Lowest least common multiple

The lowest common multiple of the first 10 positive integers is:

$2^3 3^2 5^1 7^1$

This takes into account the highest powers of each prime number that is part of the sequence (1 - 10). Now we must choose an element to remove. 7 is optimal because it will reduce the LCM by a factor of 7, whereas removing any other number would reduce the LCM by a factor of 5, 3, 2, or nothing at all.

The new LCM is $2^3 3^2 5^1 = \fbox{360}$

For the 10 positive integers, The LCD of the first 10 positive numbers is 30(7)(12) = 2520.

Inspecting the first 10 numbers in terms of its prime factors. 1 = 1 2 = 2 x 1 3 = 3 x 1 4 = 2 x 2 5 = 5 x 1 6 = 3 x 2 7 = 7 x 1 8 = 2 x 2 x 2 9 = 3 x 3 10 = 5 x 2

In order to find the least LCD of the remaining 9 elements if one of the numbers is removed, we must remove the element that has the largest prime factor and does not occur often in the factorization. So 7 fits the mark. Dividing the LCD 2520/7 = 360. Hence, the answer.

Question: Why not other numbers (other than 7)?

1. If 1 is removed, the LCD is still 2520.
2. If 2 is removed, the LCD doesn't change still because there are higher powers of 2 in 4 and 8.
3. If 3 is removed, the LCD is still 2520 (greater power of 3 in 9).
4. If 4 is removed, the LCD doesn't change (greater power of 2 in 8).
5. If 5 is removed, the LCD doesn't change because 10 has a prime factor of 5.
6. If 6 is removed, the LCD doesn't change for there are primes 2 and 3 remain in the set which produces another 6.
7. If 8 is removed, the LCD is halved but not the minimum.
8. If 9 is removed, the LCD is 1/3 of it but not still the minimum.
9. If 10 is removed, the LCD doesn't change since there are prime numbers 2 and 5 remain in the set which produces another 10.

In short, we base the number to be removed by great magnitude, a prime number, and occurs not so often which is 7.

We will get the lowest L.C.M if we remove the highest prime no.,i.e, 7. So, we need to find the LCM of (1,2,3,4,5,6,8,9,10) and we get the answer 360

lcm(1,2,3,4,5,6,7,8,9,10) = 2^3 3^2 5*7 remove 7 we get 360 as the lcm of the remaining 9 elements

Let's see how would we compute the LCM of 1st 10 positive integers:

We have the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Divide the set by 2 (the smallest positive prime number), we have the set

{1, 1, 3, 2, 5, 3, 7, 4, 9, 5}

Divide the set by 2 again, we have

{1, 1, 3, 1, 5, 3, 7, 2, 9, 5}

Divide the set by 2 again, we have

{1, 1, 3, 1, 5, 3, 7, 1, 9, 5}

Divide the set by 3 , we have

{1, 1, 1, 1, 5, 1, 7, 1, 3, 5}

Divide the set by 3 again, we have

{1, 1, 1, 1, 5, 1, 7, 1, 1, 5}

Divide the set by 5 , we have

{1, 1, 1, 1, 1, 1, 7, 1, 1, 1}

As you can see only the number 7 remains, which is a prime number.

So, the lowest possible value of L.C.M. is obtained when we remove the largest prime from the set .

Therefore if we remove 7, the L.C.M. of the remaining 9 numbers are

$2 \times 2 \times 2 \times 3 \times 3 \times 5 = \boxed{360}$

That's the answer!

7 is the largest prime number among the set of elements ..!!!

The lowest possible value of the LCM is got when the largest prime number in the group of numbers (7 in this case) is removed. Thus the required LCM is 360.