Banashree Patra
Branch Unspecified
13 Jun 2012
arrangement problem
The integers 1, 2,........., 10 are circularly arranged in an arbitrary order. Show that there are always three successive integers in this arrangement, whose sum is at least 17.
Prashanth_p@cchi
Branch Unspecified
6 years ago
Question may be confusing for many! It seems that no one is interested to make an attempt.
Could you please provide us the answer?
Could you please provide us the answer?
Ankita Katdare
Computer Science
6 years ago
That is actually a bit difficult to prove.
This a good sequence >
1, 10, 6, 2, 8, 7, 3, 5, 9, 4.
None of the 3 consecutive numbers have a summation of less than 17 and it is at most 18.
This a good sequence >
1, 10, 6, 2, 8, 7, 3, 5, 9, 4.
None of the 3 consecutive numbers have a summation of less than 17 and it is at most 18.
Prashanth_p@cchi
Branch Unspecified
6 years ago
how about the one's marked above?AbraKaDabraThat is actually a bit difficult to prove.
This a good sequence >
1, 10, 6, 2, 8, 7, 3, 5, 9, 4.
None of the 3 consecutive numbers have a summation of less than 17 and it is at most 18.
Ankita Katdare
Computer Science
6 years ago
Oh yes. In a hurry I only checked thesePrashanth_p@cchihow about the one's marked above?

1, 10, 6, 2, 8, 7, 3, 5, 9, 4.
1,10,6
10,6,2
2,8,7
8,7,3
9,5,3
5,9,4
Now, that I think about it, since it is a circular arrangement 9,4,1 is also not proper.
Prashanth_p@cchi
Branch Unspecified
6 years ago
I think its impossible! but I might be wrong.
Below is the least in terms of the entire sum, that can be used to satisfy the above criteria.
6,6,5,6,6,5,6,6,5,6 which adds up to  57
But 110 adds up to 55.😔
Below is the least in terms of the entire sum, that can be used to satisfy the above criteria.
6,6,5,6,6,5,6,6,5,6 which adds up to  57
But 110 adds up to 55.😔
Saandeep Sreerambatla
Branch Unspecified
6 years ago
Consider the numbers 1 to 10 are marked as 1 , x1, x2, ... , x9. Not necessarily in the sequence order.
So the number are arranged randomly like this: 1,x1,x2,x3,x4,x5,x6,x7,x8,x9.
We know that sum of numbers from 1 to 10 will add upto 55.
so we will prove it in a negative way, if any of the groups above like (x1+x2+x3) or (x4+x5+x6) or (x7+x8+x9) will add up to < = (less than or equal to) 17. Then the sum 1+x1+...x9 will be <=52 which is wrong.
So atleast one of the section should be greater than 17. (Hence proved 😀 )
So the number are arranged randomly like this: 1,x1,x2,x3,x4,x5,x6,x7,x8,x9.
We know that sum of numbers from 1 to 10 will add upto 55.
so we will prove it in a negative way, if any of the groups above like (x1+x2+x3) or (x4+x5+x6) or (x7+x8+x9) will add up to < = (less than or equal to) 17. Then the sum 1+x1+...x9 will be <=52 which is wrong.
So atleast one of the section should be greater than 17. (Hence proved 😀 )
Prashanth_p@cchi
Branch Unspecified
6 years ago
I agree...... I understood it wrong!EnglishScaredConsider the numbers 1 to 10 are marked as 1 , x1, x2, ... , x9. Not necessarily in the sequence order.
So the number are arranged randomly like this: 1,x1,x2,x3,x4,x5,x6,x7,x8,x9.
We know that sum of numbers from 1 to 10 will add upto 55.
so we will prove it in a negative way, if any of the groups above like (x1+x2+x3) or (x4+x5+x6) or (x7+x8+x9) will add up to < = (less than or equal to) 17. Then the sum 1+x1+...x9 will be <=52 which is wrong.
So atleast one of the section should be greater than 17. (Hence proved 😀 )
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