MathStatClub
PROBLEM OF THE MONTH
December 2016
Prove that for every positive integer k, there exist two positive integers n > m such that 10k
divides 2n − 2m (in other words, there are differences of powers of 2 ending with arbitrarily
many 0’s).
Please submit your solution to:
• Dr. Christian Avart, [email protected]
before the deadline: December 30th, 7:00pm. The WINNER will be awarded with a $15 gift
certificate and will be announced in the NEXT issue.
Solution to the November 2016 Problem of the Month
First, we can simplify the integrand as follows:
x
Z
Z
ex(e +1)
x
x(ex +1)−ex
x
dx
=
xe
dx
=
xex ee (x−1) dx.
x
e
e
Z
Setting u = ex (x − 1), we obtain du = xex dx, and the integral becomes:
Z
x (x−1)
xex ee
dx =
Z
x (x−1)
eu du = eu + C = ee
+ C.
Winner: Zachary Ritter.
Other correct submissions: Chris Lee, Saikat Nandy, and Michael Chung.
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