Find the smallest postive integer n > 1 such that 2 0 1 6 n + 1 is a multiple of n .
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Wrong. You didn't show that 2017 is minimum.
I also got 2017 in similar ways but I cant prove that it would be the smallest. Anyone help?
Substitute the corresponding values of a and b into the identity to yield 2 0 1 7 as the first factor and its prime too, so that's the smallest integer.
What values of a and b ?
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I believe a=2016 and b=1
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How does that help by showing that 2017 is the minimum?
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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.
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?
For E = {0, 2, 4, 6, 8, 10, ... to ∞ },
1 2 3 4 5 6 7 8 9 |
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This means E + 1 E E + 1 + 1 always an integer such that E E + 1 + 1 is always divisible by ( 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: 2 0 1 7
Wrong. This solution made no sense.
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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 )
The equation is 2 0 1 6 n + 1 n
Substitute the corresponding values of a and b into the identity to yield 2 0 1 7 as the first factor and its prime too, so that's the smallest integer.