2023 AMC 8 Problems/Problem 21
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5(Trial and Error)
- 7 Video Solution by CoolMathProblems
- 8 Video Solution by Math-X
- 9 Video Solution (A Clever Explanation You’ll Get Instantly)
- 10 Video Solution
- 11 Video Solution (THINKING CREATIVELY!!!)
- 12 Video Solution 1 (Using Casework)
- 13 Animated Video Solution
- 14 Video Solution by Magic Square
- 15 Video Solution by Interstigation
- 16 Video Solution by WhyMath
- 17 Video Solution by harungurcan
- 18 Video Solution by MathyWorks
- 19 Video Solution by Dr. David
- 20 See Also
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 will just see the possible groups 9 can be with. The possible groups 9 can be with 2 distinct numbers are
and
. Going down both 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
.
~CHECKMATE2021, 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
. So,there are
sets that can be formed.
~Turtwig113
Solution 3
The sum of the numbers across all equally valued sets is . The value of the numbers in each set would be
. We know that the numbers
,
, and
must belong in different sets, as putting any
numbers in
set will either pass or match the limit of
per set, and we would then still need to add
more number after that. Note that these numbers must be distinct, as Alina only has
of each number, and order does not matter in the sets. Starting with the set that includes the number
, the next two numbers must add up to
, and there are
ways of doing this
. Note we cannot use any number past
, as those numbers must be used in the other sets. The next set, which includes the number
, must have two numbers that add up to
, and there are
ways to do this
. The final set, which includes the number
, must have
numbers that sum up to
, and there are
ways to do this
. Now we have found the number of ways in which each set sums up to
. To find the number of ways in which all three sets sum up to
concurrently, we must take the minimum of
,
, and
, which gives us an answer of
triplets of sets with 3 values, in which each set sum to the same amount.
~Fernat123
Solution 4
Note that each group of numbers should sum to Thus, this is equivalent to asking, “How many ways can you fill in a three by three magic square with the integers
through
?” since we can take the three rows of the magic square as our three groups. If you have closely studied magic squares, you might know that in a three by three magic square that is to be filled in with the integers
through
, the center of the square would be
(the average of the numbers), and the numbers in the corners should be even(*). The such pairs (disregarding order) are
and
Let’s fix the position of
to be the top left corner. This would make
in the bottom right corner. We can have either
or
to be in the top right corner, for a total of
such groups of three. (The groups are
and
)
Note that if we had instead fixed the position of
,
, or
, they would correspond to one of the two cases, just in a different configuration.
(*)We can prove this using proof by contradiction. Label the nine small squares within the magic square from to
from left to right, top to bottom. Firstly, we know that
and
sum to
since the center square is
. Thus,
and
must have the same parity, and so must
and
. Suppose that
and
have different parity. Since
,
must be even. By a similar argument,
must also be even, and so must
and
. Our initial assumption is that one of
and
is odd and the other is even; however, we end up with six even numbers needed to fill in the square, but there are only four even integers from
to
. Now suppose that all of
and
are odd. This would make each of
and
odd, but clearly we do not have enough odd numbers to make all nine numbers odd. Thus, each corner square must be even.
~ Brian__Liu
Solution 5(Trial and Error)
To start, we must find the value that all three groups must add to, which is . Then, we can notice that the numbers
and
must be in different groups because two of them and any other digit 1-9 added together cannot produce 15 (For example,
which means that we need a
to sum to 15. In the same way,
and
cannot be placed in the same group because putting them in the same group would make the numbers greater than 15 (For example,
, meaning that it would be impossible to add another number with making the sum greater than 15.
Now, we can just use trial and error to solve the problem (this is feasible because there are only
ways to pair a number
and
and then add another number
in order to sum to 15. By trial and error, we see that 1 and 7 must not be paired together because to have a sum of 15, there must be another 7 in the group (The problem only lets us use 1 of each number). This eliminates two of the options we are considering. Now, we only have to check the 4 options where 7 is not in a group with 1.
By trial and error, we can see that only two cases work:
1.
and
2.
Because we only have two options, the answer is
Video Solution by CoolMathProblems
https://youtu.be/_TMRSjRWPis?feature=shared
Video Solution by Math-X
https://youtu.be/Ku_c1YHnLt0?si=S79t9BmOmSb-ACds&t=4641
~Math-X
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=2418 ~hsnacademy
Video Solution
Please like and subscribe
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution 1 (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/DBqko2xATxs&t=2747
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=872s
~harungurcan
Video Solution by MathyWorks
https://www.youtube.com/watch?v=hB7CDrVnNCs
~SlimeKnight
Video Solution by Dr. David
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.