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For each integer n ≥ 4 n\geq 4 , let a n a_n denote the base- n n number ( 0.133133133 … ) n (0.133133133\ldots)_n . The product a 4 a 5 ⋯ a 99 a_4a_5\cdots a_{99} can be expressed as m n ! \dfrac{m}{n!} , where m m and n n are positive integers and n n is as small as possible. What is the value of m m ?


The answer is 962.

Sajjan Barnwal by Sajjan Barnwal , 30, India

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1 solution

Madelyn Yu
Dec 31, 2016

Since a n a_n = ( 0.133133133 … ) n (0.133133133\ldots)_n ,

n 3 n^{3} a n a_n = ( 133.133133 … ) n (133.133133\ldots)_n

Combining the two equations, ( n 3 − 1 ) (n^{3}\, -\, 1) a n a_n = 13 3 n 133_n .

a n a_n = n 2 + 3 n + 3 ( n 3 − 1 ) \frac{n^{2}+3n+3}{(n^{3}\, -\, 1)}

Notice that: ( ( n + 1 ) 3 − 1 ) = n ( n 2 + 3 n + 3 ) ((n+1)^{3}\, -\, 1) = n(n^{2}+3n+3)

a n a_n = n 2 + 3 n + 3 ( n 3 − 1 ) \frac{n^{2}+3n+3}{(n^{3}\, -\, 1)} = ( n + 1 ) 3 − 1 n ( n 3 − 1 ) \frac{(n+1)^{3}\, -\, 1}{n(n^{3}\, -\, 1)} .

With this, a 4 a 5 ⋯ a 99 a_4a_5\cdots a_{99} = 10 0 3 − 1 ( 4 3 − 1 ) ⋅ 4 ⋅ 5 ⋅ 6 ⋅ . . . ⋅ 99 \frac{100^{3}\, -\, 1}{(4^{3}\, -\, 1)\cdot4\cdot5\cdot6\cdot...\cdot99} . = 999999 ⋅ 6 63 ⋅ 99 ! \frac{999999\cdot 6}{63 \cdot 99! } . = 95238 99 ! \frac{95238}{99! } .

We can also see that 99 | 95238. So, a 4 a 5 ⋯ a 99 a_4a_5\cdots a_{99} can be expressed as 962 98 ! \frac{962}{98! } . Now, it is obvious that 98 does not divide 962.

Therefore, m = 962.

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