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Math Expert V
Joined: 02 Sep 2009
Posts: 55277
N is a two-digit number. The sum of its digits is S and the product of  [#permalink]

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N is a two-digit number. The sum of its digits is S and the product of its digits is P. What is the largest possible value of N?

(1) N + S = 103
(2) 2N = 2S + 9P

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Re: N is a two-digit number. The sum of its digits is S and the product of  [#permalink]

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Let's take the number N as:

N = 10x+y
Hence,
S = x+y
P = xy

Statement 1:

N+S = 103
10x+y+x+y = 103
11x+2y = 103

Not sufficient

Statement 2:
2N = 2S +9P
2(10x+y) = 2(x+y) +9(xy)
solving this gives
x(18-9y) = 0

now either x = 0 or 18-9y = 0

x cannot be zero.
hence 18-9y = 0 or y = 2

But we do not know the value of x. So the number could be anything 12, 22,32, etc

But we need the largest possible value. Hence x has to be equal to 9. So the two digit number becomes 92.

Sufficient .

Answer is B
Intern  B
Joined: 18 May 2017
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N is a two-digit number. The sum of its digits is S and the product of  [#permalink]

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I think the answer is D.

Assume that N=ab. So from the question itself we can draw two equations: (1) a + b = s; (2) a x b = p. Now let's check the statements.

First statement: This statement means that the number itself (ab) + the sum of his digit (a+b) equal to 103. As we asked for the largest N i began with the largest two digit number (99) and checked whether it satisfied the statement (and if not i go backwards). After several trial and error you'll find that the largest two digit number that satisfied the statement is 92 (92 + 11 = 103). Any two digit number larger than 92 will not satisfy the statement so we can conclude that 92 is the largest N possible. Sufficient.

Second statement: 2n = 2s + 9p ----> 2(10a + b) = 2(a+b) + 9 x a x b ----> 18a-9ab=0 ----> 9a (2-b)=0. So either 9a=0------->a=0 or 2-b=0 ------> b=2. As N is a two digit number a cannot be equal to zero (as it is the first digit) and thus the only option is that b=2. The largest number that his units digit is 2 is 92. Sufficient.

Answer: D
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1530
N is a two-digit number. The sum of its digits is S and the product of  [#permalink]

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What is the source? The question makes no logical sense as a DS question. If I ask this question:

N is a positive two digit integer. What is the maximum possible value of N?
1. N is divisible by 5
2. N is divisible by 7

Then using no statements, the answer is 99, using S1, the answer is 95, using S2, the answer is 98, and using both statements the answer is 70. So what is the answer to this DS question? Is it A, B, C, D? There's no way to decide, because the question makes no logical sense.

When this question begins "N is a two digit positive integer", then N stands for a single value. It is not a variable that can have a maximum or minimum value, because there would be no way to decide what would constitute sufficient information - the more information you use, the more the maximum or minimum would be constrained. And really no matter what information you have, there will always be some maximum possible value of N (even with no information at all, the maximum is 99) so the answer is D without even looking at the statements.

If the question instead asked "what is the value of N", it becomes a more interesting question, and then the answer is A. The approach in the first reply above is good, though the analysis of S1 is incomplete - if our two digit number is XY, where X is the tens digit, then S1 tells us that

10X + Y + X + Y = 103
11X + 2Y = 103

Now Y is a digit, so it cannot be greater than 9. So 11X must be 88 or 99, or the sum can't possibly work (2Y couldn't be big enough), and if X = 8, you find that Y is a decimal (Y is 7.5) which is impossible for a digit. So the only solution is N = 92.

A different way to look at it: if N+S = 103, then since S < 20 (since S is the sum of two single digit numbers), N must be in the 80s or in the 90s. But if you take any number N in the 80s, then if N is even, the sum S of its digits is even, and if N is odd, the sum S of its digits is odd. So N+S will be even+even or odd+odd, and if N is in the 80s, N+S will always be even. N+S could never equal 103, so the only possibility is that N is in the 90s, and in that case N+S = 103 can only have one solution.

Statement 2 doesn't let you solve for N - as the above posts demonstrate, it only lets you prove the units digit is 2.
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Math Expert V
Joined: 02 Sep 2009
Posts: 55277
Re: N is a two-digit number. The sum of its digits is S and the product of  [#permalink]

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IanStewart wrote:
What is the source? The question makes no logical sense as a DS question. If I ask this question:

N is a positive two digit integer. What is the maximum possible value of N?
1. N is divisible by 5
2. N is divisible by 7

Then using no statements, the answer is 99, using S1, the answer is 95, using S2, the answer is 98, and using both statements the answer is 70. So what is the answer to this DS question? Is it A, B, C, D? There's no way to decide, because the question makes no logical sense.

When this question begins "N is a two digit positive integer", then N stands for a single value. It is not a variable that can have a maximum or minimum value, because there would be no way to decide what would constitute sufficient information - the more information you use, the more the maximum or minimum would be constrained. And really no matter what information you have, there will always be some maximum possible value of N (even with no information at all, the maximum is 99) so the answer is D without even looking at the statements.

If the question instead asked "what is the value of N", it becomes a more interesting question, and then the answer is A. The approach in the first reply above is good, though the analysis of S1 is incomplete - if our two digit number is XY, where X is the tens digit, then S1 tells us that

10X + Y + X + Y = 103
11X + 2Y = 103

Now Y is a digit, so it cannot be greater than 9. So 11X must be 88 or 99, or the sum can't possibly work (2Y couldn't be big enough), and if X = 8, you find that Y is a decimal (Y is 7.5) which is impossible for a digit. So the only solution is N = 92.

A different way to look at it: if N+S = 103, then since S < 20 (since S is the sum of two single digit numbers), N must be in the 80s or in the 90s. But if you take any number N in the 80s, then if N is even, the sum S of its digits is even, and if N is odd, the sum S of its digits is odd. So N+S will be even+even or odd+odd, and if N is in the 80s, N+S will always be even. N+S could never equal 103, so the only possibility is that N is in the 90s, and in that case N+S = 103 can only have one solution.

Statement 2 doesn't let you solve for N - as the above posts demonstrate, it only lets you prove the units digit is 2.

The source is Manhattan Review. Thank you for the feedback, Ian. The question is archived.

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________ Re: N is a two-digit number. The sum of its digits is S and the product of   [#permalink] 11 Jul 2017, 07:26
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