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#1 |
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Bronze Medalist
Jan 2004
Mumbai,India
22·33·19 Posts |
a) Find the largest possible number containing any 9 of the 10 digits (0 included) which can be divided by 11 and leaving no remainder b) Find the smallest possible number, with the same conditions, divisible fully with no remainder by 11 Eg: 896743012 (5 omitted). This is divisible by 11 but its neither the largest nor the smallest number. Mally
Last fiddled with by mfgoode on 2006-08-08 at 15:48 |
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#2 |
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Jun 2003
The Texas Hill Country
108910 Posts |
Do you consider numbers starting with "0" to be properly formed?
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#3 | |
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Nov 2003
22·5·373 Posts |
Quote:
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#4 |
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Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts |
:surprised
Not really Wacky as it would make a difference in the smallest number if we considered this permissible. Mally
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#5 | |
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Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3×5×719 Posts |
Quote:
Paul |
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#6 | |
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Nov 2003
22×5×373 Posts |
Quote:
mfgoode is on my 'ignore' list as I believe that trying to have a rational discussion with him is pointless. |
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#7 | |
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Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts |
Quote:
Thats okay with me R.D. IT suits me fine. Good riddance to bad rubbish. I dont need you anyway as you seldom solve anything in my opinion. As Dr. Peter would say 'You have reached your level of incompetence' and you intend staying there for quite awhile. The venom you tend to exude with every post will consume you. No hard feelings and I wish you luck Mally
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#8 |
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Dec 2005
19610 Posts |
102347586 987652413
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#9 |
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Bronze Medalist
Jan 2004
Mumbai,India
22·33·19 Posts |
You are absolutely right. Can you give us the rules you used to obtain this ? Mally
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#10 |
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Dec 2005
110001002 Posts |
Simple:
for the maximum you start with 98765 and then you try to obey to the eleven division rule. Same for finding the minimum, assuming that you cannot put 0 at the beginning, the best you can do is start with 10234
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#11 | |
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Bronze Medalist
Jan 2004
Mumbai,India
22×33×19 Posts |
Quote:
Thank you Kees thats great! I presume you mean the division rule that if the sum of the digits in the even places is the same as the sum of digits in the odd places then the number is divisible by 11 without a remainder. However little is known of the the other rule that if the difference between the sum of the odd and the even digits is 11 or a multiple of 11 then it can equally be applied and is valid. Regards, Mally
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