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# Every day a certain bank calculates its average daily

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Every day a certain bank calculates its average daily [#permalink]

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30 Mar 2005, 21:52
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Every day a certain bank calculates its average daily deposit for that calendar month up to and including that day. If on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100, what is the probability that the average daily deposit up to and including that day contains fewer than 5 decimal places?

(A) 1/10
(B) 2/15
(C) 4/15
(D) 3/10
(E) 11/30
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31 Mar 2005, 06:35
I think it should be D

calculating each one needs 4mn at least even with the fact that I took 200 as an easy number to calculate, found around 7-8-9 out of 30 days and by re-checking the suspicious one I've found 9 ways but I may be wrong on this one...

Of course I am interested in the algebra solution
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31 Mar 2005, 06:38
I'll let on the OA and OE in due time
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31 Mar 2005, 08:17
"D"

hmmm.....let's take 101 as the prime number....we know that all total days = 30

days multiple of 5s = 5,10,15,20,25,30......if which 15,30 will not yield finite decimals as it contains a 3....so now we have 4 possible days.

days multiple of 2s, but not any other prime number = 2,4,8,16 = 4 days

Also day 1 will always yield to whole number

total number of favorable days = 9

p(e) = 9/30 = 3/10
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31 Mar 2005, 23:43
both of you are right, and banerjeea_98 is spot on with his explanation.
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01 Apr 2005, 17:59
banerjeea_98 wrote:
"D"

hmmm.....let's take 101 as the prime number....we know that all total days = 30

days multiple of 5s = 5,10,15,20,25,30......if which 15,30 will not yield finite decimals as it contains a 3....so now we have 4 possible days.

days multiple of 2s, but not any other prime number = 2,4,8,16 = 4 days

Also day 1 will always yield to whole number

total number of favorable days = 9

p(e) = 9/30 = 3/10

terrific !
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02 Apr 2005, 16:36
banerjeea_98 wrote:
"D"

hmmm.....let's take 101 as the prime number....we know that all total days = 30

days multiple of 5s = 5,10,15,20,25,30......if which 15,30 will not yield finite decimals as it contains a 3....so now we have 4 possible days.

days multiple of 2s, but not any other prime number = 2,4,8,16 = 4 days

Also day 1 will always yield to whole number

total number of favorable days = 9

p(e) = 9/30 = 3/10

02 Apr 2005, 16:36
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