Factorials in Denominator

Number Theory Level 4

Consider the sequence $a_n=\dfrac{1}{d}$ , with $n\geq1$ of fractions:

$\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4},\cdots,\dfrac{1}{1000!}$

where $d$ is the $(n+1)$ th divisor of $1000!$ in increasing order. There is exactly one term $\dfrac{1}{d}$ for every $d \mid 1000!$ with $d>1$ .

And obviously, each term in the sequence is a rational number , where the decimal expansion of a rational number always terminates after a finite number of digits or repeats a finite sequence of digits over and over.

Find the number of terms, in the sequence, whose the decimal expansion terminates after a finite number of digits.

The answer is 248749.

1 solution

Paul Hindess
Dec 11, 2016

Only numbers of the form $2^a\times5^b$ will have terminating decimals.

Noting that $2^9$ is the largest power of 2 less than 1000, to find the highest power of 2 that divides into 1000! we need to compute:

$\displaystyle \sum_{n=1}^9 \Bigl \lfloor \frac {1000}{2^n} \Bigr \rfloor = 500+250+125+62+31+15+7+3+1 = 994$ .

Similarly, for the highest power of 5:

$\displaystyle \sum_{n=1}^4 \Bigl \lfloor \frac {1000}{5^n} \Bigr \rfloor = 200+40+8+1 = 249$ .

Consequently, for numbers of the form $2^a\times5^b$ , $a$ must be between 0 and 994 inclusive and $b$ must be between 0 and 249 inclusive. This means there are $995 \times 250 = 248750$ numbers of this form that divide into 1000!

However, the denominators in this sequence are greater than or equal to 2 which means $2^0 \times 5^0 = 1$ should be excluded and so we have 248750 - 1 = 248749 as our answer.

Factorials in Denominator

The answer is 248749.

1 solution

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