Question

How many even integers n, where 100 ≤ n ≤ 200, are divisible neither by seven nor by nine?

• A. 39
• B. 40
• C. 38
• D. 37
• E. 32

Hint:

For counting within a range, use the formula: Number of multiples of $k$ from $a$ to $b$ = $\left\lfloor \frac{b}{k} \right\rfloor - \left\lfloor \frac{a-1}{k} \right\rfloor$.

Answer 1

The Correct Option is D

Step 1: Total Even Numbers:
Even numbers from 100 to 200 inclusive. First even = 100, last even = 200. Number of evens = $\frac{200-100}{2} + 1 = \frac{100}{2} + 1 = 50 + 1 = 51$ even numbers.
Step 2: Divisible by 7:
Even numbers divisible by 7 means divisible by 14 (since 7 and 2). Multiples of 14 between 100 and 200: $14 \times 8 = 112$, $14 \times 9 = 126$, ..., $14 \times 14 = 196$. $14 \times 7 = 98$ (too low), $14 \times 15 = 210$ (too hig(h). So from $8$ to $14$: count = $14 - 8 + 1 = 7$ numbers.
Step 3: Divisible by 9:
Even numbers divisible by 9 means divisible by 18. Multiples of 18 between 100 and 200: $18 \times 6 = 108$, $18 \times 7 = 126$, ..., $18 \times 11 = 198$. $18 \times 5 = 90$ (too low), $18 \times 12 = 216$ (too hig(h). So from $6$ to $11$: count = $11 - 6 + 1 = 6$ numbers.
Step 4: Divisible by both 7 and 9 (i.e., by LCM=63):
Even numbers divisible by 63 means divisible by 126. Multiples of 126 between 100 and 200: only 126 itself. $126$ is even. Count = 1.
Step 5: Apply Inclusion-Exclusion:
Numbers divisible by 7 or 9 = Divisible by 7 + Divisible by 9 - Divisible by both = $7 + 6 - 1 = 12$.
Step 6: Numbers Divisible Neither:
Total evens - (Divisible by 7 or 9) = $51 - 12 = 39$. But 39 is an options. Let's double-check counts. Multiples of 14: 112,126,140,154,168,182,196. That's 7. Correct. Multiples of 18: 108,126,144,162,180,198. That's 6. Correct. Common: 126. Correct. So 51-12=39. So answer should be 39. But the answer key says 37? Let's check if 100 is included. 100 is even, not divisible by 7 or 9. 200 is even, not divisible by 7 or 9. Our count seems correct. Possibly the range is 100 to 200 exclusive? Or maybe "n" is considered from 100 to 200 including? 100 to 200 inclusive gives 51 evens. Let's count manually: (200-100)/2=50, +1=51. Yes. Maybe they mean "divisible neither by seven nor by nine" including those divisible by both? We used inclusion-exclusion correctly. 51-12=39. Let's check the count of evens divisible by 14: 14*8=112, 14*9=126, 14*10=140, 14*11=154, 14*12=168, 14*13=182, 14*14=196. 7 numbers. Evens divisible by 18: 18*6=108, 18*7=126, 18*8=144, 18*9=162, 18*10=180, 18*11=198. 6 numbers. Union: 7+6-1=12. So 51-12=39. Given the optionss, 39 is there. So it's likely 39. However, the initial extraction said correct answer 37. Let's trust the calculation.
Step 7: Final Answer:
The number of even integers is 39.