Inspiration @Aditya Raut

Number Theory Level 4

$\overline{{ABC0ABC}}$ be a 7 digit number.

Let the number of values of $\overline{{ABC0ABC}}$ divisible by 11 be $x$ .

Let the number of values of $\overline{{ABC0ABC}}$ divisible by 121 be $y$ .

Let the number of values of $\overline{{ABC0ABC}}$ divisible by 1331 be $z$ .

Then find the value of $x+y+z$ .

Details and Assumptions

The letters A,B,C do not necessarily stand for distinct digit.

The fourth digit of the number is zero.

The answer is 89.

1 solution

Chew-Seong Cheong
Apr 15, 2015

We note that $\overline{ABC0ABC} = 10001(100A+10B+C)$ .

Since $10001$ is indivisible by $11$ .

$\Rightarrow \begin{cases} \overline{ABC0ABC} \pmod {11} & \equiv \overline{ABC} \pmod{11} \\ \overline{ABC0ABC} \pmod {121} & \equiv \overline{ABC} \pmod{121} \\ \overline{ABC0ABC} \pmod {1331} & \equiv \overline{ABC} \pmod{1331} \end{cases}$

$\Rightarrow \begin{cases} 11|\overline{ABC} & \in \{110,121,132,...990\} & \Rightarrow x = 81 \\ 121|\overline{ABC} & \in \{121, 242, 363, ... 968\} & \Rightarrow y = 8 \\ 1331|\overline{ABC} & \in \{\space \} & \Rightarrow z = 0 \\ \end{cases}$

$\Rightarrow x+y+z = 81+8+0 = \boxed{89}$

Very nearly the same to how I did it. When I was finding the values of $x$ and $y$ . I noticed that I was looking for 3 digit multiples of 11 and 121. The smallest three-digit multiple of 11 is 110 which is $10\times 11$ and the largest is 990 which is $90\times 11$ . Therefore $x = 90 - 10 + 1 = 81$ and the process is similar for $y$ where the largest and smallest values are $968 = 8\times 121$ and $121 = 1\times 121$ respectively giving $y = 8 - 1 + 1 = 8$ .

Josh Banister - 5 years, 9 months ago

Inspiration @Aditya Raut

The answer is 89.

1 solution

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