Easy sum

For students upto class 5 (age upto 12)
User avatar
Raiyan Jamil
Posts: 138
Joined: Fri Mar 29, 2013 3:49 pm

Easy sum

Unread post by Raiyan Jamil » Sat Feb 08, 2014 12:14 pm

The product of two numbers is 1000 . If non of the numbers have a zero as their digits , what is their sum ?
A smile is the best way to get through a tough situation, even if it's a fake smile.

User avatar
asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
Location: Sylhet,Bangladesh

Re: Easy sum

Unread post by asif e elahi » Sat Feb 08, 2014 5:15 pm

The sum is $2^{3}+5^{3}=8+125=133$

User avatar
Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm
Location: Dhaka, Bangladesh.
Contact:

Re: Easy sum

Unread post by Labib » Sat Feb 08, 2014 7:10 pm

@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. :)
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.


"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

Post Reply