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# How many integers, between 100 and 150, inclusive, can be

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Intern
Joined: 19 Aug 2004
Posts: 47
Followers: 1

Kudos [?]: 3 [0], given: 0

How many integers, between 100 and 150, inclusive, can be [#permalink]  01 Jul 2005, 14:14
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0% (00:00) correct 0% (00:00) wrong based on 1 sessions
How many integers, between 100 and 150, inclusive, can be evenly divided by neither 3 nor 5?
Senior Manager
Joined: 30 May 2005
Posts: 374
Followers: 1

Kudos [?]: 7 [0], given: 0

Answer is 51 - (n(3) + n(5) - n(15)) where n represents the number of numbers divisible by the one in parantheses.

For 3:

102 = 3*34 is the first number divisible and 150 = 3*50 is the last:

n(3) = 50-34+1 = 17
n

For 5

100 = 5*20 is the first and 150 = 5*30 is last

n(5) = 30-20+1 = 11

For 15

105=15*7 and 150 = 15*10

n(15) = 10-7+1 = 4

Intern
Joined: 19 Aug 2004
Posts: 47
Followers: 1

Kudos [?]: 3 [0], given: 0

I came up with 27 as well but Princeton Review says the answer is 26
Director
Joined: 18 Apr 2005
Posts: 549
Location: Canuckland
Followers: 1

Kudos [?]: 8 [0], given: 0

((150 - 102)/3 +1) + ((150-100)/5 +1) - 1 -1 -1=25

overlaps at 105, 135 and 150 only

51-25 = 26
Intern
Joined: 19 Aug 2004
Posts: 47
Followers: 1

Kudos [?]: 3 [0], given: 0

Sparky - Overlap also occurs at 120.

So, it should be 51 - 24 = 27
Senior Manager
Joined: 30 May 2005
Posts: 374
Followers: 1

Kudos [?]: 7 [0], given: 0

Let's list it out:

101,103,104,106,107,109 = 6
112,113,116,118,119 = 5
121,122,124,127,128 = 5
131,133,134,136,137,139 =6
142,143,146,148,149 = 5

Total is 27
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