Gems of Algebra

Algebra Level 5

a 0 , a 1 , . . { a }_{ 0 }, { a }_{ 1 }, .. is a sequence of positive integers where a n = n ! a_n = n! for all n ≤ 3 n \le 3 . Moreover, for all n ≥ 4 n \ge 4 , a n a_n is the smallest positive integer such that

a n a i a n − i \large\ \frac { { a }_{ n } }{ { a }_{ i }{ a }_{ n - i } }

is an integer for all integers i , 0 ≤ i ≤ n i, 0 \le i \le n .

Find the value of a 2018 a 2014 \large\ \frac { { a }_{ 2018 } }{ { a }_{ 2014 } } .


The answer is 12.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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2 solutions

Patrick Corn
Mar 22, 2018

The formula is: a 0 = 1 , a 1 = 1 , a 2 = 2 , a 3 = 6 , a 4 = 12 , a 5 = 12 , a 6 k + r = 7 2 k a r ( 0 ≤ r ≤ 5 ) . a_0 = 1, a_1 = 1, a_2 = 2, a_3 = 6, a_4 = 12, a_5 = 12, \\ a_{6k+r} = 72^k a_r \ \ (0 \le r \le 5). This can be proved by induction--I'll leave the details to the reader.

So a 2018 = 7 2 336 ⋅ 2 a_{2018} = 72^{336} \cdot 2 and a 2014 = 7 2 335 ⋅ 12 , a_{2014} = 72^{335} \cdot 12, so a 2018 a 2014 = 12 . \frac{a_{2018}}{a_{2014}} = \fbox{12}.

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The inequality can be proven by induction

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