Gcd!

Number Theory Level 3

For all integers $n$ , $n(n+1)(n+2)$ has a greatest common divisor $a$ . Find $\dfrac a{100}$ .

The answer is 0.06.

1 solution

Md Zuhair
Aug 9, 2016

Let the no be of the form $n =6q+r$ [ $**Euclid's Division Lemma**$ ] [ $0<=r<6$ ]

Then the nos are n = $6q$ , $6q+1$ ........ $6q+5$ .

hence putting these values in all the cases of $n(n+1)(n+2)$ we get a =6 .... Hence $a/100$ = $6/100$ = $0.06$

@Md Zuhair , there are four common divisors 1, 2, 3 and 6. I have changed the wording of your problem.

Chew-Seong Cheong - 4 years, 10 months ago

Thank you Chew Seong Cheong .

Md Zuhair - 4 years, 10 months ago

$\text{Using Euclid's Division Lemma, a = bq + r, 0 } \leq \text{ r < b. Using b = 2.} \\ a = 2q + r, r \in {0, 1} \\ \fbox{1 } \implies a = 2q \\ \implies n(n+1)(n+2) = 2q(2q+1)(2q+2) = 2q(2q+1)2(q+1) = 4q(2q+1)(q+1) \\ \fbox{2 } \implies a = 2q+1\\ \implies n(n+1)(n+2) = (2q+1)(2q+2)(2q+3) = 2(2q+1)(q+1)(2q+3) \\ (\fbox{1}, \fbox{2}) = 2 \\ \text{Why can't the GCD be 2?? Elucidate?}$

Viki Zeta - 4 years, 10 months ago

You have shown that 2 is a common divisor. You have not shown that 2 is the greatest common divisor.

Calvin Lin Staff - 4 years, 10 months ago

Gcd!

The answer is 0.06.

1 solution

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