## Mymensingh higher secondary 2019#1

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

### Mymensingh higher secondary 2019#1

How many digits in the following number?

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

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

### Re: Mymensingh higher secondary 2019#1

$\fbox {2021}$

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

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

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

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

### Re: Mymensingh higher secondary 2019#1

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