How many non-negative integers less than $10000$ are there such that...

Question

I am stuck with the following two problems :

How many non-negative integers less than $10000$ are there such that the sum of digits of the number is divisible by $3$ ? The options are :

1. $\,\,1112\,\,$ 2. $\, 2213\,\,$ 3.$\, 2223\,\,$ 4.$\,3334\,$

My try: Let $I=[1,9999]$ and: $$ A = \{n\in I:n\equiv 0\pmod{3}\},$$ Then I have to count the number of elements in $A$ and I don't know how to find it?

2. Let $m$ and $n$ be two positive integers such that such that $m+n+mn=118.$ Then the value of $m+n$ is

1. not uniquely determined
2. $\,\,18$
3. $\,\,20$
4. $\,\,22$

I have to determine which of the aforementioned options is correct?

My Try: As David suggested , $(m+1)(n+1)=119=17 \times 7 \implies (m+1)=17, (n+1)=7 $ and so $m=16,n=6$ so that option 4 is the correct choice.

Can someone explain? Thanks in advance for your time.

Answer 1

For the first question, first note that an integer's digits add up to a multiple of three if and only if the integer itself is a multiple of three. Furthermore, it can be seen that:

$$\left|\left\{x~:~1\leq x\leq n~\text{and}~x~\text{is divisible by}~k\right\}\right| = \left\lfloor\frac{n}{k}\right\rfloor$$

Noting that $0$ is also considered non-negative and is not counted in the set above, you see that in your case you have $\lfloor\frac{10000}{3}\rfloor+1=3334$ non-negative integers less than (or less than or equal) to $10000$.

For the second problem, we have $m+n+mn=118$. Adding one to each side we have $m+n+mn+1=119=(m+1)(n+1)$

Noting that $m$ and $n$ are integers, that implies that $(m+1)$ and $(n+1)$ must then be two factors of $119$. Considering the prime decomposition of $119=7\cdot 17$ it must be that $(m+1)(n+1)=7\cdot 17$ implying either $\begin{cases}m+1=7\\n+1=17\end{cases}$ or implying $\begin{cases}m+1=17\\n+1=7\end{cases}$

Note: it was also possible that $(m+1)(n+1)=1\cdot 119$ implying $\begin{cases}m+1=1\\n+1=119\end{cases}$ or vice versa, however this possibility is ignored since that would imply either $m$ or $n$ will be non-positive and we were interested only in solutions with positive $m$ and $n$.

In either case we have the same result for $m+n$, namely $m+n=17-1+7-1=22$