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$$121x+11y+z=567$$ What is $x+y+z$, given $x,y,z$ are digits of a $3$-digit number $(xyz$).

I am stuck at this step:

$$ (xyz)_{11}=(567)_{10} $$ Is the above step correct or is this correct: $(xyz)_{11}=(567)_{11}$?

Also please explain the choice.

Robert Lee
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Harsh Rewari
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  • What do you think? What's the definition of a number base? – TheBestMagician Sep 24 '21 at 23:05
  • My logic says it's the first one.but according to the book, it's the second one. I want to understand why is the second one true if at all it is the correct approach. – Harsh Rewari Sep 24 '21 at 23:16
  • Are you sure the book is correct... – TheBestMagician Sep 24 '21 at 23:30
  • @HarshRewari What is the exact statement of the question in the book? You mention number systems and bases in the title, but there is no reference to either in the first line of the question as posted. – dxiv Sep 24 '21 at 23:44
  • the exact question is this: The digits x, y and z of a three-digit natural number xyz satisfy the equation 121x + 11y + z = 567. What is the sum of the digits of the three-digit natural number xyz? (1) 17 (2) 18 (3) 21 (4) Data Insufficient – Harsh Rewari Sep 24 '21 at 23:55
  • And this is the solution given: 121x + 11y + z = 567 thus is same as (xyz)base 11 = 567 convert it to base 10 and then sum up the digits of the number thus obtained i.e 6 + 7 + 8 = 21. – Harsh Rewari Sep 24 '21 at 23:57
  • @HarshRewari You wrote in another comment that "my logic says it's the first one. but according to the book, it's the second one". However, you now quote the book as saying "thus is same as $(xyz)_{11} = 567$" which is the first one, the one that you already thought was right. Then what is the question? – dxiv Sep 25 '21 at 00:08
  • @dvix The book's answer begins well enough with $(xyz){11} = 567,$ but after that point it proceeds to convert $567{11}$ to base ten, that is it acts as if the equation to solve were $567_{11} = (xyz)_{10}.$ This leads to a wrong answer. OP is right to question the book. – David K Sep 25 '21 at 00:29
  • A number $(abc)m$ in base $m$ means the number $a\times m^2 + b \times m + c$ (assuming $a,b,c$ are digits. i.e $a,b,c$ are integers $0\le a,b,c < m$). Here we have $121x + 11y +z =11^2\times x + 11\times y + z$ so $121x + 11y + z = (xyz){11}$. Now when we read "$567$" by convention that is the $567$ must of us know in our natural base as "five hundred sixty-seven". Our natural base is base $10$ and $567 =500 + 60 +7 = 5\times 10^2 + 6\times 10 + 7 = (567)_{10}$. This isn't meant to be a stumper. It's just ... language. – fleablood Sep 25 '21 at 02:24
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    Okay..... $(xyz){11} = (567){10}$ means: $121x + 11y + z = 5\times 100 + 6\times 10 + 7$. And $(xyz){10} = (567){11}$ means: $100x + 10y + z = 5\times 121 + 6\times 11 + 7$. If your book is screwing up then your book is wrong. What your write is $121 x + 11y + z =567$ so that would mean $(xyz){11} =(567){10}$. That's what it is. If your book is screwing up then your book is wrong. – fleablood Sep 25 '21 at 02:32

2 Answers2

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$$121x+11y+z=567 $$

Working in $\pmod {11}$ we get $z=6$

Thus $$121x+11y =561$$

Divide by $11$ to get $$11x+y=51$$ Working in $\pmod {11}$ we get $y=7$

Plug in the original equation to get $x=4$

As a result $$x+y+z = 17$$

Mohammad Riazi-Kermani
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Indeed $121x+11y+z = 567$ expresses $(xyz)_{11} = (567)_{10}$, under the restriction that $0 \le x,y,z \le 10$ of course.

Well $121 \times 4 =484 < 567$ while $121\times 5= 605 > 567$ so $x=4$

Then $567-484 = 83$ is left for the two final $x,y$ and so $y=7$ and $83-77=6$ so $z=6$ and so we get $(476)_{11}$ and the sum of the base $11$ "digits" is $17$.

Henno Brandsma
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