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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.
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 • Jun 12, 2012
Well, it is obvious.... But, i wonder, how can one prove it????????
Ankita Katdare
Ankita Katdare • Jun 13, 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.
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
Ankita Katdare • Jun 13, 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.
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.😔
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 😀 )
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!

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