A six-digit integer is formed by placing two 3-digit integers side by-side. If the larger of the two integers is placed ahead of the smaller integer, the number formed is six times greater than if the smaller integer is placed ahead of the larger one. What is the sum of the two 3-digit integers?
Can anyone give me any hint how to solve it.
I have assumed the bigger integer to be $b$ and the smaller to be $s$
Then the two numbers formed are $1000s + b$ and $1000b + s$
Accordingly $(1000b + s) - (1000s + b) = 6(1000s + b)$
After this I am unable to think any clue.
It should be $1000b+s=6(1000s+b)$.
That is, $994b=5999s$.
That is, $7\times142b=7\times857s$, or $142b=857s$.
Note that $142$ and $857$ are relatively prime.
Can you take it from here?