adthedaddy wrote:
If two 2-digit positive integers have their respective tens digits exchanged, the difference between the pair of integers changes by 4. What is the greatest possible difference between the original pair of integers?
A) 76
B) 80
C) 82
D) 90
E) 94
Please help solve this question
Hi adthedaddy,
I am going to show an algebraic solution to this question.
Let's say x,y,z,w represent digits. Let's say x and z both represent tens digit with values between 1 and 9 inclusive, since each number has to be 2 digit numbers, and let's say x>z. Let's say y and w represent ones digit with values between 0 and 9 inclusive. Let's say the original numbers are xy and zw, so we are trying to maximize the difference between
xy-zw or
10(x-z)+y-w=DIFFERENCE. When we reverse the tens digit of the two numbers we have zy and xw; since we want to make sure our new difference changes by 4, we have to make sure the new difference isn't negative (The difference between a negative integer and positive integer would be more than 4 in this problem). In other words we are basically changing the units digit.
So
10(x-z)+w-y=DIFFERENCE+4 or
10(x-z)+w-y=DIFFERENCE-4Now when you subtract
10(x-z)+w-y=DIFFERENCE+4 from
10(x-z)+y-w=DIFFERENCE, you are left with 2*(y-w)=-4. Subtracting the other equation from
10(x-z)+y-w=DIFFERENCE yields 2*(y-w)=4. In other words, when we want the difference to be 4 between the difference of the original set of numbers and the difference of the new set of numbers, the
absolute difference between y and w should be 2 (y-w=2 or y-w=-2). But since we are trying to maximize the difference variable in
10(x-z)+y-w=DIFFERENCE, y-w should equal 2. x can have a maximum value of 9 and z can have a minimum value of 1. So after plugging in back to that equation, you get
10*(9-1)+2=82 Correct answer choice is
C