2024 AMC 10A Problems/Problem 7
- The following problem is from both the 2024 AMC 10A #7 and 2024 AMC 12A #6, so both problems redirect to this page.
Contents
[hide]Problem
The product of three integers is . What is the least possible positive sum of the
three integers?
Solution 1
We notice that the optimal solution involves two negative numbers and a positive number. Thus we may split into three factors and choose negativity. We notice that
, and trying other combinations does not yield lesser results so the answer is
.
~eevee9406
Solution 2
We have . Let
be positive, and let
and
be negative. Then we need
. If
, then
is at least
, so this doesn't work. If
, then
works, giving
Solution 3
We can see that the most optimal solution would be positive integer and
negative ones (as seen in solution 1). Let the three integers be
,
, and
, and let
be positive and
and
be negative. If we want the optimal solution, we want the negative numbers to be as large as possible. so the answer should be
, where
... right?
No! Our sum must be a positive number, so that would be invalid! We see that -, and
are too far negative to allow the sum to be positive. For example,
, so
. For
to be the most positive, we will have
and
. Yet,
is still less than
. After
, the next factor of
would be
. if
=
,
. This might be positive! Now, if we have
, and
. It cannot be smaller because
and
would result in
being negative. Therefore, our answer would be
.
~Moonwatcher22
Video Solution by Math from my desk
https://www.youtube.com/watch?v=ptw2hYKoAIs&t=2s
Video Solution (🚀 2 min solve 🚀)
~Education, the Study of Everything
Video Solution by Pi Academy
https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
Video Solution 1 by Power Solve
https://youtu.be/j-37jvqzhrg?si=aggRgbnyZ3QjYwZF&t=806
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=_o5zagJVe1U
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 |
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.