This is a three-unknown equation. I have to get three lines there.

Let a = a three digit number, e.g 246, 371 //just for further thinking
Let x = a's ones place
Let y = a's tenths place
Let z = a's hundreds place

I've got these two lines already, just cannot get the third.

x + y + z = 11

3x = y

Here i have to put this into maths: If we flip the number like: 246 -> 642 or 371 -> 173 the new number is bigger than the old number by 297.

EDIT Added more information (because I messed something up there):
All the three digits of the number added must equal 11;
The ones place is 3 times bigger than the tens place;
Flipped number is 297 greater than the original number;

Can you help me figure this out?
Would really appreciate you help!!

EDIT2:The final equation system:
$$x+y+z=11$$ $$100x+10y+z−(100z+10y+x)=99(x−z)=297⟹x−z=3$$ $$x=3y$$ Thanks to Math Lover!


Observe that the number is $100z+10y+x$ and the flipped number is $100x+10y+z$. Consequently, $$100x+10y+z - (100z+10y+x) = 99(x-z)=297 \implies x-z = 3.$$

  • $\begingroup$ I think you've reversed the roles of $x$ and $z$. In the OP, $z$ is the hundreds digit of the original number. $\endgroup$ – G Tony Jacobs Nov 6 '17 at 19:03
  • $\begingroup$ @GTonyJacobs Thanks. Corrected! $\endgroup$ – Math Lover Nov 6 '17 at 19:05
  • $\begingroup$ I've missed a mistake I made in the second statement there. $\endgroup$ – Kerdo Nov 6 '17 at 19:06
  • $\begingroup$ It should be that the ones place is 3 times bigger than the tenths place. So is it 3x = y? $\endgroup$ – Kerdo Nov 6 '17 at 19:07
  • $\begingroup$ Because right now I get the answers (x,y,z)= (14/5;42/5;-1/5). This cannot translate to a 3 digit number :/ $\endgroup$ – Kerdo Nov 6 '17 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.