How many numbers between 10 and 100 are divisible by 3 but not 2 nor 7?

Question

My working:

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Let $|M_3|$ denote the number of integers between 10 and 100 that divides 3. So we have $$|M_3|= \lfloor \frac{100}{3} \rfloor-3=30$$, and similarly, $$|M_2|=\lfloor \frac{100}{2}\rfloor-4=46$$, $$|M_7|=\lfloor \frac{100}{7}\rfloor-1=13.$$

Thus, we want to find $|M_3 \cap (M_2 \cup M_7)'|=$. From a venn diagram, we deduce that this is equivalent to finding $$|M_3|-|M_3 \cap M_7| - |M_3 \cap M_2| + |M_3 \cap M_7 \cap M_2|=30-4-15+2=13.$$

But the tecaher said that this is incorrect, but close. After reviewing my solution, I am confused to as what I am getting wrong? Is it the interpretation of "between" 10 and 100? i.e., I minused away all integers $1≤n<10$ for each of the $M_i$, i.e., the four integers $2,3,6,8$ are not counted in $|M_2|$, but $10$ is.

Answer 1

It's easier to start by asking how many integers $k$ there are such that $10\leq 3k\leq 100$, actually. You'll quickly see that $4\leq k\leq 33$; i.e. there are $33-4=29$ numbers between $10$ and $100$ that are divsible by $3$. Beginning with $k=4$, every second one of these is divisible by $2$, i.e. $\lfloor\frac{29}{2}\rfloor=14$ of them. There are therefore $29-14=15$ numbers from $10$ to $100$ that are divisible by $3$ but not by $2$. Finally, beginning with $k=7$, every seventh number will be divisible by $7$. We therefore count the odd multiples of $7$ from $7$ to $33$ (not the even ones — we've already taken care of those by considering divisibility by $2$). There are just two: $7$ and $21$. So there are $15-2=13$ numbers from $10$ to $100$ that are divisible by $3$ but not by $2$ or $7$.

Looks like your answer is correct after all! I'd have a polite word with your teacher if I were you.

Answer 2

The teacher may have wanted you to fill in some missing steps.

1. When you render $|M_2|,|M_3|,|M_7|$ you use the floor function for the first term but skip that for the second term. Better form is to render

$|M_3|=\lfloor\dfrac{100}{3}\rfloor-\lfloor\dfrac{10-1}{3}\rfloor=33-3=30$

and similarly for $|M_2|$ and $|M_7|$. Note that the limits are set to include both ends of the "between" range. If you want to exclude the ends you would use $100-1$ and $10$ as the numerators instead; that does not matter here since the end values $10$ and $100$ are not divisible by $3$. Also you don't actually need $|M_2|$ or $|M_7|$ in the next step.

2. Your Venn diagram relation is correct, except that "Venn" is a proper name and so is capitalized. But you may want to work out the sizes of the intersection sets you identify using the same method as in Section 1. Using your $M$ notation, $|M_2\cap M_3|=|M_{2×3}|=|M_6|$ and similarly the other intersection quantities are $|M_{21}|$ and $|M_{42}|$.

You then have

$|M_6|=\lfloor\dfrac{100}{6}\rfloor-\lfloor\dfrac{10-1}{6}\rfloor=16-1=15$

$|M_{21}|=\lfloor\dfrac{100}{21}\rfloor-\lfloor\dfrac{10-1}{21}\rfloor=4-0=4$

$|M_{42}|=\lfloor\dfrac{100}{42}\rfloor-\lfloor\dfrac{10-1}{42}\rfloor=2-0=2$

This together with $|M_3|$ as calculated in section 1 properly gives $13$, with all the steps filled in and thus (if the teacher is competent) earning full credit.