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

As for this question, trying different values is the most straightforward approach.

For maximizing most of 2 digit integer questions, you will have numbers closes to 90s and 10s.

Start with 10 and 99 you see that the differences are 89 (=99-10) and 71 (=90-19) giving you a change of 89-71=18, a lot more than 4.

Next check with 15 and 99, you get 84 (=99-15) and 76 (=95-19), you get 84-76=8. Thus the difference has been decreasing.

Finally check with 99 and 17, you get 82 (=99-17) and 78 (=97-19), giving you 82-78 = 4 . This is the answer.

Thus, the maximum difference is 99-17=82. C is the correct answer.

Although it does look like a time consuming approach, once you realize that the difference between the pairs will decrease and that you can not have differences \(\geq\) 90 (as both the integers are 2 digit integers), it become straightforward to find 99 and 17 as the pair.

Hope this helps.

Hi,
you do not have to check too many integers....
when the difference between the pairs decreases/increases by 4, the units digit will have a difference of two.....
so if you are taking 99, only numbers with 7 in it...
so the second integer will be 17...

Hi,
few points to be noted..
1) the largest 2 digit number is 99 and smallest is 10.. so the difference has to be less than 89... D and E will be out
2) now the two 2 digit numbers interchange their tens place... one is 90s and other 10s.. so the number should be greater than 80.... 82 fits in

We may have two choices between 80 and 90, so the method to pinpoint the answer will be...
for max difference , one number should be in 90s and other in 10s..
take the smallest two digit number, 10.. now for the difference in the differences of two numbers to be 4 , the units digit should differ by 2...
if the difference is 6, the units digit will be differ by 3 and so on
so 92 comes as the second number...
diff=92-10=82.... when we interchange the tens digit 90-12=78... so satisfies the condition..
if we take the largest number 99... the other number will be 17.... again diff=82
ans C

Let the number be ab & cd. As per the given question,

10a + b - 10c - d = k
10a + d - 10c - b = k -4

Subtracting above two equations we have,

b - d = 2. Now, In any case the last digit difference will be 2. Only one option is there containing 2 as last digit.

Hence, C.

Am I missing something? The question says they exchange their tens digit. Would that not be their ones digit that is being exchanged?

ScottTargetTestPrep Bunuel

Yes and no. Their ones digits remain the same. Only their tens digits are exchanged. For example, if the old pair of the numbers are 15 and 24, then the new pair of numbers are 25 and 14.

Now, as you can see, one might argue that their ones digits are also exchanged. However, that is only true because their tens digits are exchanged. That is, if you consider that the tens digits are being exchanged, then you must consider that the units digits are being fixed, and vice versa, if you consider that the units digits are being exchanged, then you must consider that the tens digits are being fixed.

I was a bit confused with the last step here.

What are the constraints on the numbers we must choose to determine that the original difference would be 78?

I also solved as (10a + b) - (10c +d) = 10c+b - (10a+d) + 4 to produce (10a -10c) = 2

According to our calculations, as long as the difference between the units digits is 2, the condition of "the difference between the original difference and the difference of the exchanged numbers is 4" is satisfied. That's the constraint of the units digit. For instance, 75 and 43 also satisfy that condition, but those numbers will not give you the greatest possible difference. In order to find "the greatest possible difference", the tens digit of the greater number must be 9 and the tens digit of the smaller number must be 1. As long as these two consitions are satisfied, you can choose any numbers you want (such as 92 and 10, 93 and 11, 94 and 12 etc.) The original difference is always 82 and the new difference is always 78.

In your equation, if 10a + b is the greater number, then after the switch is made, 10a + d will be the greater number; hence the right hand side of your equation should be (10a + d) - (10c + b) + 4.

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

Went with option elimination here

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

E - If 94 is the difference : Max one number can be 99 and other will be 5 which is not two digit. Eliminate

D. If 90 is the difference: Max one number can be 99 and other will be 9 which is not two digit. Eliminate

C. If 82 is the difference: Max one number can be 99 and other will be 17. Reversing tens digit will give difference of 78 which is 4 less than original. Select and move on.

This can be helpful when approaches dont click