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

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

by Max@Math Revolution » Thu Mar 08, 2018 1:17 am
[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
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Source: — Data Sufficiency |
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by GMATinsight » Fri Mar 09, 2018 6:23 am
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

Answer: option D
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by Vincen » Fri Mar 09, 2018 7:48 am
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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by Max@Math Revolution » Sun Mar 11, 2018 5:23 pm
=>
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.

Therefore, D is the answer.

Answer: D

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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