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$X$ 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?

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    Equations are nice, but why not do a brute force search? You will be able to eliminate possibilities quite quickly.2011-11-11
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    Probably $X=(10a+5)_{10} \ne (8a+b)_8$ because the a's can be different.2011-11-11
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    @AndréNicolas: I find 15, 25, or 35 for X with 32, 52, or 72 for Y (respectively) all work.2011-11-11
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    @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.2011-11-11

2 Answers 2

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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.

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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.

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    How$\space Y<64$?2011-11-11
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    @MaX: It isn't! Answer corrected.2011-11-11
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    +1,I like your approach but actually the answer is what I and Ross got there is not unique solution.Thanks :)2011-11-12
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    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.2011-11-12
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    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.2011-11-12
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    I agree,and this is something me and my CAT-instructor don't often end up debating...what I meant it is in pure/applied mathematics we *always* try to develop a *general* approach so that when the constraints vary, (cont'd)2011-11-12
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    we could be able to solve using the same strategy, but as much I can see now quantitative aptitude is all about *particularize* and when I (often) try to *generalize* the problem I end up with problems like [this](http://math.stackexchange.com/questions/65862) and end up flustering myself.Really I spend lots of time in generalization (which may be I shouldn't ?!), but that's how we learn more, doesn't it?2011-11-12
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    *Two-digit* number problems, if not absolutely trivial tend to be particular, and don't readily generalize. Other digit problems, such as what is total number of $0$'s used in writing the numbers $1$ to $999$, do generalize.2011-11-12
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    Thanks, and I think we have [recent question](http://math.stackexchange.com/questions/80918/) on the similar topic.2011-11-12