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The digits of a positive integer, having three digits, are in A.P and their sum is $15$. The number obtained by reversing the digits is $594$ less than the original number. Find the original number.

My Attempt:

Let the three digits number be $100x+10y+z$ where $x$, $y$ and $z$ are in A.P.

Then,

$y=\frac {x+z}{2}$

$2y=x+z$.

Then, what should I do??

pi-π
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6 Answers6

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So, $$x-2y+z = 0$$ $$x+y+z = 15$$ $$100z+10y+x +594= 100x+10y+z \implies 99x-99z=594$$ i.e. $$\left[ \begin{array}{ccc} 1 & -2 & 1 \\\ 1 & 1 & 1 \\\ 99 & 0 & -99 \end{array} \right] \left[ \begin{array}{ccc} x \\\ y \\\ z \end{array} \right] = \left[ \begin{array}{ccc} 0 \\\ 15 \\\ 594 \end{array} \right]$$ Solving for $x,y,z$, our number is 852.

florence
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  • How do I find the determinant and inverse of a $3\times 3$ matrix? I am not familiar with that? – pi-π Dec 27 '16 at 05:58
  • You can solve the system with whatever method you know. I just wrote it in matrix form to be explicit. – florence Dec 27 '16 at 06:00
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I wouldn't bother using algebra on such an easy problem. The three digits are in arithmetic progression and add up to $15$ so the middle digit is $5.$ Since the number is bigger than its reversal, the only possibilities are $654,753,852,$ and $951.$ Let's see, $654-456=198,$ nope. $753-357=396,$ nope. $852-258=594,$ we have a winner. The answer is $852.$

bof
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Suppose the number is $abc$, notice that $594=abc-cba=99(a-c)$, so $a-c=6$.

Since the digits are in arithmetic progression they must be $a,a-3,a-6$. Since they add $15$ we have $a=8$.

So the number is $852$

Asinomás
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You have used the fact that the digits are in arithmetic progression. You have not used the fact that the difference between the original number and the reversed number is $594$. You need to do so. Now you have $(100x+10y+z)-(100z+10y+x)=594$. Does this help?

Ross Millikan
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You are off to a good start, but you need to use all of the given information.

The fact that $x,y,z$ are an arithmetic progression means that $y = \dfrac{x+z}{2}$.

The problem also gives that the sum of the digits is $15$, i.e. $x+y+z = 15$.

Finally, reversing the digits gives the number $100z+10y+x$. So we have $100z+10y+x = (100x+10y+z)-594$.

Now, you have $3$ linear equations and $3$ unknowns. Is this enough to solve the problem?

JimmyK4542
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You know that $x+y+z=15$, so if you replace $x+z$ with $2y$ in this equation, you'll find $y$. Also, using the fact that this is an arithmetic progression, you can say that $x=y+d$ and $z=y-d$, where $d$ is the difference of this progression (and $y$ already has a known numerical value). That leaves you with only one unknown quantity $d$. Set the difference of the original and reversed numbers to be equal $594$, and you'll be able to find $d$.

zipirovich
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  • Then, can we directly assume three digits to be $a-d$, $a$ and $a+d$, since they are in A.P., instead of assuming $100x+10y+z$? – pi-π Dec 27 '16 at 05:55
  • @user354073: Yes, of course we can! I wanted to show how you can proceed from what you already did rather than starting over from scratch. But what you're saying now is very true, and probably more efficient too. I just wouldn't say "instead"; rather, you need to use everything together. Represent the digits in this form, and using the fact that they add up to $15$, find $a$. Then put them into the three-digit number as $100(a-d)+10a+(a+d)$, and continue solving from there. – zipirovich Dec 27 '16 at 06:56
  • which method would look beautiful? Or, which one is standard? – pi-π Dec 27 '16 at 07:05
  • Oh, I don't think there's a definitive answer to that! It's rather a matter of taste, although I'd say that efficiency is certainly one of the factors here. Just look at this page: you can see multiple ways to solve this same problem (which is great!), and I like some of them even better than my own. – zipirovich Dec 27 '16 at 07:11