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Suppose we have the integral of the form and then we do a substitution of the form . what will happen for the integral limits, 0 and ? —Preceding unsigned comment added by Re444 ( talk • contribs) 15:32, 18 September 2009 (UTC)
My friend sent me this question via mail and i tried various options for getting this sum.Even after 3 hrs of futile trying I failed pls help.The question as follows
A farmer was dividing his property among his children. This was what the farmer told to his first son “Take as many number of cattle as you can care for and your wife may take one-ninth of the remaining number.” To his second son he said “Take one more than what the first son took and your wife will have one ninth of the cattle remaining after you have taken.” This applied to the remaining sons too i.e. each son would take one more than the next oldest brother and their wives would have one ninth of the remaining after their husbands have got their share.
After the cattle were divided, the farmer proceeded to divide the gold bars he had among the children. Each gold bar was valued at 3.5 times each cow's value. The gold bars were divided such that each couple had equally valued inheritance. What are the number of cows, gold bars and the number of sons the farmer had?
My deductions :
so pls help...3 to 4 hrs of working on this didnt help much :-( 164.100.170.4 ( talk) 16:14, 18 September 2009 (UTC)
what does that mean???
Of course, if there's only 1 son, it's trivial, and (2 sons, 16 cows, no gold) works, but isn't very interesting. There are other solutions, with 2 sons. Not sure about 3 or more. - GTBacchus( talk) 20:13, 18 September 2009 (UTC)
(after ec) The problem has several possible solutions. For example:
I am confident that there are other solutions with more sons, but I haven't worked them out yet. Note though, that the number of gold bars will always be non-unique, because for any solution with n sons, one can always add np to the number of gold bars while meeting all the conditions of the problem. Abecedare ( talk) 20:15, 18 September 2009 (UTC)
There are no solutions with 3 sons. The number of cows the first son takes would have to be congruent to both 6 and 0, modulo 64, which is impossible. - GTBacchus( talk) 20:24, 18 September 2009 (UTC)
Hmmm, I found a few solutions with three sons!
I am guessing there are solutions with even more number of progeny, but I'm donw with this family :-) Abecedare ( talk) 20:40, 18 September 2009 (UTC)
Mathematics desk | ||
---|---|---|
< September 17 | << Aug | September | Oct >> | September 19 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Suppose we have the integral of the form and then we do a substitution of the form . what will happen for the integral limits, 0 and ? —Preceding unsigned comment added by Re444 ( talk • contribs) 15:32, 18 September 2009 (UTC)
My friend sent me this question via mail and i tried various options for getting this sum.Even after 3 hrs of futile trying I failed pls help.The question as follows
A farmer was dividing his property among his children. This was what the farmer told to his first son “Take as many number of cattle as you can care for and your wife may take one-ninth of the remaining number.” To his second son he said “Take one more than what the first son took and your wife will have one ninth of the cattle remaining after you have taken.” This applied to the remaining sons too i.e. each son would take one more than the next oldest brother and their wives would have one ninth of the remaining after their husbands have got their share.
After the cattle were divided, the farmer proceeded to divide the gold bars he had among the children. Each gold bar was valued at 3.5 times each cow's value. The gold bars were divided such that each couple had equally valued inheritance. What are the number of cows, gold bars and the number of sons the farmer had?
My deductions :
so pls help...3 to 4 hrs of working on this didnt help much :-( 164.100.170.4 ( talk) 16:14, 18 September 2009 (UTC)
what does that mean???
Of course, if there's only 1 son, it's trivial, and (2 sons, 16 cows, no gold) works, but isn't very interesting. There are other solutions, with 2 sons. Not sure about 3 or more. - GTBacchus( talk) 20:13, 18 September 2009 (UTC)
(after ec) The problem has several possible solutions. For example:
I am confident that there are other solutions with more sons, but I haven't worked them out yet. Note though, that the number of gold bars will always be non-unique, because for any solution with n sons, one can always add np to the number of gold bars while meeting all the conditions of the problem. Abecedare ( talk) 20:15, 18 September 2009 (UTC)
There are no solutions with 3 sons. The number of cows the first son takes would have to be congruent to both 6 and 0, modulo 64, which is impossible. - GTBacchus( talk) 20:24, 18 September 2009 (UTC)
Hmmm, I found a few solutions with three sons!
I am guessing there are solutions with even more number of progeny, but I'm donw with this family :-) Abecedare ( talk) 20:40, 18 September 2009 (UTC)