Tricky Division

Algebra Level 4

For what integer $a$ , does $x^{2} - x +a$ divide $x^{13} + x + 90$ ?

15 3 90 45 26 2

3 solutions

Naheem Ahmed
Jan 31, 2015

When x=1 then 92 must be divisible by a. When x=0 then 90 must be divisible by a. Therefore the only common factor of 90 and 92 is 2. Therefore 2 is the answer :)

but there may be other three integers -2,-1,1 but they conradicts to the statemet so the answer is 2 !!! :-)

Shivram Badhe - 6 years, 4 months ago

This question is from Putnam 1963!

Nihar Mahajan - 6 years, 4 months ago

Kenneth Tay
Jan 30, 2015

My solution here: https://beyondmathsolutions.wordpress.com/2015/01/06/soln-condition-for-divisibility/

Hey there Kenneth!

Calvin Lin Staff - 6 years, 2 months ago

Lu Chee Ket
Jan 30, 2015

Fortunately not completing a tedious division for the remainder to answer correctly; the thing is a remainder obtainable not likely to have integer 'a' to plug in to become zero. By searching for a = 0.5 with step of 0. 5 to 100, while x from -10 to 10 with step of 0.5, it is found that a = 2 makes all integer x all right to (x^13 + x + 90) MOD (x^2 - x + a) = 0. This does not happen to all other 'a'.

Actually, if (x^2 - x + 2) does divide (x^13 + x + 90), then not only integer x can be true but all. Although (x^2 - x + 2) has not been a factor to (x^13 + x + 90) in this sense, only a = 2 can be the best answer. Nevertheless, I think we should NOT think that (x^2 - x + 2) does divide (x^13 + x + 90).

Note: Remainder, R (a) for a = 2 is not mentioned by https://beyondmathsolutions.wordpress.com/2015/01/06/soln-condition-for-divisibility/

After checking, I found that (x^2 - x + 2) is indeed a factor of (x^13 + x + 90). They share the same x = [1 +/- j Sqrt (7)]/ 2 as zeroing values. R (x) = 0.

Lu Chee Ket - 6 years, 4 months ago

Tricky Division

3 solutions

0 pending reports