Easy sum

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Raiyan Jamil
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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 ?
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asif e elahi
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Re: Easy sum

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

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

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Labib
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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. :)
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Qsolver
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Re: Easy sum

Unread post by Qsolver » Sat Jan 21, 2023 1:09 pm

the sum of the numbers is: 2^3 + 5^3 = 133

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