If n is the least of 3 consecutive positive integers and

Question

If n is the least of 3 consecutive positive integers and n is odd, what is the least common multiple of the 3 integers, in terms of n?

A. n(n+1)(n+2)
B. n^3 + 3
C. n(n^2 + 2)
D. n^2 + 3
E. 3(n +1)

Bunuel · Accepted Answer

Two consecutive integers are co-prime, meaning that they do not share any common factor but 1. For example 4 and 5 are consecutive integers and they do not share any common factor but 1. This means that the leas common multiple of two consecutive integers is always their product: LCM(4, 5)=20.

So, for n, n+1 and n+2 each pair consecutive terms is co-prime (n and n+1, as well as n+1 and n+2). Now, since n is odd, then n and n+2 are also co-prime (consecutive odd integers are also co-prime: 3 and 5, 5 and 7, 7 and 9, ...), which means that all 3 integers are co-prime to each other, hence the least common multiple of the 3 integers is n(n+1)(n+2).

Answer: A.

Else you can pick some number for n, say 3. Then three consecutive integers will be 3, 4, and 5, their LCM is 3*4*5=60. Now, just plug 3 in the answer choices to see which yields 60: only A.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only. For example if you pick n=1 then you get two "correct" choices A and E.

Hope it helps.

stonecold · Answer

Excellent Question
here i choose N=1 and discarded B,C,D then choose N=3 and Marked A as the option.
Nice One

sjackson411 · Answer

It seems like A is the answer whether N is even or odd. Are they just throwing in that N is odd to throw us off?

ArjunJag1328 · Answer

n = odd & least of 3 consecutive positive integers(n,n+1,n+2)

I took n as 3 and the other 2 numbers are 4 & 5

LCM of 3,4,5 = 60

Plug in 3 on each options and the answer should equal to LCM.

A. n(n+1)(n+2)
=>3.4.5
=>60

Hence A is the answer

Archit3110 · Answer

let no be 1,2,3
LCM ; 1*2*3 ; 6
n=1
option A ; is valid

Abhishek009 · Answer

Plug in any value and check , let the 3 nos be 1 , 2 & 3

LCM of ( 1,2,3) = 6

Check the options -

(A) 1*2*3 = 6 (Possible)
(B) 1^3 + 3 = 4 (Not possible)
(C) 1(1^2 + 2) = 3 (Not possible)
(D) 1^2 + 3 = 4 (Not possible)
(E) 3 ( 1 + 1 ) = 6 (POssible)

Now check with

Here is why we need tp check all the options carefully as (E) will yeild the same result in this case of 1 ,2 & 3 and hence we need to plug in once again to confirm with 3,4 & 5 as  ArjunJag1328 
has done...

Correct Answer will definitely be (A)

dcummins · Answer

Testing x=1 will eliminate answers B,C,D but A and E will both produce the same result, so test x=3 to determine that A is correct.

CrackverbalGMAT · Answer

Since the smallest of the 3 consecutive positive integers is odd, the 3 consecutive integers will be co-prime to each other. 
Therefore, LCM of the 3 consecutive positive integers = n (n+1) (n+2).

We can take a few examples to be sure.
If n =1, n+1 = 2 and n+2 = 3. LCM = 6
If n = 3, n+1 = 4 and n+2 = 5. LCM = 60.

The correct answer option is A.

This can now become a property that you can add to the repository of Number properties.

Hope that helps!
Aravind BT

siddhantvarma · Answer

Given three consecutive positive integers where n is the smallest and is odd, we need to determine the least common multiple (LCM) of these three integers in terms of n.

Let the three consecutive integers be n, n+1, and n+2.

First, let's consider the prime factorizations of these three integers:

n : Since n is odd, it does not include the prime factor 2. 
  n+1 : Since n is odd, n+1 is even. Therefore, n+1 includes the prime factor 2. 
  n+2 : Since n is odd, n+2 is odd and does not include the prime factor 2.  
For the least common multiple, we need the highest power of all prime factors that appear in the factorizations of n, n+1, and n+2

LCM:

The prime factor 2 will appear because n+1 is even. 
 Any odd prime factors will come from n, n+1, and n+2.  
Let's look at the product of n, n+1, and n+2:

n(n+1)(n+2)Since n, n+1, and n+2 are three consecutive numbers, their product includes all necessary factors, but we are interested in the LCM, not just the product.

Simplification:

We should check if there is any redundancy in the factors:

If n or n+2 is a multiple of 3, the highest power of 3 will be included. 
 If n is odd, then one of n or n+2 might have additional factors which could reduce to lower powers. 
 The number n+1 will include the factor 2 and other odd primes.  
Because n is odd and n is not a multiple of 2, n+1 is the only even number contributing the highest power of 2.

Conclusion:

The LCM of n, n+1, and n+2 is the product of the highest powers of all prime factors in these three numbers. Given the properties of consecutive numbers, there are no common factors beyond those explicitly present in each number.

Thus, the LCM is:

n(n+1)(n+2)This ensures we include all unique prime factors at their highest powers.

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If n is the least of 3 consecutive positive integers and

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