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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
Each of three consecutive positive integers is less than 100. Their su  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 46% (03:07) correct 54% (02:32) wrong based on 88 sessions

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[GMAT math practice question]

Each of three consecutive positive integers is less than 100. Their sum of is a multiple of 10. What is the smallest of the three integers?

1) Their median is a multiple of 9
2) Two of the integers are prime numbers.

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Manager  B
Joined: 26 Sep 2017
Posts: 95
Re: Each of three consecutive positive integers is less than 100. Their su  [#permalink]

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89 90 91 is the only combination

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Retired Moderator P
Joined: 22 Aug 2013
Posts: 1417
Location: India
Re: Each of three consecutive positive integers is less than 100. Their su  [#permalink]

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MathRevolution wrote:
[GMAT math practice question]

Each of three consecutive positive integers is less than 100. Their sum of is a multiple of 10. What is the smallest of the three integers?

1) Their median is a multiple of 9
2) Two of the integers are prime numbers.

(1) Let the median integer be 9x, where x is a positive integer. So these three consecutive integers are: 9x-1, 9x, 9x+1. Their sum = 9x-1 + 9x + 9x+1 = 27x. This is given to be a multiple of 10. For 27x to be a multiple of 10, x has to be a multiple of 10. So, in this case, 9x can only be 90 and nothing else (all three numbers have to be less than 100). So these three integers are 89, 90, 91. Sufficient.

(2) Two are prime. We can have 29,30,31 as three consecutive integers (29 and 31 are prime). We could also have 59,60,61 as three consecutive integers (59 and 61 are prime). More than one cases, so not sufficient.

Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: Each of three consecutive positive integers is less than 100. Their su  [#permalink]

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Let the integers be x – 1, x and x + 1 where x is a positive integer. Since their sum, x – 1 + x + x + 1 = 3x is a multiple of 10, x must be a multiple of 10.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
The median, x, of the three integers is a multiple of 9 and a multiple of 10.
Since x < 100, we must have x = 90.
Therefore, the smallest of the integers is x – 1 = 89.
Thus, condition 1) is sufficient.

Condition 2)
If x = 30, the integers are 29, 30, 31 and the smallest integer is 29.
If x = 90, the integers are 89, 90, 91 and the smallest integer is 89.
Since we do not obtain a unique answer, condition 2) is not sufficient.

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Director  P
Joined: 02 Oct 2017
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Re: Each of three consecutive positive integers is less than 100. Their su  [#permalink]

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I) sum of three number is 10 for consecutive can happen when one is multiple of 10 and other numbers contain 1 and 9 in unit place
i.e. 9,10,11
19,20,21etc

So if median is multiple of 9
Then only pair qualify is
89,90,91
Sufficient

II) many pair come under this

29,30,31
59,60,61
N others as well
Multiple cases so insufficient

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Retired Moderator V
Joined: 27 Oct 2017
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Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: Each of three consecutive positive integers is less than 100. Their su  [#permalink]

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1
MathRevolution,
As per second statement 89,90,91 can not be required the integers as the two numbers are not prime.
Only 89 is prime.
91 is 7*13.
This seems a flawed question.

MathRevolution wrote:
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Let the integers be x – 1, x and x + 1 where x is a positive integer. Since their sum, x – 1 + x + x + 1 = 3x is a multiple of 10, x must be a multiple of 10.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
The median, x, of the three integers is a multiple of 9 and a multiple of 10.
Since x < 100, we must have x = 90.
Therefore, the smallest of the integers is x – 1 = 89.
Thus, condition 1) is sufficient.

Condition 2)
If x = 30, the integers are 29, 30, 31 and the smallest integer is 29.
If x = 90, the integers are 89, 90, 91 and the smallest integer is 89.
Since we do not obtain a unique answer, condition 2) is not sufficient.

_________________ Re: Each of three consecutive positive integers is less than 100. Their su   [#permalink] 14 Oct 2018, 10:17
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