Difference between revisions of "2005 AMC 10B Problems/Problem 25"
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~GentleTiger | ~GentleTiger | ||
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+ | ==Solution 4== | ||
+ | We can put all odd numbers into subset B, or we can put all even numbers into subset B, so now there are 50 numbers in the set. I will use all even numbers in this solution. Now, we need to add other odd(or even!) numbers possible in this subset, which is all odd(or even) numbers that can be added so that the sum with 100(or 99) plus the biggest possible odd number(or even) to get 123. This will get us the numbers 1,3,5...21,23(or numbers 2,4,6...22,24), which gives us 12 more numbers, and adding that to the 50 original numbers, we get <math>B =\boxed{\textbf{(C)}\ 62}</math>. | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2005|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:46, 14 August 2023
Contents
Problem
A subset of the set of integers from
to
, inclusive, has the property that no two elements of
sum to
. What is the maximum possible number of elements in
?
Solution 1
The question asks for the maximum possible number of elements. The integers from to
can be included because you cannot make
with integers from
to
without the other number being greater than
. The integers from
to
are left. They can be paired so the sum is
:
,
,
,
,
. That is
pairs, and at most one number from each pair can be included in the set. The total is
.
Also, it is possible to see that since the numbers to
are in the set there are only the numbers
to
to consider. As
gives
, the numbers
to
can be put in subset
without having two numbers add up to
. In this way, subset
will have the numbers
to
, and so the answer is
.
Solution 1 Alternate Solution
Since there are numbers that sum to
, there are
numbers not summing to
~mathboy282
Solution 2 (If you have no time)
"Cut" into half. The maximum integer value in the smaller half is
. Thus the answer is
.
Solution 3
The maximum possible number of elements includes the smallest numbers. So, subset where n is the maximum number of elements in subset
. So, we have to find two consecutive numbers,
and
, whose sum is
. Setting up our equation, we have
. When we solve for
, we get
.
~GentleTiger
Solution 4
We can put all odd numbers into subset B, or we can put all even numbers into subset B, so now there are 50 numbers in the set. I will use all even numbers in this solution. Now, we need to add other odd(or even!) numbers possible in this subset, which is all odd(or even) numbers that can be added so that the sum with 100(or 99) plus the biggest possible odd number(or even) to get 123. This will get us the numbers 1,3,5...21,23(or numbers 2,4,6...22,24), which gives us 12 more numbers, and adding that to the 50 original numbers, we get .
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.