New year problems

Number Theory Level 3

$\large A=\underbrace{20162016\dots 2016}_{\text{2016 (2016)'s}}$

A number " $A$ " is made by concatenating the integer 2016 by 2016 times .What is the remainder when $A$ is divided by 999?

The answer is 0.

2 solutions

Otto Bretscher
Dec 20, 2015

We can write $A=\sum_{k=0}^{671}A_k$ where $A_k=2016\times 10^{12k}(1+10^4+10^8)$ . We claim that $A_k$ is divisible by both 27 and 37, for all $k$ , so that it will be divisible by $999=27\times 37$ . Thus $A$ will be divisible by 999 as well.

Since 9 divides 2016 and 3 divides $1+10^4+10^8$ , we see that 27 divides $A_k$ .

Noting that $1000\equiv 1 \pmod{37}$ , we find that $1+10^4+10^8\equiv 1+10+100= 111 \equiv 0 \pmod{37}$ .

Thus the remainder we seek is $\boxed{0}$ .

Moderator note:

Great observation that 201620162016 is a multiple of 999.

Surya Sharma , This problem is copied from DPS RK Puram Math's Symposium - Crusade's Problem Solving event.

Rajdeep Dhingra - 5 years, 5 months ago

Ahsanul Habib
Dec 21, 2015

sum of the digit of 999 is 27 and sum of the digit of 20162016.........2016 =18144 which is divisible by 27. so reminder is 0.

I fail to understand your logic here. Are all numbers whose digit sum is 27 divisible by 999? E.g. 9099.

Calvin Lin Staff - 5 years, 5 months ago

New year problems

The answer is 0.

2 solutions

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