GATE CSE 2017 | Set 1 | Question: 47

Question

The number of integers between $1$ and $500$ (both inclusive) that are divisible by $3$ or $5$ or $7$ is ____________ .

Comments

@Sanjay Sharma can you give an example to support your point and why we should not use floor??

---

@abir_banerjee e.g total numbers  divisible by 7 between 6 and 498 are 71  (first if 7 and last is 497 ) but  floor[(498-6)/7] will give 70 

and if we use a.p,. formula

7+(n-1)7=497 gives n=71

---

this Question can also be solved using a probability approach. See this variant https://gateoverflow.in/314348/gate-cse-1995-question-25b, and check out the answer by @techbd123 and read in his comments, to know.

Answer 1

Here, we can apply the property of set. Let $D_n$ denote divisibility by $n,$ $D_{n_1,n_2}$ denote divisibility by both $n_1$ and $n_2$ and so on.

$N(D_3 \cup D_5 \cup D_7)=N(D_3)+N(D_5)+N(D_7) -N(D_{3,5})-N(D_{ 5,7})-N(D_{3,7})+N(D_{3,5,7})$
$\quad \quad =166+100+71-33-14-23+4$
$\quad \quad =271$

Comments

got it

---

It is important to consider the floor value of each term, otherwise calculation mistake is obvious.

Answer 2

EASY way to solve using GATE interface CALCULATOR;

$P(3\cup5\cup7=P(3)+P(5)+P(7)-P(3\times5)-P(5\times7)-P(3\times7)+P(3\times7\times5)$

• $P(3)=\frac{500}{3}=\lfloor166.66\rfloor=166$
• $P(5)=\frac{500}{5}=100$
• $P(7)=\frac{500}{7}=\lfloor71.42\rfloor=71$
• $P(3\times5)=\frac{500}{15}=\lfloor33.33\rfloor=33$
• $P(7\times5)=\frac{500}{35}=\lfloor14.28\rfloor=14$
• $P(3\times7)=\frac{500}{21}=\lfloor23.8\rfloor=23$
• $P(3\times5\times 7)=\frac{500}{105}=\lfloor4.76\rfloor=4$

$\therefore P(3\cup5\cup7)= 166+100+71-33-14-23+4 = 271$

Comments

its ok shiva

---

I made the exact same approach :)

---

nice explanation..

Answer 3

solution

Answer 4

| AUBUC |= | A | + | B |+| C | - | A ⋂ B | - | A ⋂ C | -| B ⋂ C | + | A ⋂ B ⋂ C |
166 + 100 + 71 - 33 - 14 -23 + 4  = 271

Answer 5

Here ,we can find:

$|A| = \left \lfloor \frac{500}{3} \right \rfloor = 166$

$|B| = \left \lfloor \frac{500}{5} \right \rfloor = 100$

$|C| = \left \lfloor \frac{500}{7} \right \rfloor = 71$

$|A\cap B| = \left \lfloor \frac{500}{LCM(3,5)} \right \rfloor =\left \lfloor \frac{500}{15} \right \rfloor = 33 $

$|B\cap C|  = \left \lfloor \frac{500}{LCM(5,7)} \right \rfloor =\left \lfloor \frac{500}{35} \right \rfloor = 14 $

$|A\cap C| = \left \lfloor \frac{500}{LCM(3,7)} \right \rfloor =\left \lfloor \frac{500}{21} \right \rfloor = 23 $

$|A\cap B \cap C| = \left \lfloor \frac{500}{LCM(3,5,7)} \right \rfloor =\left \lfloor \frac{500}{105} \right \rfloor = 4 $

Now we,can Apply Principle of Inclusion - Exclusion:

$(|A| \cup |B| \cup |C|) = |A| + |B| + |C|- | A\cap B| - | B\cap C| - | A\cap C| + | A\cap B\cap C|$

 

Put the values:

$(|A| \cup |B| \cup |C|) = 166+100+71-33-14-23+4$

$(|A| \cup |B| \cup |C|) = 341-70$

$(|A| \cup |B| \cup |C|) = 271$

 Number 1 to 500 is not divisible by either 2,3 or 5:

${(|A| \cup |B| \cup |C|)}' = N(U) - (|A| \cup |B| \cup |C|)$

${(|A| \cup |B| \cup |C|)}' = 500-271$

${(|A| \cup |B| \cup |C|)}' =229$

So, the number of integers between $1$ and $500$ (both inclusive) that are divisible by $3$ or $5$ or $7$ is $:271$ 

Answer 6

by applying the Principle of Inclusion and Exclusion. Note that all divisions are to be rounded down to the nearest integer.

N = [ 500/3 + 500/5 + 500/7 ] - [ 500/15 + 500/35+ 500/21 ] + [ 500/(105) ]

= 166+100+71 - (33+14+23) + 4 = 271

if the question is of divisible then ans is 271.

if the question is of not divisible then its

500-271=229

Comments

but here u need to take lowershield function also

---

In both ways ans is coming same....