BdMO Online Forum

The Official Online Forum of BdMO
https://matholympiad.org.bd/forum/

Easy sum

https://matholympiad.org.bd/forum/viewtopic.php?f=18&t=3000

Page 1 of 1

Easy sum

Posted: Sat Feb 08, 2014 12:14 pm
by Raiyan Jamil
The product of two numbers is 1000 . If non of the numbers have a zero as their digits , what is their sum ?

Re: Easy sum

Posted: Sat Feb 08, 2014 5:15 pm
by asif e elahi
The sum is $2^{3}+5^{3}=8+125=133$

Re: Easy sum

Posted: Sat Feb 08, 2014 7:10 pm
by Labib
@Asif Please do elaborate how you eliminated other possible answers when posting solutions like these.
Here is a explanation to why the only possible answer is $125$.
Notice that - $1000 = 2^3*5^3$.
Assume that one of the two factors is $a = 2^n*5^m$ where $0\leq n,m\leq 3$.
If $n \geq 1$ and $m \geq 1$, then the last digit of $a$ will be $0$ which would be contradictory to the statement.
Therefore $n$ and $m$ cannot simultaneously be greater than $1$. So whether it is a factor of $5$ or a factor of $2$.
The same is true for its compliment as well.
I'd leave the rest for you to deduce. :)

Re: Easy sum

Posted: Sat Jan 21, 2023 1:09 pm
by Qsolver
the sum of the numbers is: 2^3 + 5^3 = 133
All times are UTC+06:00
Page 1 of 1
Powered by phpBB® Forum Software © phpBB Limited
https://www.phpbb.com/