Find all whole numbers such that the number increased by the sum of its digits equals 73.

Question

I'm really lost on how to figure this out. Work shown would help.

Answer 1

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?

Answer 2

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?

Answer 3

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 $$a=5, b=9 \\ 59$$

Answer 4

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.