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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Is at least one of 3 consecutive integers a multiple of 5? tagged by: Max@Math Revolution ##### This topic has 1 expert reply and 2 member replies ### GMAT/MBA Expert ## Is at least one of 3 consecutive integers a multiple of 5? [GMAT math practice question] Is at least one of 3 consecutive integers a multiple of 5? 1) The sum of the integers is divisible by 5 2) The product of the integers is divisible by 5 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$99 for 3 month Online Course
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Max@Math Revolution wrote:
[GMAT math practice question]

Is at least one of 3 consecutive integers a multiple of 5?

1) The sum of the integers is divisible by 5
2) The product of the integers is divisible by 5
Question : Is at least one of 3 consecutive integers a multiple of 5?

Statement 1: The sum of the integers is divisible by 5

For any Arithmetic Progression (Sequence in which terms are equally spaced)

Mean = Median = (First Term+Last Term)/2

The addition of 3 consecutive integers = 3* Middle Term = Multiple of 5 (As given in statement)

i.e. Middle term must be a multiple of 5 for this to be true

SUFFICIENT

Statement 2: The product of the integers is divisible by 5

For product of Integers to be a multiple of 5, one of those 3 consecutive integers must be a multiple of 5

SUFFICIENT

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I think that the correct option is D.

(1) The sum of the integers is divisible by 5.

Let's take 3 consecutive integers k, k+1 and k+2. Hence $$\frac{k+k+1+k+2}{5}=\frac{3k+3}{5}=\frac{3\left(k+1\right)}{5}=T\ \ \Leftrightarrow\ \ 3\left(k+1\right)=5\cdot T,\$$ where T is an integer.
Now, since 3(k+1) is a multiple of 5, then k+1 must be multiple of 5. Therefore, one of the 3 integers is multiple of 5. SUFFICIENT.

(2) The product of the integers is divisible by 5.

We have that $$k\cdot\left(k+1\right)\cdot\left(k+2\right)=5\cdot T,\$$ where T is an integer. Therefore, one of the 3 integers must be multiple of 5. SUFFICIENT.

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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.

Write the 3 consecutive integers as n-1, n, n+1.

Since we have 1 variable (n) 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)
Since ( n - 1 ) + n + ( n + 1 ) = 3n is divisible by 5, and 3 and 5 are prime numbers, n is a multiple of 5, since 3 is not divisible by 5.
Condition 1) is sufficient.

Condition 2)
Since (n-1)n(n+1) is divisible by 5 and 5 is a prime number, one of n-1, n and n+1 is a multiple of 5.
Condition 2) is sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

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