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The multiples are 12,15....198

Thus, Nth= a1+ (n-1)d
198= 12 + (n-1)d
198= 12 + (n-1)3
N= 63 ans.

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MathRevolution
[Math Revolution GMAT math practice question]

\(3, 6, 9, 12, …\) are multiples of \(3\). How many multiples of \(3\) lie between \(10\) and \(200\), inclusive?

\(A. 54\)
\(B. 57\)
\(C. 60\)
\(D. 63\)
\(E. 66\)

The difference between the endpoints of the range = 200-10 = 190.
Since neither endpoint is a multiple of 3, we can count the multiples of 3 simply by dividing 3 into 190:
190/3 ≈ 63

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=>

\(12\) is the smallest multiple of \(3\) and \(198\) is the largest multiple of \(3\) between \(10\) and \(200\), inclusive.
Thus, the number of multiples of \(3\) between \(10\) and \(200\), inclusive is \(\frac{( 198 – 12 )}{3 + 1} = \frac{186}{3} + 1 = 62 + 1 = 63\)

Therefore, the answer is D.
Answer: D
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MathRevolution
[Math Revolution GMAT math practice question]

\(3, 6, 9, 12, …\) are multiples of \(3\). How many multiples of \(3\) lie between \(10\) and \(200\), inclusive?

\(A. 54\)
\(B. 57\)
\(C. 60\)
\(D. 63\)
\(E. 66\)

The smallest multiple of 3 within the range is 12 and the largest is 198. So the number of multiples of 3 within the range is:

(198 - 12)/3 + 1 = 186/3 + 1= 62 + 1 = 63

Answer: D
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MathRevolution
[Math Revolution GMAT math practice question]

\(3, 6, 9, 12, …\) are multiples of \(3\). How many multiples of \(3\) lie between \(10\) and \(200\), inclusive?

\(A. 54\)
\(B. 57\)
\(C. 60\)
\(D. 63\)
\(E. 66\)

Last digit- First/spacing between digits + 1

198-12/3 +1
62+1=63
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3,6,9,12,…3,6,9,12,… are multiples of 33. How many multiples of 33 lie between 1010 and 200200, inclusive?

A.54A.54
B.57B.57
C.60C.60
D.63D.63
E.66

I used AP formula, which is an=a1+(n-1)d. an=198, nearest integer divisible by 3, and a1=12, the smallest integer 10 divisible by 3 greater than 10 as well as d=3.
198=12+(n-1)*3, we will divide both sides by 3
66=4+n-1
n=63
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number of integers : 200-10=190 integers.
every third integer is a multiple of 3 so 190/3=63
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