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How many of the integers $1,2, \dots, 120,$ are divisible by none of $2,5$ and $7 ?$

1. $40$
2. $42$
3. $43$
4. $41$

1 Answer

We know that,

$n( A \cup B \cup C) = n(A) + n(B) + n(C) – n(A \cap B) – n(B \cap C) – n(A \cap C) + n ( A \cap B \cap C)$

We have $n(U) = 120$

Now,

• $n(2) = \left \lceil \frac{120}{2} \right \rceil = 60$
• $n(5) = \left \lceil \frac{120}{5} \right \rceil = 24$
• $n(7) = \left \lceil \frac{120}{7} \right \rceil = 17$
• $n(2 \cap 5) = \left \lceil \frac{120}{\text{LCM}(2,5)} \right \rceil = \left \lceil \frac{120}{10} \right \rceil= 12$
• $n(5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(5,7)} \right \rceil = \left \lceil \frac{120}{35} \right \rceil = 3$
• $n(2 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,7)} \right \rceil = \left \lceil \frac{120}{14} \right \rceil= 8$
• $n(2 \cap 5 \cap 7) = \left \lceil \frac{120}{\text{LCM}(2,5,7)} \right \rceil= \left \lceil \frac{120}{70} \right \rceil = 1$

So, $n(2 \cup 5 \cup 7) = 60 + 24 + 17 – 12 – 3 – 8 + 1 = 79$

Therefore, number not divisible by $2,5,$ and  $7 = n \overline{(2 \cup 5 \cup 7)} = n(U) – n(2 \cup 5 \cup 7) = 120 – 79 = 41.$

Correct Answer$: \text{D}$

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