Question

P =1234567891011.............9989991000

Q.1 Find the remainder when first 1000 digits of the number P are divided by 16

Q.2 Find the remainder when P is divided by 9

Answer 1

Q1 :
Let $P'$ be the number trucated after $1000$ digits.

Since $16\,\mid\,10^4$, it is sufficient to check the last four digits of $P'$ :
1-9     : 1*9 = 9 digits
10-99   : 90*2 =180 digits
100-370 : 271*3= 813 digits

So $P'$ ends in $\ldots 3683693$ and the last four digits are $3693$
$3693=16\times 230+\boxed{13}$

So the remainder is $13$ when $P'$ is divided by $16$.

Q2 :
The remainder will be same as the remainder when sum of digits of $P$ is divided by $9$(we don't need to worry about placevalue here as $10^n\equiv 1\pmod{9}$ for all $n\in\mathbb{Z^+}$) :
$1+2+3+\cdots+1000=\dfrac{1000(1000+1)}{2}=500500\equiv 5+5=10\equiv\boxed{1}$

Therefore the remainder is $1$ when $P$ is divided by $9$.

Answer 2

Answer:

the correct answer is option no 3