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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?
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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
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eybrj2
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)

let no be 1,2,3
LCM ; 1*2*3 ; 6
n=1
option A ; is valid
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eybrj2
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)

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
Archit3110
eybrj2
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)

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

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)
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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.
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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
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­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:


  1. n: Since n is odd, it does not include the prime factor 2.
  2. n+1: Since n is odd, n+1 is even. Therefore, n+1 includes the prime factor 2.
  3. 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:


  1. If n or n+2 is a multiple of 3, the highest power of 3 will be included.
  2. If n is odd, then one of n or n+2 might have additional factors which could reduce to lower powers.
  3. 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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