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The natural numbers from 1 to 9 are divided into 3 groups - not necessarily of equal size. Do you agree that in any pattern of grouping the product of the numbers in at least one group will exceed 71
Hi;
I would say that is true.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Logic?
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Hi;
Computer. Just providing verification. The OP did not mention he needed a proof.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Sorry i didnt get logic behind your GM logic
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Last edited by thickhead (2016-10-15 15:19:04)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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If the numbers are not limited to single digits , say from
1 to 16 , and divided into say 4 groups other than 3 .
( with at least 1 number in each group ) Then the related
value will be ( 16 ! ) ^ 1/4 .
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To solve the original problem is to minimize the greatest ( maximum )
value of the 3 groups . As ( 9 ! ) ^ 1/3 = 71.327 and the values of
the groups must be whole numbers . Thus the group with smallest
value will be ≥ 71 , which is a prime no. So its value will be
reduced to 70 = 2 * 5 * 7 .
The value of any one of the remaining groups cannot be 70 also , to minimize the value of the greater one of the remaining 2 groups we take the geometric mean again .
Since √1* 3* 4* 6* 8* 9 = 72 , therefore the 3 groups will be
70 , 72 and 72 .
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