Competition problem — Number theory

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Number theory Difficulty 3.2 AMC 10/12 Find the answer

There are $F$ fractions $\frac{m}{n}$ with the properties: $m$ and $n$ are positive integers with $m<n$ , $\frac{m}{n}$ is in lowest terms, $n$ is not divisible by the square of any integer larger than 1, and the shortest sequence of consecutive digits that repeats consecutively and indefinitely in the decimal equivalent of $\frac{m}{n}$ has length 6. We define $G=F+p$ , where the integer $F$ has $p$ digits. What is the sum of the squares of the digits of $G$ ?

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

We start with 1111109 fractions, as above, and want to remove all of the fractions in $U, V$ and $W$ . Since each fraction in $W$ is in $U$ and $V$ , it is enough to remove those $U$ and $V$ only. The total number of fractions in $U$ and $V$ (that is, in $U \cup V$ ) equals the number of fractions in $U$ plus the number of fractions in $V$ minus the number of fractions in their overlap (that is, in $U \cap V=W)$ . This is because any fraction in the overlap is "counted twice" when include all fractions in $U$ and all fractions in $V$ . Therefore, we need to remove $1009+329-29$ fractions from the set of 1111109. Therefore, $F$ , the number of fractions having the desired properties, is $F=1111109-(1009+329-29)=1109700$ . Since $F$ has 7 digits, then $G=F+7=1109707$ . The sum of the squares of the digits of $G$ is $1^{2}+1^{2}+0^{2}+9^{2}+7^{2}+0^{2}+7^{2}=1+1+81+49+49=181$ .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.