Re: [algogeeks] Re: Highest reminder
Hi Ankit, If that is the case, I can very well say, 23 = 2 X 1 + 21 If you divide 23 by 11, remainder would be 1 and not 12. On Thu, May 30, 2013 at 1:16 PM, Ankit Agarwal wrote: > Hi, > > 23 = 11 X 1 + 12. Thus 12 would the highest remainder. Not 11 > > > > On Thu, May 30, 2013 at 10:24 AM, Sreenivas Sigharam > wrote: > >> Dave's explanation was clear..and informative.. Thank you Dave.. >> >> Thank you , Soumya Prasad, for a simple but nice topic.. >> >> Thank you, >> Sigharam. >> >> >> On Thu, May 30, 2013 at 10:16 AM, Sanjay Rajpal wrote: >> >>> Hi Ankit, >>> >>> for 23, how can the remainder be 12 ? Can you elaborate more ? >>> >>> *Regards,* >>> *Sanjay Kumar* >>> *Software Engineer(Development)* >>> *Winshuttle Softwares(India) Pvt. Ltd.* >>> *Mobile +91-89012-36292, +91-80535-66286* >>> *Email: sanjay.ku...@winshuttle.com* >>> * >>> *** >>> >>> * * >>> >>> ** >>> * >>> * >>> >>> >>> On Thu, May 30, 2013 at 9:40 AM, Ankit Agarwal >>> wrote: >>> @Dave: For N = 23, the highest remainder is 12, not 11 On Thu, May 30, 2013 at 5:02 AM, Dave wrote: > The highest remainder when dividing n by a number less than n is > floor((n-1)/2). > For n = 11, floor((11-1)/2) = floor(10/2) = floor(5) = 5. > For n = 17, floor((17-1)/2) = 8 > For n = 23, floor((23-1)/2) = 11 > > For n = 12, floor((12-1)/2) = floor(11/2) = floor(5.5) = 5. > Etc. > > Dave > > > On Wednesday, May 29, 2013 1:36:13 PM UTC-5, Ankit wrote: > >> Hi, >> >> Number 23: = 11 * 1 + 12 Number/2 = 11.5 >> >> Number 17: = 9 * 1 + 8 Number/2 = 8.5 >> >> So, its neither floor(n/2) +- 1, nor ceil(n/2) +- 1 >> >> >> On Wed, May 29, 2013 at 2:19 PM, Ankit Sambyal > > wrote: >> >>> Hi Nikhil, >>> >>> Highest remainder can't be floor(n/2) - 1. >>> If n = 11, highest remainder would be 5 when it is divided by 6, but >>> your formula gives 4. >>> >>> >>> >>> On Mon, May 27, 2013 at 8:16 PM, Nikhil Kumar >>> wrote: >>> Since we need to divide so the quotient should be at least 1, and we need greatest remainder, so we need the least no. which will give the quotient 1 upon dividing and that would be the no. you described. Also you would have noted the greatest remainder would be floor(n/2)-1 . On Thursday, 16 May 2013 13:56:40 UTC+5:30, Soumya Prasad Ukil wrote: > > > For a given number when divided by a number between 1 and n. I > figured out that highest reminder can be got if I divide the number by > (⌊(n/2)⌋+1) .Can anyone give me pointers ? > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+...@**googlegroups.com. >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To unsubscribe from this group and stop receiving emails from it, >>> send an email to algogeeks+...@**googlegroups.com. >>> >>> >>> >> >> >> >> -- >> >> *Ankit Agarwal* >> >> >> -- > You received this message because you are subscribed to the Google > Groups "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send > an email to algogeeks+unsubscr...@googlegroups.com. > > > -- *Ankit Agarwal* *Software Engineer* *Datacenter & Cloud Division* *Citrix R&D India Pvt. Ltd.* *Ph. No. +91-8095470278* -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com. >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to algogeeks+unsubscr...@googlegroups.com. >>> >>> >>> >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to algogeeks+unsubscr...@googlegroups.com. >> >> >> > > > > -- > > *Ankit Agarwal* > > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to algogeeks+unsubscr...@googlegroups.com. > > > -- You received this message because you are subscribe
Re: [algogeeks] Re: Highest reminder
oh.. sorry.. my bad... On Thu, May 30, 2013 at 2:50 PM, Sanjay Rajpal wrote: > 23 = 11 * 2 + 1 > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to algogeeks+unsubscr...@googlegroups.com. > > > -- *Ankit Agarwal* *Software Engineer* *Datacenter & Cloud Division* *Citrix R&D India Pvt. Ltd.* *Ph. No. +91-8095470278* -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] Re: Highest reminder
23 = 11 * 2 + 1 -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] Re: Highest reminder
Hi, 23 = 11 X 1 + 12. Thus 12 would the highest remainder. Not 11 On Thu, May 30, 2013 at 10:24 AM, Sreenivas Sigharam wrote: > Dave's explanation was clear..and informative.. Thank you Dave.. > > Thank you , Soumya Prasad, for a simple but nice topic.. > > Thank you, > Sigharam. > > > On Thu, May 30, 2013 at 10:16 AM, Sanjay Rajpal wrote: > >> Hi Ankit, >> >> for 23, how can the remainder be 12 ? Can you elaborate more ? >> >> *Regards,* >> *Sanjay Kumar* >> *Software Engineer(Development)* >> *Winshuttle Softwares(India) Pvt. Ltd.* >> *Mobile +91-89012-36292, +91-80535-66286* >> *Email: sanjay.ku...@winshuttle.com* >> * >> *** >> >> * * >> >> ** >> * >> * >> >> >> On Thu, May 30, 2013 at 9:40 AM, Ankit Agarwal wrote: >> >>> @Dave: >>> >>> For N = 23, the highest remainder is 12, not 11 >>> >>> >>> On Thu, May 30, 2013 at 5:02 AM, Dave wrote: >>> The highest remainder when dividing n by a number less than n is floor((n-1)/2). For n = 11, floor((11-1)/2) = floor(10/2) = floor(5) = 5. For n = 17, floor((17-1)/2) = 8 For n = 23, floor((23-1)/2) = 11 For n = 12, floor((12-1)/2) = floor(11/2) = floor(5.5) = 5. Etc. Dave On Wednesday, May 29, 2013 1:36:13 PM UTC-5, Ankit wrote: > Hi, > > Number 23: = 11 * 1 + 12 Number/2 = 11.5 > > Number 17: = 9 * 1 + 8 Number/2 = 8.5 > > So, its neither floor(n/2) +- 1, nor ceil(n/2) +- 1 > > > On Wed, May 29, 2013 at 2:19 PM, Ankit Sambyal > wrote: > >> Hi Nikhil, >> >> Highest remainder can't be floor(n/2) - 1. >> If n = 11, highest remainder would be 5 when it is divided by 6, but >> your formula gives 4. >> >> >> >> On Mon, May 27, 2013 at 8:16 PM, Nikhil Kumar wrote: >> >>> Since we need to divide so the quotient should be at least 1, and we >>> need greatest remainder, so we need the least no. which will give the >>> quotient 1 upon dividing and that would be the no. you described. >>> Also you would have noted the greatest remainder would be >>> floor(n/2)-1 . >>> >>> >>> On Thursday, 16 May 2013 13:56:40 UTC+5:30, Soumya Prasad Ukil wrote: For a given number when divided by a number between 1 and n. I figured out that highest reminder can be got if I divide the number by (⌊(n/2)⌋+1) .Can anyone give me pointers ? >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To unsubscribe from this group and stop receiving emails from it, >>> send an email to algogeeks+...@**googlegroups.com. >>> >>> >>> >> >> -- >> You received this message because you are subscribed to the Google >> Groups "Algorithm Geeks" group. >> To unsubscribe from this group and stop receiving emails from it, >> send an email to algogeeks+...@**googlegroups.com. >> >> >> > > > > -- > > *Ankit Agarwal* > > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com. >>> >>> >>> >>> -- >>> >>> *Ankit Agarwal* >>> *Software Engineer* >>> *Datacenter & Cloud Division* >>> *Citrix R&D India Pvt. Ltd.* >>> *Ph. No. +91-8095470278* >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to algogeeks+unsubscr...@googlegroups.com. >>> >>> >>> >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to algogeeks+unsubscr...@googlegroups.com. >> >> >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to algogeeks+unsubscr...@googlegroups.com. > > > -- *Ankit Agarwal* -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] Re: Highest reminder
Dave's explanation was clear..and informative.. Thank you Dave.. Thank you , Soumya Prasad, for a simple but nice topic.. Thank you, Sigharam. On Thu, May 30, 2013 at 10:16 AM, Sanjay Rajpal wrote: > Hi Ankit, > > for 23, how can the remainder be 12 ? Can you elaborate more ? > > *Regards,* > *Sanjay Kumar* > *Software Engineer(Development)* > *Winshuttle Softwares(India) Pvt. Ltd.* > *Mobile +91-89012-36292, +91-80535-66286* > *Email: sanjay.ku...@winshuttle.com* > * > *** > > * * > > ** > * > * > > > On Thu, May 30, 2013 at 9:40 AM, Ankit Agarwal wrote: > >> @Dave: >> >> For N = 23, the highest remainder is 12, not 11 >> >> >> On Thu, May 30, 2013 at 5:02 AM, Dave wrote: >> >>> The highest remainder when dividing n by a number less than n is >>> floor((n-1)/2). >>> For n = 11, floor((11-1)/2) = floor(10/2) = floor(5) = 5. >>> For n = 17, floor((17-1)/2) = 8 >>> For n = 23, floor((23-1)/2) = 11 >>> >>> For n = 12, floor((12-1)/2) = floor(11/2) = floor(5.5) = 5. >>> Etc. >>> >>> Dave >>> >>> >>> On Wednesday, May 29, 2013 1:36:13 PM UTC-5, Ankit wrote: >>> Hi, Number 23: = 11 * 1 + 12 Number/2 = 11.5 Number 17: = 9 * 1 + 8 Number/2 = 8.5 So, its neither floor(n/2) +- 1, nor ceil(n/2) +- 1 On Wed, May 29, 2013 at 2:19 PM, Ankit Sambyal wrote: > Hi Nikhil, > > Highest remainder can't be floor(n/2) - 1. > If n = 11, highest remainder would be 5 when it is divided by 6, but > your formula gives 4. > > > > On Mon, May 27, 2013 at 8:16 PM, Nikhil Kumar wrote: > >> Since we need to divide so the quotient should be at least 1, and we >> need greatest remainder, so we need the least no. which will give the >> quotient 1 upon dividing and that would be the no. you described. >> Also you would have noted the greatest remainder would be >> floor(n/2)-1 . >> >> >> On Thursday, 16 May 2013 13:56:40 UTC+5:30, Soumya Prasad Ukil wrote: >>> >>> >>> For a given number when divided by a number between 1 and n. I >>> figured out that highest reminder can be got if I divide the number by >>> (⌊(n/2)⌋+1) .Can anyone give me pointers ? >>> >> -- >> You received this message because you are subscribed to the Google >> Groups "Algorithm Geeks" group. >> To unsubscribe from this group and stop receiving emails from it, >> send an email to algogeeks+...@**googlegroups.com. >> >> >> > > -- > You received this message because you are subscribed to the Google > Groups "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send > an email to algogeeks+...@**googlegroups.com. > > > -- *Ankit Agarwal* -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to algogeeks+unsubscr...@googlegroups.com. >>> >>> >>> >> >> >> >> -- >> >> *Ankit Agarwal* >> *Software Engineer* >> *Datacenter & Cloud Division* >> *Citrix R&D India Pvt. Ltd.* >> *Ph. No. +91-8095470278* >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to algogeeks+unsubscr...@googlegroups.com. >> >> >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to algogeeks+unsubscr...@googlegroups.com. > > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] Re: Highest reminder
Hi Ankit, for 23, how can the remainder be 12 ? Can you elaborate more ? *Regards,* *Sanjay Kumar* *Software Engineer(Development)* *Winshuttle Softwares(India) Pvt. Ltd.* *Mobile +91-89012-36292, +91-80535-66286* *Email: sanjay.ku...@winshuttle.com* * *** * * ** * * On Thu, May 30, 2013 at 9:40 AM, Ankit Agarwal wrote: > @Dave: > > For N = 23, the highest remainder is 12, not 11 > > > On Thu, May 30, 2013 at 5:02 AM, Dave wrote: > >> The highest remainder when dividing n by a number less than n is >> floor((n-1)/2). >> For n = 11, floor((11-1)/2) = floor(10/2) = floor(5) = 5. >> For n = 17, floor((17-1)/2) = 8 >> For n = 23, floor((23-1)/2) = 11 >> >> For n = 12, floor((12-1)/2) = floor(11/2) = floor(5.5) = 5. >> Etc. >> >> Dave >> >> >> On Wednesday, May 29, 2013 1:36:13 PM UTC-5, Ankit wrote: >> >>> Hi, >>> >>> Number 23: = 11 * 1 + 12 Number/2 = 11.5 >>> >>> Number 17: = 9 * 1 + 8 Number/2 = 8.5 >>> >>> So, its neither floor(n/2) +- 1, nor ceil(n/2) +- 1 >>> >>> >>> On Wed, May 29, 2013 at 2:19 PM, Ankit Sambyal wrote: >>> Hi Nikhil, Highest remainder can't be floor(n/2) - 1. If n = 11, highest remainder would be 5 when it is divided by 6, but your formula gives 4. On Mon, May 27, 2013 at 8:16 PM, Nikhil Kumar wrote: > Since we need to divide so the quotient should be at least 1, and we > need greatest remainder, so we need the least no. which will give the > quotient 1 upon dividing and that would be the no. you described. > Also you would have noted the greatest remainder would be > floor(n/2)-1 . > > > On Thursday, 16 May 2013 13:56:40 UTC+5:30, Soumya Prasad Ukil wrote: >> >> >> For a given number when divided by a number between 1 and n. I >> figured out that highest reminder can be got if I divide the number by >> (⌊(n/2)⌋+1) .Can anyone give me pointers ? >> > -- > You received this message because you are subscribed to the Google > Groups "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send > an email to algogeeks+...@**googlegroups.com. > > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+...@**googlegroups.com. >>> >>> >>> >>> -- >>> >>> *Ankit Agarwal* >>> >>> >>> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to algogeeks+unsubscr...@googlegroups.com. >> >> >> > > > > -- > > *Ankit Agarwal* > *Software Engineer* > *Datacenter & Cloud Division* > *Citrix R&D India Pvt. Ltd.* > *Ph. No. +91-8095470278* > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to algogeeks+unsubscr...@googlegroups.com. > > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] Re: Highest reminder
@Dave: For N = 23, the highest remainder is 12, not 11 On Thu, May 30, 2013 at 5:02 AM, Dave wrote: > The highest remainder when dividing n by a number less than n is > floor((n-1)/2). > For n = 11, floor((11-1)/2) = floor(10/2) = floor(5) = 5. > For n = 17, floor((17-1)/2) = 8 > For n = 23, floor((23-1)/2) = 11 > > For n = 12, floor((12-1)/2) = floor(11/2) = floor(5.5) = 5. > Etc. > > Dave > > > On Wednesday, May 29, 2013 1:36:13 PM UTC-5, Ankit wrote: > >> Hi, >> >> Number 23: = 11 * 1 + 12 Number/2 = 11.5 >> >> Number 17: = 9 * 1 + 8 Number/2 = 8.5 >> >> So, its neither floor(n/2) +- 1, nor ceil(n/2) +- 1 >> >> >> On Wed, May 29, 2013 at 2:19 PM, Ankit Sambyal wrote: >> >>> Hi Nikhil, >>> >>> Highest remainder can't be floor(n/2) - 1. >>> If n = 11, highest remainder would be 5 when it is divided by 6, but >>> your formula gives 4. >>> >>> >>> >>> On Mon, May 27, 2013 at 8:16 PM, Nikhil Kumar wrote: >>> Since we need to divide so the quotient should be at least 1, and we need greatest remainder, so we need the least no. which will give the quotient 1 upon dividing and that would be the no. you described. Also you would have noted the greatest remainder would be floor(n/2)-1 . On Thursday, 16 May 2013 13:56:40 UTC+5:30, Soumya Prasad Ukil wrote: > > > For a given number when divided by a number between 1 and n. I figured > out that highest reminder can be got if I divide the number by (⌊(n/2) > ⌋+1) .Can anyone give me pointers ? > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+...@**googlegroups.com. >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to algogeeks+...@**googlegroups.com. >>> >>> >>> >> >> >> >> -- >> >> *Ankit Agarwal* >> >> >> -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to algogeeks+unsubscr...@googlegroups.com. > > > -- *Ankit Agarwal* *Software Engineer* *Datacenter & Cloud Division* *Citrix R&D India Pvt. Ltd.* *Ph. No. +91-8095470278* -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] Re: Highest reminder
The highest remainder when dividing n by a number less than n is floor((n-1)/2). For n = 11, floor((11-1)/2) = floor(10/2) = floor(5) = 5. For n = 17, floor((17-1)/2) = 8 For n = 23, floor((23-1)/2) = 11 For n = 12, floor((12-1)/2) = floor(11/2) = floor(5.5) = 5. Etc. Dave On Wednesday, May 29, 2013 1:36:13 PM UTC-5, Ankit wrote: > Hi, > > Number 23: = 11 * 1 + 12 Number/2 = 11.5 > > Number 17: = 9 * 1 + 8 Number/2 = 8.5 > > So, its neither floor(n/2) +- 1, nor ceil(n/2) +- 1 > > > On Wed, May 29, 2013 at 2:19 PM, Ankit Sambyal > > > wrote: > >> Hi Nikhil, >> >> Highest remainder can't be floor(n/2) - 1. >> If n = 11, highest remainder would be 5 when it is divided by 6, but your >> formula gives 4. >> >> >> >> On Mon, May 27, 2013 at 8:16 PM, Nikhil Kumar >> >> > wrote: >> >>> Since we need to divide so the quotient should be at least 1, and we >>> need greatest remainder, so we need the least no. which will give the >>> quotient 1 upon dividing and that would be the no. you described. >>> Also you would have noted the greatest remainder would be floor(n/2)-1 . >>> >>> >>> On Thursday, 16 May 2013 13:56:40 UTC+5:30, Soumya Prasad Ukil wrote: For a given number when divided by a number between 1 and n. I figured out that highest reminder can be got if I divide the number by (⌊(n/2)⌋ +1) .Can anyone give me pointers ? >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to algogeeks+...@googlegroups.com . >>> >>> >>> >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to algogeeks+...@googlegroups.com . >> >> >> > > > > -- > > *Ankit Agarwal* > > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] Re: Highest reminder
Hi, Number 23: = 11 * 1 + 12 Number/2 = 11.5 Number 17: = 9 * 1 + 8 Number/2 = 8.5 So, its neither floor(n/2) +- 1, nor ceil(n/2) +- 1 On Wed, May 29, 2013 at 2:19 PM, Ankit Sambyal wrote: > Hi Nikhil, > > Highest remainder can't be floor(n/2) - 1. > If n = 11, highest remainder would be 5 when it is divided by 6, but your > formula gives 4. > > > > On Mon, May 27, 2013 at 8:16 PM, Nikhil Kumar wrote: > >> Since we need to divide so the quotient should be at least 1, and we need >> greatest remainder, so we need the least no. which will give the quotient 1 >> upon dividing and that would be the no. you described. >> Also you would have noted the greatest remainder would be floor(n/2)-1 . >> >> >> On Thursday, 16 May 2013 13:56:40 UTC+5:30, Soumya Prasad Ukil wrote: >>> >>> >>> For a given number when divided by a number between 1 and n. I figured >>> out that highest reminder can be got if I divide the number by (⌊(n/2)⌋+ >>> 1) .Can anyone give me pointers ? >>> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to algogeeks+unsubscr...@googlegroups.com. >> >> >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to algogeeks+unsubscr...@googlegroups.com. > > > -- *Ankit Agarwal* -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
Re: [algogeeks] Re: Highest reminder
Hi Nikhil, Highest remainder can't be floor(n/2) - 1. If n = 11, highest remainder would be 5 when it is divided by 6, but your formula gives 4. On Mon, May 27, 2013 at 8:16 PM, Nikhil Kumar wrote: > Since we need to divide so the quotient should be at least 1, and we need > greatest remainder, so we need the least no. which will give the quotient 1 > upon dividing and that would be the no. you described. > Also you would have noted the greatest remainder would be floor(n/2)-1 . > > > On Thursday, 16 May 2013 13:56:40 UTC+5:30, Soumya Prasad Ukil wrote: >> >> >> For a given number when divided by a number between 1 and n. I figured >> out that highest reminder can be got if I divide the number by (⌊(n/2)⌋+1 >> ) .Can anyone give me pointers ? >> > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to algogeeks+unsubscr...@googlegroups.com. > > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com.
