Let-n-N-Using-the-formula-lcm-a-b-ab-gcd-a-b-and-lcm-a-b-c-lcm-lcm-a-b-c-find-all-the-possible-value-of-6-lcm-n-n-1-n-2-n-3-n-n-1-n-2-n-3-

Question Number 110565 by Aina Samuel Temidayo last updated on 29/Aug/20

Let n \in N . Using the formula lcm (a, b)

= \frac{ab}{\gcd (a, b)} and lcm (a, b, c)

= lcm (lcm (a, b), c), find all the possible

value of \frac{6 ∙ lcm (n, n + 1, n + 2, n + 3)}{n (n + 1) (n + 2) (n + 3)}

Commented by kaivan.ahmadi last updated on 29/Aug/20

l e t m = \frac{6 . l c m (n, n + 1, n + 2, n + 3)}{n (n + 1) (n + 2) (n + 3)}

l c m (n, n + 1) = n (n + 1) =

l c m (n + 2, n + 3) = (n + 2) (n + 3)

i f n = 2 k \Rightarrow l c m (n (n + 1), (n + 2) (n + 3)) =

l c m (2 k (2 k + 1), 2 (k + 1) (2 k + 3)) = 2 k (k + 1) (2 k + 1) (2 k + 3)

\Rightarrow m = \frac{12 k (k + 1) (2 k + 1) (2 k + 3)}{4 k (2 k + 1) (k + 1) (2 k + 3)} = 3

i f n = 2 k + 1 \Rightarrow l c m (n (n + 1), (n + 2) (n + 3)) =

l c m (2 (2 k + 1) (k + 1), 2 (2 k + 3) (k + 2)) =

2 (k + 1) (k + 2) (2 k + 1) (2 k + 3)

\Rightarrow m = \frac{12 (k + 1) (k + 2) (2 k + 1) (2 k + 3)}{4 (2 k + 1) (k + 1) (2 k + 3) (k + 2)} = 3

s o m = 3

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Thanks

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

If anyone has any other solution .

Please post it .

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Why does lcm (n, n + 1, n + 2, n + 3) =

lcm ((n (n + 1), (n + 2) (n + 3)) ?

@ kaivan . ahmadi

Commented by floor(10²Eta[1]) last updated on 29/Aug/20

because lcm (a, b, c) = lcm (lcm (a, b), c)

the same logic works for 4 numbers :

lcm (a, b, c, d) = lcm (lcm (a, b), lcm (c, d))

since n and n + 1 are coprimes and

n + 2 and n + 3 also are coprimes so

lcm (n, n + 1) = n (n + 1) and

lcm (n + 2, n + 3) = (n + 2) (n + 3)

so

lcm (n, n + 1, n + 2, n + 3) =

lcm ((n (n + 1), (n + 2) (n + 3))

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

lcm (a, b, c, d) = lcm (lcm (a, b, c), d) = lcm (lcm (lcm (a, b), c), d)

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

I {don}^{'} t think

lcm (a, b, c, d) = lcm (lcm (a, b), lcm (c, d))

is a good argument .

Commented by floor(10²Eta[1]) last updated on 29/Aug/20

well \dots {it}^{'} s true even you think

that is not a^{″} good {argument}^{″}

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Can you prove it ?

Commented by floor(10²Eta[1]) last updated on 29/Aug/20

are you serious ? ? just think about for

a second, {it}^{'} s the same logic of why

lcm (a, b, c) = lcm (lcm (a, b), c)

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

lcm (a, b, c, d) = lcm (lcm (a, b, c), d) = lcm (lcm (lcm (a, b), c), d)

This is what is correct . (Using that

logic) .

Commented by floor(10²Eta[1]) last updated on 30/Aug/20

yeah this is also correct

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

The answer is 3 or 1 . I^{'} ve been able

to solve it .

Commented by floor(10²Eta[1]) last updated on 30/Aug/20

congratulations

Commented by kaivan.ahmadi last updated on 30/Aug/20

l e t l c m (a, b) = x, l c m (c, d) = y \Rightarrow

l c m (a, b, c, d) = l c m (l c m (a, b, c), d) =

l c m (l c m (l c m (a, b), c), d) = l c m (l c m (x, c), d) =

l c m (x, c, d) = l c m (x, l c m (c, d)) = l c m (x, y)