Arick wrote:
P and Q are two two-digit numbersTheir product equals the product of the numbers obtained on reversing them. None of the digits in P or Q is equal to the other digit in it or any digit in the other number. The product of tens digits of the two numbers' is a composite single digit number. How many ordered pairs (P, Q) satisfy these conditions?
(A) 8
(B) 16
(C) 12
(D) 4
(E) 9
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P = xy ⇒ 10x + y
Q = pq ⇒ 10p + q
The reverse numbers are
Reverse(P) = yx = 10y + x
Reverse(Q) = qp = 10q + p
This question can be solved using a two-step process -
- Identify the relationship between x, y, p, and q.
- Identify the pairs of values that satisfy that relationship
Step 1: Identify the relationship between x, y, p, and q.Given: Their product equals the product of the numbers obtained on reversing them
\((10x + y) (10p+q) = (10q + p)(10y +x)\)
\(100xp + 10xq + 10yq + yq= 100qy + 10qx + 10py + px\)
\(99(xp - yq) = 0\)
\(x*p =y*q \)Step 2: Identify the pairs of values that satisfy that relationshipConstriants:
- None of the digits in P or Q is equal to the other digit in it or any digit in the other number
- The product of tens digits of the two numbers' is a composite single digit number
Inferences:
- None of the digits in P or Q is equal to the other digit in it or any digit in the other number
\(x \neq y \neq p \neq q\) - The product of tens digits of the two numbers' is a composite single digit number
\(x *p\) is composite number
If x = 1, p \(\neq\) prime and vice-versa
Let's try to understand the given conditions a bit better, let's say x = 2 and p = 3
2 * 3 = y * q
So (y,q) can be (1,6) or (6,1) but not (2,3) or (3,2). So for one combination of (x,p) we can have four possible values
(x,p) = (2,3)
- (y,p) = (6,1)
xy = 26 | pq = 31
- (y,p) = (1,6)
xy = 21 | pq = 36
(x,p) = (3,2)- (y,p) = (6,1)
xy = 36 | pq = 21
- (y,p) = (1,6)
xy = 31 | pq = 26
Possible values of (x,p) = (1,6), (1,8), (2,3), and (4,2).
Total possible combinations = 4 * 4 =16
Option B
P.S. - This is not a GMAT Style question.