Mymensingh higher secondary 2019#1

Forum rules
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
User avatar
samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

Mymensingh higher secondary 2019#1

Unread post by samiul_samin » Sat Feb 23, 2019 1:20 pm

How many digits in the following number?

$2^{2017}\times 7^3\times 5^{2018}$

User avatar
samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

Re: Mymensingh higher secondary 2019#1

Unread post by samiul_samin » Sun Mar 10, 2019 4:06 pm

Answer
$\fbox {2021}$

$Sol^n$
\[2^{2017}\times 5^{2018}\times7^3\]
\[=10^{2017}\times 5\times 343\]
\[=10^{2017}\times 1715\]
\[=1715\underbrace{000\cdots 0}_{2017}\]

So total digit$=2017+4=\fbox {2021}$
Last edited by samiul_samin on Sun Mar 10, 2019 4:24 pm, edited 1 time in total.

vyper47
Posts: 4
Joined: Fri Mar 30, 2018 4:56 pm

Re: Mymensingh higher secondary 2019#1

Unread post by vyper47 » Sun Mar 10, 2019 4:20 pm

Shouldn't it be $5^{2018}$ instead of $5^{2017}$?

User avatar
samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

Re: Mymensingh higher secondary 2019#1

Unread post by samiul_samin » Sun Mar 10, 2019 4:25 pm

vyper47 wrote:
Sun Mar 10, 2019 4:20 pm
Shouldn't it be $5^{2018}$ instead of $5^{2017}$?
Yes.

Post Reply