Nine two-digit positive integers form an arithmetic progression. When

Question

Nine two-digit positive integers form an arithmetic progression. When the digits of each integer are reversed, the sum of the integers remains the same. The sum is

A. 481
B. 486
C. 490
D. 495
E. 499

WiziusCareers1 · Accepted Answer

Nine two-digit positive integers form an arithmetic progression. When the digits of each integer are reversed, the sum of the integers remains the same. The sum is

A. 481
B. 486
C. 490
D. 495
E. 499

Alternate approach:
As the sum remains the same when the digits are reversed, think about the numbers which have the same tens and units digits.
Let the AP be 11, 22, 33, 44, 55, 66, 77, 88, 99
The sum of the AP = 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 = 495

The answer is (D).

WiziusCareers1 · Answer

Nine two-digit positive integers form an arithmetic progression. When the digits of each integer are reversed, the sum of the integers remains the same. The sum is

A. 481
B. 486
C. 490
D. 495
E. 499

Sol: 
To answer such a question, use answer choices. 
As the nine numbers are in an AP, the mean of the numbers must equal the median of the 9 numbers.

Let's check the mean of each the answer choices
A) 481/9 = 53.44 (Not possible as the median)
B) 486/9 = 54
C) 490/9 = 54.44 (Not possible as the median)
D) 495/9 = 55
E) 499/9 = 55.44 (Not possible as the median)

So, the median is either 54 or 55.
Let's try 54
Hence, the 9 numbers could be
50 51 52 53 54 55 56 57 58
The sum of these numbers = 486
When the digits are reversed
05 15 25 35 45 55 65 75 85
The sum of these numbers is 405
Not equal

Hence, the answer must be (D).
If the median is 55
then a possible set of numbers is 
51 52 53 54 55 56 57 58 59
The sum is 495

When the digits are reversed
15 25 35 45 55 65 75 85 95
The sum is 495

fmcgee1 · Answer

10A + B + (10A + 1 + B + 1) + (10A + 2 + B + 2).... (10A + 8 + B + 8) = 10B + A + (10B + 1 + A + 1) + (10B + 2 + A + 2)... (10B + 8 + A + 8)

9(10A) + 9B + 72 = 9(10B) + 9A + 72

90A + 9B = 90B + 9A

81A = 81B

A = B

This means that each number has the same tens and units digits.

So the progression can only be 11, 22, 33, 44, 55, 66, 77, 88, 99

= 495

MathRevolution · Answer

Let 'a' be the ten's digit and 'b' be the unit's digit: 2-digit number: 10a + b

Nine two-digits positive integers: 10a + b ,10 (a+1) + (b+1) ,........, 10 (a+8) + (b+8)

Digits reversed : 10b + a ,10 (b+1) + (a+1) ,........, 10 (b+8) + (a+8)

Both sum equal:

=> 10a + b + 10 (a+1) + (b+1) +........+ 10 (a+8) + (b+8) = 10b + a + 10 (b+1) + (a+1) +........+ 10 (b+8) + (a+8)

=> 9(10a) + 9b + 72 = 9(10b) + 9A + 72

=> 90a + 9b + 72 = 90b + 9a + 72

=> 81a = 81b

=> a = b

This means both digits should be the same.

A.P. : 11, 22, 33, 44, 55, 66, 77, 88, 99.

Sum:  \frac{n}{ 2}

=> Sum:  \frac{9}{2}

=> Sum:  \frac{9}{2}  *

=> Sum: 9 * 55 = 495

Answer D

chacinluis · Answer

Not really. Once you are told that if you flip the digits the sum of all the terms should remain the same it should pop up immediately that when the digit repeats flipping will not change the sum b/c the number remain unchanged. Also it would be an arithmetic progression b/c the difference is always 10.

kajaldaryani46 · Answer

—-
Hi,

Im not sure why we’re taking the common difference as 1? It could be 10, 20 or any other number right? Also, we wouldn’t know that when the digits are reversed, the two digits have to be the same because of rhe following scenario:

What if the two digits are AB and the first AP is:
AB, AB+d, AB+2d...AB+8d (d is the common difference)
The next AP would be:
BA, BA+s, BA+2s,....BA+8s (s is the common difference that would not be equal to d here, )- this would be the case where A and B are two different digits

I setup the sum of both of these APs and equated them and so couldnt get anywhere with it. But I think the key question here is how do we determine that both the digits are equal?

Posted from my mobile device

bshn · Answer

I imagined the series to be 19, 28, 37, 46, 55, 64, 73, 82, 91. (When reversed the first number becomes last and your sum remains same)
Since this is AP with a = 19, d=9 and then calculate Sum of AP = 495.

Brindac2 · Answer

The solution is to simply divide each number from the option with 11 and the only number which is divisible by 11 is 495. So, that's the answer.

Now the logic of dividing by 11 is as below.

Any 2 digit number N=10x+y
If you reverse the order of digits in N, it becomes, 10y+x

The question tells that these 9 numbers first they took the sum of 9 number in original order and then reverse order.

So, 10x+y+10y+x=11x+11y

So the sum must be divisible by 11.

It can also be true if you divide by 9

Same logic

10x-x-10y-y=9x-9y

Adit_ · Answer

The sum is not necessarily 495 though. I can also have 9 11s or 9 of the same 2 digit numbers as part of the set too right? d=0 is also considered an AP or am I missing something?

Bunuel · Answer

Yes, the question should have mentioned  different integers .

Nine two-digit positive integers form an arithmetic progression. When

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Nine two-digit positive integers form an arithmetic progression. When

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