# Smallest positive integer that starts with 2016 and is divisible by 2017

I have the following quiz but I don't really know how to solve it.

Find the smallest positive integer that starts (in decimals) with 2016 and is divisible by 2017.

Why to they specify the decimals in brackets?

First thought was to divide 2016 by 2017. 2016/2017 = 0.999504. Should the integer start with 2016? I find it confusing. Can someone help?

• Just for an experiment. Multiply $2017$ by $9$ or $99$ or $999$ or $9999$. The last works. Nov 4 '20 at 17:42

Your first thought is actually quite spot-on.

Since $$2016/2017 \approx 0.999504$$, we have $$0.9995 \times 2017 < 2016 < 1\times 2017$$

Multiplying this by various powers of $$10$$ yields

$$9\times 2017 < 10 \times 2016 < 10 \times 2017$$ $$99\times 2017 < 100 \times 2016 < 100 \times 2017$$ $$999\times 2017 < 1000 \times 2016 < 1000 \times 2017$$

The above shows that there are no $$5,6,7$$-digit multiple of $$2017$$ starting with $$2016$$. Finally:

$$9995\times 2017 < 10000 \times 2016< 9996 \times 2017 = 20161932$$

• Thanks a lot for your answer. Could you please explain why you started multiplying by various powers of 10? I don't get the underlying logic of the solution. I would never thought to solve it like that... Nov 5 '20 at 7:24
• This amounts to the observation that $2016\times 10^k$ is the smallest $(k+4)$-digit number starting with $2016$. Any number smaller than it cannot start with "$2016$". Nov 5 '20 at 7:26

Another solution:

20160 mod 2017 = 2007 No solution with 1 digit because 2017 - 2007 = 10

201600 mod 2017 = 1917 No solution with 2 digits because 2017 - 1917 = 100

2016000 mod 2017 = 1017 No solution with 3 digits because 2017 - 1017 = 1000

20160000 mod 2017 = 85 This leads to the last 4 digits solution 2017 - 85 = 1932