arrangement problem

@banashree-patra-m0OJwR • Oct 25, 2024

Oct 25, 2024

1.1K

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.

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Prashanth_p@cchi

@prashanth-p-at-cchi-rg0z63 • May 23, 2012

Question may be confusing for many! It seems that no one is interested to make an attempt.
Could you please provide us the answer?
Sagar07

@sagar07-LecI1i • Jun 12, 2012

Well, it is obvious.... But, i wonder, how can one prove it????????
Ankita Katdare

@abrakadabra • Jun 12, 2012

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.
Prashanth_p@cchi

@prashanth-p-at-cchi-rg0z63 • Jun 12, 2012

AbraKaDabra
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.
how about the one's marked above?
Ankita Katdare

@abrakadabra • Jun 12, 2012

Prashanth_p@cchi
how about the one's marked above?
Oh yes. In a hurry I only checked these
-
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

@prashanth-p-at-cchi-rg0z63 • Jun 12, 2012

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 1-10 adds up to 55.😔
Saandeep Sreerambatla

@saandeep-sreerambatla-hWHU1M • Jun 12, 2012

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 😀 )
Prashanth_p@cchi

@prashanth-p-at-cchi-rg0z63 • Jun 12, 2012

English-Scared
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 😀 )
I agree...... I understood it wrong!