Welcome, %1$s. Please login or register.

August 22, 2017, 11:02:58 AM
: 1
: try out this...??  ( 4044 )
« : March 13, 2008, 04:55:09 PM From Rajesh»



try out this...??


In a shop, a burger can be purchased in quantities: 6, 9 and 20 only. Looking at this, we can say one can purchase exactly 15 pieces by buying a 6 and a 9, but cant buy exactly 10 burgers. What is the maximum number of burgers that a customer cannot purchase?

 
Liked It? Share it!

              


« #1 : March 13, 2008, 09:04:25 PM From Shrinidhi»

There are infinite number of burgers that he cannot purchase.

For any number greater than 15, he can purchase all burgers with numbers that are divisible by 3.
i.e. He can purchase 21, 24, 27, 30... 99, 102, etc burgers.

Now since 20 can be added to these numbers, he can purchase any burger which are "multiples of 3 + 2" (2 because 20 mod 3 is 2). i.e. He can purchase 26, 29, 32,... 101, 104 etc. burgers.

However he cannot purchase burgers with numbers which are "multiples of 3 + 1". i.e. He cannot purchase burgers of number 25, 28, 31, 100, etc. burgers... which are infinite of them.
« #2 : March 17, 2008, 04:28:41 PM From vishwakarma.deepak»

@ Rajesh , really nice question rajesh..sounds simple but tough one to solve...

@srinidhi...u have good analysis..but there are some issues...see all ther numbers greater than 9 and multiple of 3 can be purchases for obvious reason...numbers will consitute of either multiple of 9 or 6...

ofcourse 20 mod 3 is 2 but not all the multiple of 3 plus 2 is multiple of 20...

finally u said (multiple of 3) +1 can not be purchased...but 46, 49,..100 all number of burger can ber purchases...


question needs more attentetion....

No offence intended..
Deepak
« #3 : March 17, 2008, 09:07:43 PM From Shrinidhi»


@ Rajesh , really nice question rajesh..sounds simple but tough one to solve...
Even I agree (learn't the harder way though  :P)

Quote
ofcourse 20 mod 3 is 2 but not all the multiple of 3 plus 2 is multiple of 20...
If you subtract 20 from a number which is a multiple of 3 + 2, then the number becomes a multiple of 3 (of course only for numbers greater than 22).

Quote
finally u said (multiple of 3) +1 can not be purchased...but 46, 49,..100 all number of burger can ber purchases...
I agree I was wrong here. If one subtracts 20 from a number which is a multiple of 3 + 1, then it will be another number which is a multiple of 3 + 2, and these many burgers can be bought as explained above.

So the solution is in figuring out the maximum number which
1. can be subtracted by 20 and not 40. i.e. the answer lies between 20 & 40+6.
2. is a multiple of 3 + 1.


So I feel the answer is 43.


Thanks Deepak for putting me on the right track :)

« #4 : March 17, 2008, 10:50:57 PM From Rajesh»

Great job Deepak.... n thanx for appreciation....  ;) !!
n Srinidhi finally u have done it....  :)
« #5 : March 20, 2008, 06:57:41 PM From nirav parikh»

lcm of 6 9 and 20 is 180 and 180n+1 will be the form that he cannot purchase. so should be infinity
: 1
« previous next »

 

Best RatedList All>>



Latest
Random



SMF 2.0.10 | SMF © 2015, Simple Machines | Contact Webmaster | OnlineFunDb.com © 2009/10 | Legal Disclaimer