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CTK Exchange
Suresh Reddy
Member since Nov-2-05
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Nov-02-05, 03:25 PM (EST) |
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"Reminder Problem"
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Hi, Can anyone pl try to solve the following problem: What is the reminder when ( 1!+2!+3!+.....+95! ) is divided by 15 ? I could not apply reminder theorm or Patten method to the above sum. Thank you, Suresh First they laugh at you. Then they ignore. After that they fight. And then you win. |
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alexb
Charter Member
1682 posts |
Nov-02-05, 03:29 PM (EST) |
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1. "RE: Reminder Problem"
In response to message #0
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>Can anyone pl try to solve the following problem: There is no try. Do! >What is the reminder when ( 1!+2!+3!+.....+95! ) is divided >by 15 ? > >I could not apply reminder theorm or Patten method to the >above sum. I am not sure about what are those, but you may want to note that, for some sufficiently large n, n! is divisible by 15. I am certain that 95! has this property, and so does 94!, and also 93!, etc. You should think of the smallest number with that property.
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Deepak Gupta
guest
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Nov-05-05, 10:55 AM (EST) |
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2. "RE: Reminder Problem"
In response to message #0
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>Hi, > >Can anyone pl try to solve the following problem: > >What is the reminder when ( 1!+2!+3!+.....+95! ) is divided >by 15 ? > >I could not apply reminder theorm or Patten method to the >above sum. > >Thank you, >Suresh >Well Suresh, Let us write the sum as follows - S = 1! + 2! + 3! + 4! + X; Note now that X is divisible by 15 (Always !!!) becoz each n! term in X when written down as a product 1*2*3*4*5*...*n can be seen to have 3 and 5 in the product :o) So the reminder must be 3. What say?
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iliaden
Member since Aug-14-05
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Nov-13-05, 09:00 PM (EST) |
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5. "RE: Reminder Problem"
In response to message #3
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We can simple count as follows: 1 + 2 + (2*3) + (2*3*4) = 1 + 2 + 6 + 24 = 33. 33/15 = 2, remainder 3. so the answer must be 3, not 5. If you have some other way to get to the answer, i'd like to see it. |
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