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el1981
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A..

198=102+(n-1)3 where n is number of terms

solve and you gt n=33
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fresinha12
A..

198=102+(n-1)3 where n is number of terms

solve and you gt n=33


dude..where have you been ? long time..
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el1981
What is the total number of integers beween 100 and 200 that are divisible by 3?
A. 33
B. 32
C. 31
D. 30
E. 29


BTWN 200 and 100, so we dont count 200 and 100. Thus it should be 199-101 +1 /3 = 33
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el1981
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fresinha12
A..

198=102+(n-1)3 where n is number of terms

solve and you gt n=33

how did you come up with this 198=102+(n-1)3 ??? Could you please clarify.
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GMATBLACKBELT
el1981
What is the total number of integers beween 100 and 200 that are divisible by 3?
A. 33
B. 32
C. 31
D. 30
E. 29


BTWN 200 and 100, so we dont count 200 and 100. Thus it should be 199-101 +1 /3 = 33

I like your version better than mine - it's much faster. kudos!
My version is similar finding the closest starting/ending # divisible by 3
102/3 = 34 (start) and 198/3 (last) = 66
66 - 34 = 32 +1 = 33



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