Inspired by Dhruv Bhasin

Logic Level 3

A positive integer is said to be trams if it is divisible by the number formed by its first two digits.

For example, 100 is trams as $10 \mid 100$ .

Find the sum of the first 5 trams numbers, which have the first two digits equal to 17.

Inspiration

The answer is 5338.

1 solution

Arulx Z
Nov 6, 2015

We can rewrite this as

$f\left( n \right) = 17\cdot10^n+x$

for some integer $x$ . Since $17|\left( 17\cdot 10^{ n } +x \right)$ and $17|17\cdot 10^{ n }$ , it implies that $17|x$ . Additionally since the first 2 digits need to be 17, $x$ needs to be smaller than $10^{n-2}$ .

Since we want the tram numbers as small as possible, we try to keep the value of $n$ as small as possible. After that, we test different small values of $x$ based on above restrictions to find out tram numbers.

For $n = 0$ , we have $f\left( n \right) = 17 + x$ . Only possible value of $x$ satisfying the restrictions is 0.

For $n = 1$ , we have $f\left( n \right) = 170 + x$ . Again only possible value of $x$ satisfying the restrictions is 0.

For $n = 2$ , we have $f\left( n \right) = 1700 + x$ . Now $x$ can take the values of 0, 17 and 34.

So the smallest 5 tram numbers are 17, 170, 1700, 1717 and 1734. Sum of them is 5338.

Moderator note:

Good observations. Being able to work comfortably in modulus helps us clearly express our thoughts for such divisibility questions.

You should clarify that "the restriction" is the number of digits in $x$ . In particular, $x < 10 ^ n$ . Otherwise, we would have $17 \times 10 ^1 + 17$ as an answer.

Calvin Lin Staff - 5 years, 7 months ago

I have modified the answer. Please check it.

Arulx Z - 5 years, 7 months ago

Inspired by Dhruv Bhasin

The answer is 5338.

1 solution

Moderator note:

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