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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?  [#permalink]

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New post 08 Mar 2018, 01:24
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[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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Is at least one of 3 consecutive integers a multiple of 5?  [#permalink]

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New post 08 Mar 2018, 01:39
MathRevolution 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


Problems invoving consecutive integers can often be solved with number properties and very little calculation
This is a Logical approach.

Say our 3 consecutive integers have remainder r, r+1, r+2 when divided by 5.
(that is, they can be [0,1,2] [1,2,3] [2,3,4] [3,4,0] or [4,0,1])

(1) since the sum of the 3 integers is divisible by 5, then the sum of the 3 remainders has to be equal to 5.
Looking at our possibilites above, only 4,0,1 works out. That is, the only time 3 consecutive integers have a sum that is divisible by 5 is when the middle number is divisible by 5.
Then our answer is YES.
Sufficient!

(2) If the product of 3 numbers is divisible by 5 then one of the 3 numbers must have 5 as its prime factor.
Once again, our answer is YES.
Sufficient.

(D) is our answer.
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Re: Is at least one of 3 consecutive integers a multiple of 5?  [#permalink]

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New post 11 Mar 2018, 17:29
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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.

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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Re: Is at least one of 3 consecutive integers a multiple of 5?  [#permalink]

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New post 01 Apr 2018, 04:24
1
Let 3 cons. integers be n , n+1, n+2
1) n+(n+1)+(n+2) is divisible by 5
3n+3 is divisible by 5
3(n+1) to be divisible by 5, (n+1) needs to be divisible by 5
Statement 1 is Suff.

2) n*(n+1)*(n+2) is divisible by 5, so it means at least on integer is divisible by 5.
Statement 2 is Suff.
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Re: Is at least one of 3 consecutive integers a multiple of 5? &nbs [#permalink] 01 Apr 2018, 04:24
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