If two 2-digit positive integers have their respective tens digits exc

Hi All,

We're told that if two 2-digit positive integers have their respective TENS digits exchanged, the difference between the pair of integers changes by 4. We're asked for the GREATEST possible difference between the original pair of integers. There are a few different ways to approach this question (and some of them are incredibly "step heavy"). Thankfully, we can solve it rather easily by TESTing THE ANSWERS.

Since we're looking for the GREATEST possible difference between the two 2-digit integers, we should start with Answer E. However, we can use some Number Properties to quickly eliminate a couple of the possibilities first.

The smallest 2-digit number is 10 and the largest is 99, so the difference between those two numbers is 89 (and can only get smaller). Thus, Answers D and E are NOT possible, so we do not have to consider them. Let's TEST Answer C first....

Answer C: 82
Let's pick two 2-digit numbers that have a difference of 82... the easiest would be 10 and 92.
We already know that the difference between these numbers is 82.
When we switch the TENS digits, we have 90 and 12. The difference between these numbers is 90 - 12 = 78.
The two results (82 and 78) differ by 4 - and this is an exact match for what we were told, so this MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

GIVEN:

1. Two 2-digit positive integers.
   
   • Let us suppose the two integers are ‘ab’ and ‘cd’ such that ab > cd.
2. If the tens digits of the two integers are exchanged, the difference between the pair of integers changes by 4. This means that:
   
   • (ab – cd) = (ad – cb) + 4, or
   • (ab – cd) = (ad – cb) – 4. (Because change by 4 can mean either +4 or –4.)

TO FIND:

• Greatest possible value of (ab – cd).

SOLUTION:
Before we move to trying to find (ab – cd), let’s understand some more about the numbers in consideration.

SOME IMPORTANT INFERENCES:
Note that we took ab > cd. This is possible only in two ways:

1. When a > c: In this case, whatever may be the units’ digits, b and d, ab > cd always remains true because of the higher tens digit in ab.
   
   • After interchanging the tens digits, the new numbers, ad and cb, also have a > c, thus, guaranteeing that ad > cb.
2. When a = c and b > d: In this case, the original pair of 2-digit numbers is ab and ad (cd = ad since a = c). This time, interchanging the tens digits will have NO impact since both numbers have the same tens digit, a. - This is IMPOSSIBLE!
   
   • Since the 2-digit numbers did not change, their difference cannot change by 4. And this will contradict the question stem itself.

Now, with this inference let’s start working out the solution.


WORKING OUT:
Our initial pair of integers is ab and cd, and a > c.
In ab, a = tens digit and b = units’ digit. Thus, we can write ab as (10a + b). Similarly, ‘cd’ can be written as (10c + d).

• So, the difference between the initial pair of integers = 10a + b – 10c – d ----(I)

Now, after the exchange of the tens digits, the new pair become ad and cb. These can again be written as (10a + d) and (10c + b), respectively.

• So, the difference between this new pair of integers = 10a + d – 10c – b ----(II)

Now, according to the question: (I) - (II) = +4 or –4. That is,

1. (10a + b – 10c – d) - (10a + d – 10c – b) = 4, ----(III), OR
2. (10a + b – 10c – d) - (10a + d – 10c – b) = - 4 ----(IV)

Let’s solve them one by one.


Analysis from (III):
(10a + b – 10c – d) - (10a + d – 10c – b) = 4
⇒ 10a + b – 10c – d – 10a – d + 10c + b = 4
⇒ 2b – 2d = 4
⇒ b – d = 2 ----(V)

So, a and c can take any values, while b and d will be such that b – d = 2.
Now, recall that we need to find the greatest possible value of (ab – cd), and we just saw that there are no restrictions on a and c:

• To maximize (ab – cd), we need to maximize ab and minimize cd, keeping (V) in mind.
  
  • ab will be maximum when a = 9, and cd will be minimum when c = 1. (Observe, c can’t be zero because cd will become a single digit number in that case.)

Finally, using (I), the required difference = 10a + b – 10c – d = 90 + b – 10 – d = 80 + b – d

• Since (V) gives b – d = 2, the required difference becomes (80 + 2).

Thus, 82 is the maximum possible difference in this case.


Analysis from (IV):
(10a + b – 10c – d) - (10a + d – 10c – b) = - 4
⇒ 10a + b – 10c – d – 10a – d + 10c + b = - 4
⇒ 2b – 2d = - 4
⇒ b – d = - 2 ----(VI)

Doing the same analysis, as we did in the previous case, the greatest possible value of (ab – cd) will be when a = 9 and c = 1.
Finally, using (I), the required difference = 10a + b – 10c – d = 90 + b – 10 – d = 80 + b – d

• Since (VI) gives b – d = -2, the required difference becomes (80 – 2)

So, 78 is the maximum possible difference in this case.


CONCLUSION:
So, overall, the greatest possible difference between ab and cd is 82.

Correct answer: Choice C


Hope this helps!


Best,
Aditi Gupta
Quant expert, e-GMAT

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