solution:

(a , b , c , d) or (a , b , d , c)

Taking a*b as Common from the Expression:

ab * (a^2 - b^2) = c * d

ab * (a + b) * (a - b) = c * d

2nd) Analysis of the Equation

Since the R.H.S. is the product of 2 Prime Nos., the L.H.S. can NOT produce a (-)Negative Result (Prime Numbers are never (-)Negative)

Therefore ---- a > b

Testing the Smallest Possible (+)Positive Integers possible:

Case 1:
a = 2 ---- b = 1

2 * 1 * (2 + 1) * (2 - 1) = 
2 * 3 * 1 = 2 * 3

or

2 * 3 * 1 = 3 * 2

1st Solution to (a, b, c, d) ----- (a=2 --- b=1 --- c=2 --- d=3)
2nd Solution to (a, b, c, d) ---- (a =2 --- b=1---- c=3 ---- d=2)

if we try any larger numbers, such as: a= 3 ---- b = 1

the Factor of (a + b) = 4

and the Factor of a*b = 3

This will NEVER Result in the Product of 2 Prime Bases = c * d

Therefore, there is only 2 Solutions

-C-

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Find the number of ordered quadruples of positive integers (a,b,c,d) s

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Find the number of ordered quadruples of positive integers (a,b,c,d) s

Prep Toolkit

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