Following number is divisible by

Question

If $ n = 2009$ , then $N = 2009^n -1982^n -1972^n + 1945^n $ is not divisible by

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

2

HINT:

Using Why $a^n - b^n$ is divisible by $a-b$? $$a^m-b^m-c^m+d^m=a^m-b^m-(c^m-d^m)$$ will be divisible by $(a-b,c-d)=P$

Similarly, $$a^m-b^m-c^m+d^m=a^m-c^m-(b^m-d^m)$$ will be divisible by $(a-c,b-d)=Q$

So, $a^m-b^m-c^m+d^m$ will be divisible by $(P,Q)$

As $2009-1982=27=1972-1945,$ our $N$ will be divisible by $(27,27)=?$

As $2009-1972=37=1982-1945,$ our $N$ will be divisible by $(37,37)=?$

So, $N$ will be divisible by $(37,27)=?$

Finally, here $N$ is also even.

$$N=2009^n-1982^n-1972^n+1945^n=2009^n-(-1945)^n-\{1982^n-(-1972)^n\}$$

will be divisible by $(2009+1945,1982+1972)=?$

edited Apr 13 '17 at 12:20

Community

answered Jul 02 '16 at 16:26

lab bhattacharjee

I got P =27 and Q=37 but they are not divisible by any option – Aakash Kumar Jul 02 '16 at 16:38
@AakashKumar, So, N is divisible by $(37,27)=999$ Again, $N$ is even, hence is divisible by $(999,2)=1998$ – lab bhattacharjee Jul 02 '16 at 16:40
How N is even,I cannot understand this – Aakash Kumar Jul 02 '16 at 16:51
@AakashKumar, Divide each term by $2$ to find each remainder – lab bhattacharjee Jul 02 '16 at 16:52
Which term you are talking about – Aakash Kumar Jul 02 '16 at 16:57
@Aakash, $2009,1982,1975,1945$ – lab bhattacharjee Jul 02 '16 at 17:44

score 0 · Answer 2 · answered Jul 02 '16 at 16:17

0

Hint: Compute modulo 7. (And the main "miracle" is that only one of the four answers is correct)

answered Jul 02 '16 at 16:17

Hagen von Eitzen

2 Answers2