Find the number of three-digit integers that contain at least one $0$ or $5$. The leading digit of the three-digit integer cannot be zero

Question

Here, I tried guessing possibilities of the following problem. Consider the set $\{100, 101, \ldots, 110\}$. All these are desirable and there are $11$ elements in this set. Now consider $\{115, 116, \ldots, 195\}$, here if we consider the finite arithmetic progression $115,120,125,\ldots,195$ of $17$ terms, none of them is desirables. So from $100$ to $195$ there are $28$ desirable numbers. Similarly, from $200$ to $295$ there are $28$ desirable numbers. Between $300$ and $395$ there are also $28$ desirable numbers.

And so on... But every integer between $500$ and $599$ is also desirable, so there are $100$ more desirable numbers.

So we can say that the required answer would be $28 \cdot 8+100=324$.

But in answer key, the correct answer is $388$. Where did I go wrong?

Answer 1

Where did I go wrong?

"Now consider $\{115,116,…,195\}$ , here if we consider the finite arithmetic progression $115,120,125,…,195$ of $\color{red}{17}$ terms"

Wrong. Because, $\color{green}{\{151,152,153,154\}} \, \& \, \color{green}{\{156,157,158,159\}}$ i.e $8$ terms are also eligible. So. there is total $17+8=25$ eligible terms not $17$ in $\{115,116,…,195\}$.

Now, you can count $(25+11)\cdot 8+100=388$

Answer 2

Inclusion Exclusion works here.

There are exactly $~900~$ numbers (i.e. elements) between $~100~$ and $~999,~$ inclusive. Let $~S~$ denote the set $~\{1,2,\cdots,999\}.$

For $~k \in \{1,2,3\},~$ let $~S_k~$ denote the subset of such elements that have a [0 or 5] in column $~k,~$ of the 3 digit number.

Then, you want $~|S_1 \cup S_2 \cup S_3|.~$

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See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

So, the desired computation is

$$|S_1| + |S_2| + |S_3| \\ - \left\{ ~|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3| ~\right\} \\ + |S_1 \cap S_2 \cap S_3|.$$

Clearly $~S_3~$ which focuses on the rightmost column has $~\dfrac{900}{5} = 180~$ elements, since a number has $~0~$ or $~5~$ in its rightmost column if and only if it is divisible by $~5.~$

By symmetrical considerations, $~|S_2|~$ is also equal to $~180.~$

Similarly, since each element in $~S~$ can not have a $~0,~$ in its leftmost column, $~|S_1|~$ must equal $~\dfrac{|S|}{9} = 100.~$

Therefore, $~|S_1| + |S_2| + |S_3| = 460.~$

In considering the subset $~S_1 \cap S_2,~$ the coordination between $~0,5~$ in column-3 with $~0,5~$ in column-2 is random. So, you must have that $~|S_1 \cap S_2| = |S| \times \frac{1}{5} \times \frac{1}{5} = 36.$

Similarly, both $~|S_1 \cap S_3|~$ and $~|S_2 \cap S_3|~$ must equal $~|S| \times \frac{1}{5} \times \frac{1}{9} = 20.$

Therefore, $~|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3| = 76.~$

Similarly, $~|S_1 \cap S_2 \cap S_3| = |S| \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{9} = 4.~$

Therefore, the final computation is

$$[460 - 76] + 4 = 388.$$

Answer 3

What you're doing is equivalent to considering three positions

$$—~—~—$$

and asking yourself how many possibilities are there where you can place digits from $0$ to $9$ in each position with the mentioned restrictions.

Well, we can't fix $0$ on the first position, but you can fix $1$ (as you did in your first and second sets), so now we have to count how many desirable possibilities are there for the other two digits with the first fixed at $1$. Well, if we fix $0$ on position two, we are left with $10$ possibilities for the last one, and each is desirable because we have at least $0$, and the same occurs if you fix $5$ ($10$ more possibilities).

$$\overset{1}{—} ~ \overset{0}{—} ~ \overset{p_3}{—} ~~~ : ~~~ 10 ~ \text{choices}$$

$$\overset{1}{—} ~ \overset{5}{—} ~ \overset{p_3}{—} ~~~ : ~~~ 10 ~ \text{choices}$$

For the rest of the choices of the middle position, we always get two desirable choices for the last one ($0$ or $5$), and since there are $8$ more such choices for the middle (we already used $0$ and $5$), we have $8\cdot 2=16$ remaining possibilities. Adding them gives $2\cdot 10+16=36$, not $28$.

Now, if you choose any other digit for the first position (other than $5$), you'll get the same number of possibilities as you got with $1$, but since $0$ is not allowed (and $5$ a special one), we have $8$ choices, and therefore $8\cdot 36=288$ possibilities.

To address $5$ in the first position is easy, because any choice of the other two digits is desirable, and we have $10$ choices for the second and third positions, totalizing $10\cdot 10=100$ possibilities when $5$ is fixed.

Adding everything gives $288+100=388$.