2016 is absolutely awesome 7

Find the smallest postive integer n > 1 n>1 such that 201 6 n + 1 2016^n+1 is a multiple of n n .


This is a part of the Set .


The answer is 2017.

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2 solutions

William Isoroku
Dec 20, 2015

Use the factoring identity:

a n + b n = ( a + b ) ( a n 1 a n 2 b + a n 3 b 2 . . . a b n 2 + b n 1 ) a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...-ab^{n-2}+b^{n-1})

The equation is 201 6 n + 1 n 2016^n+1^n

Substitute the corresponding values of a a and b b into the identity to yield 2017 2017 as the first factor and its prime too, so that's the smallest integer.

Wrong. You didn't show that 2017 is minimum.

Pi Han Goh - 5 years, 5 months ago

I also got 2017 in similar ways but I cant prove that it would be the smallest. Anyone help?

Shreyash Rai - 5 years, 4 months ago

Substitute the corresponding values of a a and b b into the identity to yield 2017 2017 as the first factor and its prime too, so that's the smallest integer.

What values of a a and b b ?

Pi Han Goh - 5 years, 5 months ago

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I believe a=2016 and b=1

Department 8 - 5 years, 5 months ago

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How does that help by showing that 2017 is the minimum?

Pi Han Goh - 5 years, 5 months ago

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@Pi Han Goh I think when you keep them in second bracket and start factorising it will always give values bigger than 2017.

Khang Nguyen Thanh can help us.

Department 8 - 5 years, 5 months ago

This makes a progress to what I have quoted, I think. We simply can have them with integers! Yes, I have noticed that 2017 is a prime number. Can you imply from this that 2017 is smallest possible answer?

Lu Chee Ket - 5 years, 5 months ago
Lu Chee Ket
Dec 19, 2015

For E = {0, 2, 4, 6, 8, 10, ... to \infty },

1
2
3
4
5
6
7
8
9
0   1
2   3
4   205
6   39991
8   14913081
10  9090909091
12  8230246567621
14  10371206370520815
16  17361641481138401521

This means E E + 1 + 1 E + 1 \displaystyle \frac{E^{E + 1} + 1}{E + 1} always an integer such that E E + 1 + 1 E^{E + 1} + 1 is always divisible by ( E + 1 ) . (E + 1).

Here, E = 2016 and therefore the expression is divisible by E + 1 = 2017. An expression added by 1 is not easily divisible by integers. Therefore, I were to guess that 2017 is the minimum possible value as answer, as I tried one by with integers from 2 to over 130 and spot checked to 170 but none is found!

Answer: 2017 \boxed{2017}

Wrong. This solution made no sense.

Pi Han Goh - 5 years, 5 months ago

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