A very interesting math puzzle indeed?

Question

Let x[1], x[2], x[3], ..., x[100], denote 100 positive numbers whose sum is 100.

a) Explain why there must be at least one suffix 'a', such that x[a] is less than or equal to 1.

b) Explain why there must be at least one suffix 'a', such that x[a] + x[a+1] is less than or equal to 2.

c) Find 100 such positive numbers, all different, such that x[a] + x[a+1] is less than or equal to 2 for all a.

PLEASE NOTE: I have used the square brackets [ ] to show what is underscript, i.e. x[a] would normally look like x with an underscript a next to it. This is simply due to the limitations of posting a question here.

The correct answer with the clearest explanation gets 10 points! Thank you for your help, and have fun!

Answer 1

a. assume that all numbers are more than 1
x[1]>1
x[2]>1
...
x[100]>1
-----------
x[1]+...+x[100]>100 (it can not)
so at least 1 number must be less than or equal to 1
b. assume that x[a]+x[a+1]>2 with all a
similarly,
x[1]+x[2]>2
x[3]+x[4]>2
...
x[99]+x[100]>2
-------------------
x[1]+...+x[100] > 2*50= 100 (it can not)
So it must be ....
c. (0.01+ 1.99)+ (0.02+ 1.98)+ (0.03+ 1.97)+ ... + (0.5+ 1.5)
= 2+ 2+ 2+ ...+ 2 (50 times)= 100
Good luck!

Answer 2

Ok - A) because there are 100 positive numbers, all different - which means that automatically, you can NOT choose 100, right? (If you don't know that this is right, then boy do we have our work cut out for us! ;) ) So, let's think about this logically. You have to have 100 numbers, all that add up to 100. They have to (obviously) be between 0 and 100. How many numbers are there between 0 and 100? 101. We can't use 100 - so that leaves us with 100 numbers (0-99). The third condition C says that any TWO numbers must be less than or equal to 2.

Sooo...we have to use all decimal numbers. Your answer toA & B)
At least 1 number that is less than 1 must be used because the sum of a + (a+1) must be less than or equal to 2. Because all numbers are positive, all integers must not exceed the value of two in order to make the sum of two numbers 2.

For C - this is how you find all 100 numbers. You are going to need to open up Microsoft Excel and do some playing - type .1, .2, .3, .4 in different cells, and keep going. Then sum up all of the numbers until you get to 2. Find out what all of that equals when added together (DO NOT FORGET THAT ZERO IS ONE OF YOUR NUMBERS!!) Then, add the numbers of 1.1, 1.2, 1.3, etc to your .1, .2, .3 total. Keep working it that way until the sum equals 100.

Make sense?

Answer 3

a.) suppose all 100 numbers are greater than 1. then the average of all numbers will be greater than 1. and 100 * the average will be greater than 100. but the numbers need to sum to 100, so all numbers cannot be greater than 1 and therefore at least one must be less than or equal to 1.

b) same as a. consider 50 pairs of numbers all greater than 2. average > 2. 50 * average > 2 is greater than 100, etc, etc.

the answer to c is wikked easy. consider 100 positive numbers sum to 100. pair up the numbers so you have 50 pairs of numbers, if each pair = 2. then 50 * 2 = 100. so, how do you get 50 pairs who's sum equals two?

take .9 and 1.1 for the first pair.

then subtract .01 from .9 and add .01 to 1.1 for the next pair
.89 and 1.11, .88 and 1.12, .87 and 1.13

x[a] and x[a+1] always equal 2, and 50 pairs of 2 = 100

Answer 4

Let k be any positive number
Let all x[a] = 1 + k
Therefore these will add up to 100 + 100k. As k > 0, 100k > 0, so 100 + 100k must be greater than 100.

Answer 5

a) If all the numbers were greater than 1, then the sum would be greater than 100. This contradicts the given information.

b) If all these pairs-wise sums were greater than 2, then the sum would be greater than 100.

c) 1-ep, 1+ep, 1-2ep, 1+2ep, ... 1-50ep, 1+50ep.
... where ep is any positive non-zero number. The pair-wise sums are 2, 1-ep, 2, 1-ep, ...

Answer 6

I'll just do c..

Yake any strictly decreasing sequence starting below 1 and staying above 0. Call it b_n and define a_{2n-1}=b_n and a_{2n} =2-b_n.
You can check it works and that any solution has this form.

Answer 7

Ewwww - I was right with you till the ten point dangle. I hate that - we're not donkeys you know that need carrots dangled.