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A number is chosen at random from the first 1,000 positive integers. What is the probability that it's divisible by 3,5, or 7?

So I started off by breaking the problem up and having:
divisible by 3: p(a)
divisible by 5: p(b)
divisible by 7 p(c)

I know I'm going to apply the exclusion inclusion principle, but how do I find out how many numbers are divisible by each without going through all the numbers between 1 and 1000?

hardmath
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Lil
  • 2,529

3 Answers3

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Hint The largest multiple of $3$ in $\{1, \ldots, 1000\}$ is $3 \lfloor \tfrac{1000}{3} \rfloor = 3 \cdot 333 = 999$ and so there are $\lfloor \frac{1000}{3} \rfloor = 333$ mutiples of $3$ in that range. Similarly, there are $\lfloor \frac{1000}{5} \rfloor = 200$ multiples of $5$, and there are $\lfloor \frac{1000}{15} \rfloor = 66$ multiples of $3 \cdot 5 = 15$.

Travis Willse
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From Excel

=INT(1000/3)+INT(1000/5)+INT(1000/7)-INT(1000/15)-INT(1000/21)-INT(1000/35)+INT(1000/105)=543

p=0.543

kmitov
  • 4,731
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Use inclusion/exclusion principle:

  • Include the amount of numbers divisible by $3$, which is $\Big\lfloor\frac{1000}{3}\Big\rfloor=333$
  • Include the amount of numbers divisible by $5$, which is $\Big\lfloor\frac{1000}{5}\Big\rfloor=200$
  • Include the amount of numbers divisible by $7$, which is $\Big\lfloor\frac{1000}{7}\Big\rfloor=142$
  • Exclude the amount of numbers divisible by $3$ and $5$, which is $\Big\lfloor\frac{1000}{3\cdot5}\Big\rfloor=66$
  • Exclude the amount of numbers divisible by $3$ and $7$, which is $\Big\lfloor\frac{1000}{3\cdot7}\Big\rfloor=47$
  • Exclude the amount of numbers divisible by $5$ and $7$, which is $\Big\lfloor\frac{1000}{5\cdot7}\Big\rfloor=28$
  • Include the amount of numbers divisible by $3$ and $5$ and $7$, which is $\Big\lfloor\frac{1000}{3\cdot5\cdot7}\Big\rfloor=9$

Hence the amount of numbers divisible by $3$ or $5$ or $7$ is:

$$333+200+142-66-47-28+9=543$$

barak manos
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