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

We want numbers that avoid all three divisors: 2, 5, and 7 between the numbers 1 to 120. This is a perfect application of the inclusion-exclusion principle.

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The strategy is to find how many numbers ARE divisible by at least one of these, then subtract from the total.

Total numbers: 120

Numbers divisible by each factor:

Divisible by 2: $\lfloor 120 ÷ 2 \rfloor = 60$

Divisible by 5: $\lfloor 120 ÷ 5 \rfloor = 24$

Divisible by 7: $\lfloor 120 ÷ 7 \rfloor = 17$ (since $120 ÷ 7 = 17.14...$)

Numbers divisible by pairs:

Divisible by both 2 and 5 (i.e., by 10): $\lfloor 120 ÷ 10 \rfloor = 12$

Divisible by both 2 and 7 (i.e., by 14): $\lfloor 120 ÷ 14 \rfloor = 8$

Divisible by both 5 and 7 (i.e., by 35): $\lfloor 120 ÷ 35 \rfloor = 3$

Numbers divisible by all three:

Divisible by 2, 5, and 7 (i.e., by 70): $\lfloor 120 ÷ 70 \rfloor = 1$

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Inclusion-Exclusion Formula:

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

Numbers divisible by at least one of 2, 5, or 7:

$60 + 24 + 17 - 12 - 8 - 3 + 1 = 79$

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Numbers divisible by NONE of 2, 5, or 7:

$120 - 79 = 41$

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Alternative approach: We can also solve this by finding numbers that are coprime to 2 and 5 first, then removing those divisible by 7.

Numbers not divisible by 2 or 5:

$120 \times (1 - \tfrac{1}{2}) \times (1 - \tfrac{1}{5}) = 120 \times \tfrac{1}{2} \times \tfrac{4}{5} = 48$

Among these 48, how many are divisible by 7?

The multiples of 7 from 1 to 120 that are NOT divisible by 2 or 5 are:

$7, 21, 49, 63, 77, 91, 119$ (that's 7 numbers)

Final answer: $48 - 7 = 41$

Therefore, 41 integers from 1 to 120 are divisible by none of 2, 5, and 7.