Find the remainder in the following case?

Question

Find the remainder when $987987987$...(up to $132$ digits) is divided by $1001$

I don't know to how go for this please some help?

Answer 1

Since it is clear that $987\cdot 1001 = 987987$, and you are looking at a combination of such numbers multiplied by powers of $10$, the answer is that $1001$ exactly divides your large number with no remainder.

Answer 2

Express our number is something a little more "algebraic"

132 digits long...

$\frac {132}{3} = {44}$ cycles of 3 ditits.

Each cycle adds $987\cdot 1000^n$

Summing the series.

$987 \cdot \sum_\limits{n=0}^{43} 1000^n\\ \frac {987 (1000^{44} - 1)}{1000-1}$

What is the remainder when we divide $(1000^{44} - 1)$ by $1001$

$1000\equiv -1 \pmod{1001}\\ 1000^{44}\equiv 1 \pmod{1001}\\ 1000^{44}-1\equiv 0 \pmod{1001}$

Every number composed of an even number of 3-digit cycles is divisible by $1001$

Answer 3

Hint:

Since $987 \times 1001 = 987987$, every two blocks of $987$ is a multiple of $1001$. How many blocks of $987$ do you have?

Answer 4

observe:

$$987=987\cdot 1$$

$$987987=987000+987=987\cdot 1001$$

$$987987987=987000000+987000+987=987\cdot 1001001$$

if there are $n$ blocks of $987$ on the left side, you'll have $n$ 1's at the right side.