Maths question...

DirtyJester

DirtyJester

DirtyJester

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Three minutes to complete.

5aouvq.jpg


Answer and reasoning.

GO..
 
I'll venture at 71. I'm sure the reasoning is something to do with prime numbers but I'll be honest, at 7.30 pm on a Monday evening I just brute forced it in Excel :p

Awaiting the "it's a riddle" solution ... I'm an engineer. I don't deal with that!
 
seems striaght forward enough

firstly the number is less than 100

secondly it is one greater than a multiple of 5 so from that we can deduce it ends in a 1 or a 6

it is also one less than a common multiple of both 6 and 8... so we're looking for a multiple of 6 and 8 ending in a 2 (we can't have one ending in a 7!)

there is only one common multiple of both 6 and 8 that ends in a 2 and is less than 100

our number is that multiple - 1

also it is a prime number so our OCD book packer isn't going to find a way to pack his books in equal numbers - Katie needs to tell Jack to get a grip
 
Did it like Dowie but didnt spot the 2/7 business as it was fairly obvious once you know its one below a number that has the factors 6 and 8

ps3ud0 :cool:
 
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dowie said:
seems striaght forward enough

firstly the number is less than 100

secondly it is one greater than a multiple of 5 so from that we can deduce it ends in a 1 or a 6

it is also one less than a common multiple of both 6 and 8... so we're looking for a multiple of 6 and 8 ending in a 2 (we can't have one ending in a 7!)

there is only one common multiple of both 6 and 8 that ends in a 2 and is less than 100

our number is that multiple - 1

also it is a prime number so our OCD book packer isn't going to find a way to pack his books in equal numbers - Katie needs to tell Jack to get a grip
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are you sure :confused:
 
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ps3ud0 said:
Did it like Dowie but didnt spot the 2/7 business as it was fairly obvious once you know its one above a number that has the factors 6 and 8

ps3ud0 :cool:
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eh There are 4 multiples of 6 and 8 that are less than 100 what made you pick the answer you did pick and disregard the other three if you didn't spot that the multiple would have to end in a 2? Was it more trial an error - subtract 1 then divide all 4 possible answers by 5?
 
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dowie said:
eh? But there are 4 multiples of 6 and 8 that are less than 100 what made you pick the answer you did pick and disregard the other three if you didn't spot that it would have to end in a 2?
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Sorry thats not what I meant, I just didnt bother to methodise beyond it being a one below a multiple of 6 and 8 and knowing it was near 100 I just hit the answer at the first try and let someone else waste time replying with a proper method :p

And I typoed above, meant below in your quote of mine...

ps3ud0 :cool:
 
X is number of books in box, because there is a remainder you divide x by 6 and 8, x must be odd.

There is one left over when you divide x by 5, so the last digit is either 1 or 6. 6 is even, so the last digit must be one

So now you need to find a multiple of both 6 and 8 that ends in 2. This can be done quickly through trail and error to get the final answer of 71

Or, you could break 6 and 8 into their prime factors of 2, 3, and 4, multiply them together to get 12, and then do:

x = 12y - 1
12y mod 10 = 2

To get y = 6 and therefore x = 71
 
I went through my 6 and 8 multiples till I found the first number that they had in common, 72 and 5 goes in to 70 and that fits the question so 71. Took maybe 45 seconds doing in that way in my head.
 
defo didn't brute force it but i've got the following:

11, 23, 31, 41, 47, 71 all satisfy the criteria, given the girl's original estimate of 100 books it'd make sense to go for the largest number and go with 71, although as dave pointed out i'm an engineer, i don't do riddles
 
OliP said:
I went through my 6 and 8 multiples till I found the first number that they had in common, 72 and 5 goes in to 70 and that fits the question so 71. Took maybe 45 seconds doing in that way in my head.
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erm no... you've missed a couple of numbers there, but you got lucky
 
adolf hamster said:
defo didn't brute force it but i've got the following:

11, 23, 31, 41, 47, 71 all satisfy the criteria, given the girl's original estimate of 100 books it'd make sense to go for the largest number and go with 71, although as dave pointed out i'm an engineer, i don't do riddles
Click to expand...

missed your spoiler.. they don't all fit the criteria
 
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