Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is an exercise from a textbook in Portuguese.

Choose a symbol (number) $abc$ in the decimal system ($a$ being the hundreds digit, $b$ the tens digit, and $c$ the units digit), in a way that the hundreds digit $a$ and the units digit $c$ differ by at least $2$ units. Consider the numbers $abc$ and $cba$ and subtract the smaller from the bigger in order to get a number $xyz$. Show that the sum of $xyz$ with $zyx$ is 1089.

I know that $abc=a10^{2}+b10+c$, $cba=c10^{2}+b10+a$, and $c=a+i$, $i\geq 2$ (or $a=c+i$, $i\geq 2$). So if $c=a+i$, $i\geq 2$, then $$cba=(a+i)10^{2}+b10+(c-i)=a10^{2}+b10+c+(i10^{2}-i)=a10^{2}+b10+c+(99i)\;.$$ So $xyz=cba-abc=99i$, or $9 \mid xyz$ wich give me that $x+y+z=9k$, and $11 \mid xyz$ wich give me a condition, but I still can use this.

I would appreciate your help! Sorry for the lousy title.

share|improve this question
1  
See also my answer here, which proves it using grade-school arithmetic. –  Bill Dubuque Sep 5 '12 at 20:24

3 Answers 3

up vote 4 down vote accepted

You're almost there. $xyz = 99i = 100i - i$ which has digits $x = i-1$, $y = 9$, $z = 10-i$ and so $$xyz + zyx = 100(i-1) + 90 + (10-i) + 100(10-i) + 90 + (i-1)$$$$= 100(9) + 90 + 90 + 9 = 1089.$$

share|improve this answer

Assume without loss of generality that $c=a+i$, where $i\ge2$. Then

$$cba-abc=\Big(100(a+i)+10b+a\Big)-\Big(100a+10b+a+i\Big)=100i-i=99i\;.$$

Now $2\le i\le 9$, so $99i\in\{198,297,396,495,594,693,792,891\}$. Note that the reversal of $99i$ is $99(11-i)$ for $i=2,\dots,8$, so $xyz+zyx=99i+99(11-i)=99\cdot11=1089$.

share|improve this answer

Wlog $a>c$ (i.e. $abc>cba$), Then $c-a$ will need a borrow, which causes in the second column $b-b-\mathrm{borrow}$ also needing a borrow, therefore with $abc-cba = xyz$ we have $x=(a-c-1)$, $y=9$ and $z=(10+c-a)$. The condition of $\lvert a-c\rvert>1$ ensures that $x\neq0$. Now $xyz+zyx = 101(x+z)+10\cdot 2y = 101(a-c-1+10+c-a)+10\cdot 18=101\cdot9+180=1089$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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