Difference between revisions of "2023 AMC 8 Problems/Problem 21"
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==Video Solution by Magic Square== | ==Video Solution by Magic Square== | ||
https://youtu.be/-N46BeEKaCQ?t=2853 | https://youtu.be/-N46BeEKaCQ?t=2853 | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/1bA7fD7Lg54?t=2062 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=20|num-a=22}} | {{AMC8 box|year=2023|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:47, 16 February 2023
Contents
Problem
Alina writes the numbers on separate cards, one number per card. She wishes to divide the cards into
groups of
cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?
Solution 1
First we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. . Then dividing by
we have
so each group of
must have a sum of 15. To make the counting easier we we will just see the possible groups 9 can be with. The posible groups 9 can be with with 2 distinct numbers are
and
. Going down each of these avenues we will repeat the same process for
using the remaining elements in the list. Where there is only 1 set of elements getting the sum of
,
needs in both cases. After
is decided the remaining 3 elements are forced in a group. Yielding us an answer of
as our sets are
and
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
Solution 2
The group with must have the two other numbers adding up to
, since the sum of all the numbers is
. The sum of the numbers in each group must therefore be
. We can have
,
,
, or
. With the first group, we have
left over. The only way to form a group of
numbers that add up to
is with
or
. One of the possible arrangements is therefore
. Then, with the second group, we have
left over. With these numbers, there is no way to form a group of
numbers adding to
. Similarly, with the third group there is
left over and we can make a group of
numbers adding to
with
or
. Another arrangement is
. Finally, the last group has
left over. There is no way to make a group of
numbers adding to
with this, so the arrangements are
and
. There are
sets that can be formed.
~Turtwig113
Video Solution 1 by OmegaLearn (Using Casework)
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=2853
Video Solution by Interstigation
https://youtu.be/1bA7fD7Lg54?t=2062
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.