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Bunuel
Find the number of integers divisible by 3 in the interval from 10 to 200 inclusive.

(A) 61
(B) 62
(C) 63
(D) 64
(E) 65

Multiples of 3 from 10 to 200 are
12, 15, 18, . . . . . 198

The above is AP series. To find number of terms, use last term
198 = 12 + (n - 1)3
—>186 = (n - 1)3
—> n - 1 = 62
—> n = 63

IMO Option C.

Pls Hit kudos if you like the solution

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b/w 10 and 200:
first no divisible by 3 is 12=3*4
last no divisible by 3 is 198=3*66

Therefore,total numbers divisible by 3 b/w 10 and 200 = 66-4+1=63

Hence C
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Between 10 and 200:
- the first number divisible by 3 is 12
- the last number divisible by 3 is 198

The total number divisible by 3 is: (198-12)/3+1 = 63

IMO option C

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Bunuel
Find the number of integers divisible by 3 in the interval from 10 to 200 inclusive.

(A) 61
(B) 62
(C) 63
(D) 64
(E) 65

In the given interval, the smallest multiple of 3 is 12, and the greatest multiple of 3 is 198. Thus, the number of multiples of 3 is:

(198 - 12)/3 + 1 = 63

Answer: C
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Bunuel
Find the number of integers divisible by 3 in the interval from 10 to 200 inclusive.

(A) 61
(B) 62
(C) 63
(D) 64
(E) 65

No of integers divisible between \(1 - 200 = \frac{200}{3} = 66\)
No of integers divisible between \(1 - 10 = \frac{200}{3} = 3\)

Thus, no of integers divisible by 3 in the interval from 10 - 200 inclusive \(= 66 - 3 = 63\), Answer must be (C)
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Hello from the GMAT Club BumpBot!

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