I'm really lost on how to figure this out. Work shown would help.
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Brute force is also an acceptable answer. – copper.hat Aug 02 '14 at 04:03
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Hint: the number must have two digits, so it is 10a+b. What is an expression for the number plus the sum of its digits?
rogerl
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Hint: You can immediately rule out any single digit number, and all whole numbers with three or more digits. (Why?)
Consequently, you're looking at two-digit whole numbers $n$, which can all be expressed in the form $$n=d_1\cdot 10+d_2,$$ where $d_1,d_2\in\{0,1,2,3,4,5,6,7,8,9\}$ and $d_1\ne0.$ (Why?)
So what you want is that $n+d_1+d_2=73.$ Can you rewrite $n+d_1+d_2$ in terms of $d_1$ and $d_2,$ and take it from there?
Cameron Buie
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It would have to be two digits, say $ab$. Then the sum is $$(10a+b)+(a+b)=11a+2b=73$$ For this to work, $a$ must be odd, and $$b=\frac{73-11a}{2}\in [1,9]\implies a=5,6$$ So the only solution is $$a=5, b=9 \\ 59$$
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Let $n$ be the number in question. Clearly, the standard decimal representation of $n$ must have two digits, so $n = 10a + b$. Then the condition becomes $73 = n + a + b$. Substituting in, we have $$11a + 2b = 73.$$ Now, we need to find $b, 0\leq b\leq 9$ such that $73-2b$ is divisible by $11$. I will leave the rest to you.
William Stagner
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$$11a+2b=73\iff11(a-5)=2(9-b)\iff\frac{a-5}2=\frac{9-b}{11}=c$$ $$\implies 0\le a=2c+5<10\iff c=0,1,2$$ and $$0\le9-11c=b<10\implies c=0$$ – lab bhattacharjee Aug 02 '14 at 05:13