MEI Maths Item of the Month
April 2014
Divisibility of consecutive integers
Let N   n  1 n  2  ...  2n  , i.e. the product of n consecutive integers from n+1 to 2n.
Prove that, for any positive integer n, N is divisible by 2n but not a higher power of 2.
For example, for n = 3:
N = 4×5×6
= 120.
120 is divisible by 8 but not divisible by 16.
Solution
For n = 1:
N=2
2 is divisible by 2 but not divisible by 4, therefore the result is true for n = 1.
Assume the result is true for n = k :
Nk   k  1 k  2 ...  2k  is divisible by 2k but not by 2k+1 , i.e. N k  2k m , where m is an
odd number.
For the case n  k  1 :
N k 1  (k  2)(k  3)...(2k )(2k  1)(2k  2)
 2(k  1)(k  2)(k  3)...(2k )(2k  1)
 2  2k m(2k  1)
 2k 1 m(2k  1)
As both m and 2k+1 are an odd numbers, N k 1 is divisible by 2k+1 but not by 2k+2 .
So if the result is true n = k, then it is true n  k  1 . As the result is true for n = 1, then
it is result is true for all n ≥ 1.
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