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What is the sum of the digits of two-digit positive integer M? (1) If

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What is the sum of the digits of two-digit positive integer M? (1) If  [#permalink]

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New post 22 Apr 2015, 03:35
2
9
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

43% (02:21) correct 57% (02:15) wrong based on 261 sessions

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Re: What is the sum of the digits of two-digit positive integer M? (1) If  [#permalink]

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New post 22 Apr 2015, 03:53
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Bunuel wrote:
What is the sum of the digits of two-digit positive integer M?

(1) If the digits of M are reversed, the resulting integer is 27 greater than M.

(2) The difference between the squares of the digits of M is 39.


Kudos for a correct solution.


1) let's our number be \(ab\) (a = tens and b = units), than we can write this number as \(a*10 + b*1\)
and as we know that difference \(ba-ab = 27\) we can write this statement in such way:
\((b*10+a)-(a*10+b) = 27\)
\(9b-9a=27\)
\(b-a=3\)
and we see that there is more than one combination: \(4-1\); \(5-2\) etc
Insufficient

2) Picking numbers which squares more than 39: 49, 64, 81 the only possible difference of squares that gives us result \(39\) is \(64-25=39\)
Sufficient

And answer is B
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Re: What is the sum of the digits of two-digit positive integer M? (1) If  [#permalink]

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New post 22 Apr 2015, 05:07
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What is the sum of the digits of two-digit positive integer M?

(1) If the digits of M are reversed, the resulting integer is 27 greater than M.

(2) The difference between the squares of the digits of M is 39.


Took me 2 minutes 50 seconds, but what the heck.

1) [x] [y] + 27 = [y] [x] where x and y represent digits of a two number digits
I thought x was 0 and y was 3, 3+27 = 30, y = 3 and x = 0. But after working on the second statement, I realised 5 and 8 was also a possibility.

I.

2) I just wrote down the squares of all integers from 1 - 9
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81

The only possibility is 64 - 25 = 39. S.

B.
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Re: What is the sum of the digits of two-digit positive integer M? (1) If  [#permalink]

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New post 23 Apr 2015, 14:50
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Hi All,

This DS question can be solved with a combination of Number Properties and TESTing VALUES.

We're told that M is a two-digit positive integer. We're asked for the SUM of the two digits.

Fact 1: If the digits of M are reversed, the resulting integer is 27 greater than M.

This Fact is actually based on an 'accounting error' that happens when digits are accidentally reversed. You'll find that it's not too difficult to find a few examples that differ by 27.

IF...
M is 41
41 - 14 = 27
The answer to the question is 4+1=5

IF....
M is 52
52 - 25 = 27
The answer to the question is 5+2=7
Fact 1 is INSUFFICIENT

Fact 2: The difference between the squares of the digits of M is 39.

Since we're dealing with digits, there are a finite number of values that we need to consider:

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81

We have to 'play around' a bit with the values, but for the DIFFERENCE in the digits to = 39, the ONLY values that fit are 8 and 5 (64-25 = 39).

Whether M is 58 or 85, the answer to the question is the SAME: 5+8=13. The answer is ALWAYS 13.
Fact 2 is SUFFICIENT

Final Answer:

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Re: What is the sum of the digits of two-digit positive integer M? (1) If  [#permalink]

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New post 27 Apr 2015, 03:01
Bunuel wrote:
What is the sum of the digits of two-digit positive integer M?

(1) If the digits of M are reversed, the resulting integer is 27 greater than M.

(2) The difference between the squares of the digits of M is 39.


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Let’s call the digits A and B. If M is written as two-digit number AB, then algebraically M may be expressed as M =10A+B. The question asks for the value of A+B.

Statement (1) reverses the digits of M. This may be written as two-digit number BA, expressible algebraically as 10B+A. Combining this expression with the previous expression and the given information from statement (1), we get the equation 10B+A=10A+B+27. Moving the variables to the left, this becomes 9B–9A=27, or B–A=3. This is nice, but it’s not sufficient to solve for A+B.

Statement (2) can be written algebraically as A^2–B2= ±39. This is a common algebraic equation, the difference of squares, and it factors as (A+B)(A–B) = ±39. Since A and B are integers, (A+B) and (A–B) must be integer factors of ±39. Putting aside any possible negative side for a moment, the options are 39 * 1 and 13 * 3. But since A and B are digits, they cannot add or subtract to ±39, so we must be looking at (A+B)(A–B) = ±13 * 3. And since A and B are positive and since no two digits can be 13 apart, we must have A+B=13 and A–B = ±3. At this point it doesn’t really matter whether A–B is 3 or –3, since the question asks only for A+B. We know that this equals 13, so statement (2) alone is sufficient.
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Re: What is the sum of the digits of two-digit positive integer M? (1) If  [#permalink]

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New post 07 Sep 2017, 04:11
Ans is B:
1) ab is number
10b+a- 10a-b =27
9(b-a) =27
b-a =3
too many numbers AD eliminate

2) Diff in squares of digit =39
|b^2-a^2|=39
8^2-5^2 =39
20^2-19^2 =39
but a and b cant be 19 and 20 so 5 and 8 will be the digits.
58 is the number or 85 , sum =13.
B is Answer
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Re: What is the sum of the digits of two-digit positive integer M? (1) If   [#permalink] 07 Sep 2017, 04:11
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