$X$ and $Y$ are two-digit numbers. If $Y=2X+2$ and $Y=2X$ in decimal and octal system respectively, and unit digits of $X$ and $Y$ are $5$ and $2$ respectively, then how to find $X+Y$ in decimal number system?

My attempt:

I tried representing the two numbers in decimal as $(10a+5,10b+2)$ and in octal as $(8a+5,8b+2)$ and then tried to manipulate with according to the conditions $Y=2X+2$ and $Y=2X$, but they only give me one equation $b-2a=1$, how to get $10a+10b+7$(the sum of $X+Y$ in decimals) from these?

  • $\begingroup$ Equations are nice, but why not do a brute force search? You will be able to eliminate possibilities quite quickly. $\endgroup$ – André Nicolas Nov 11 '11 at 0:18
  • $\begingroup$ Probably $X=(10a+5)_{10} \ne (8a+b)_8$ because the a's can be different. $\endgroup$ – GarouDan Nov 11 '11 at 0:30
  • $\begingroup$ @AndréNicolas: I find 15, 25, or 35 for X with 32, 52, or 72 for Y (respectively) all work. $\endgroup$ – Ross Millikan Nov 11 '11 at 0:40
  • $\begingroup$ @GarouDan: $b$ it the leading digit of $Y$, so the relations are $10b+2=2(10a+5)+2$ and $8b+2=2(8a+5)$, which are redundant as MaX says. $\endgroup$ – Ross Millikan Nov 11 '11 at 0:42

Do you mean "unit digits of $X$ and $Y$ are $5$ and $2$ respectively"? I agree with you that $a$ can be any of $1,2, \text{or } 3$ (no higher or $Y$ will carry in octal) and there is no single answer.


Edit: (shortened) We are told that the (decimal) units digit of $X$ is $5$. So the only candidates for $X$ are $15$, $25$, and $35$. (Anything bigger, when expressed in octal, then doubled, is not a two-digit octal number.) The specification that $Y$ has units digit $2$ is superfluous.

Check which ones of $15$, $25$, and $35$ work. They all do.

For a problem in which the numbers are so nearly pinned down, trying to use "algebra" can be a waste of time. Before introducing symbols, it is useful to play with the numbers to get a concrete grip on the problem.

  • $\begingroup$ How$\space Y<64$? $\endgroup$ – Quixotic Nov 11 '11 at 0:52
  • $\begingroup$ @MaX: It isn't! Answer corrected. $\endgroup$ – André Nicolas Nov 11 '11 at 1:01
  • $\begingroup$ +1,I like your approach but actually the answer is what I and Ross got there is not unique solution.Thanks :) $\endgroup$ – Quixotic Nov 12 '11 at 5:11
  • $\begingroup$ Sorry I didn't clarified myself before, so yes I noticed that, but you are not supporting the algebraic approach which according to the official solution is the best method. However, your answer has it's own value as it invokes lateral thinking than the usual algebra. $\endgroup$ – Quixotic Nov 12 '11 at 5:51
  • $\begingroup$ To deal with a big problem, we may need to devise a "general" approach. But this is a very small problem. Analogy: to solve the problem roughly how many primes are there less than $10^{30}$, we need to develop general theory. To solve the problem of roughly how many primes are less than $40$, we get our hands dirty and count. $\endgroup$ – André Nicolas Nov 12 '11 at 6:12

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