Question

How many even integers 'n', where, $100\le n\le200$ are divisible neither by 7 nor by 9?

• A. 37
• B. 38
• C. 39
• D. 44
• E. 40

Hint:

When asked for numbers "divisible by neither A nor B", first find the total numbers, then subtract the union of sets A and B using the Principle of Inclusion-Exclusion: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

Answer 1

The Correct Option is C

Step 1: Find the total number of even integers.
Even integers from 100 to 200 form an arithmetic progression: 100, 102, ..., 200.
Total number of even integers = $\frac{200 - 100}{2} + 1 = 51$.

Step 2: Find even integers divisible by 7 and 9.
An even integer divisible by 7 must be divisible by 14. (Multiples of 14 between 100 and 200: 112 to 196 $\rightarrow$ 7 integers).
An even integer divisible by 9 must be divisible by 18. (Multiples of 18 between 100 and 200: 108 to 198 $\rightarrow$ 6 integers).

Step 3: Apply Principle of Inclusion-Exclusion.
Even integers divisible by both 7 and 9 must be divisible by LCM(14, 18) = 126.
There is only 1 such integer (126) in the given range.
Total even integers divisible by 7 or 9 = $7 + 6 - 1 = 12$.
Therefore, integers divisible by neither = $51 - 12 = 39$.