Difference between revisions of "2024 AMC 10A Problems/Problem 7"
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We have <math>abc = 60</math>. Let <math>a</math> be positive, and let <math>b</math> and <math>c</math> be negative. Then we need <math>a > |b + c|</math>. If <math>a = 6</math>, then <math>|b + c|</math> is at least <math>7</math>, so this doesn't work. If <math>a = 10</math>, then <math>(b,c) = (-6,-1)</math> works, giving <math>10 - 7 = \boxed{\textbf{(B) }3}</math> | We have <math>abc = 60</math>. Let <math>a</math> be positive, and let <math>b</math> and <math>c</math> be negative. Then we need <math>a > |b + c|</math>. If <math>a = 6</math>, then <math>|b + c|</math> is at least <math>7</math>, so this doesn't work. If <math>a = 10</math>, then <math>(b,c) = (-6,-1)</math> works, giving <math>10 - 7 = \boxed{\textbf{(B) }3}</math> | ||
| − | + | ~[[User:pog|pog]],~[[User:Mathkiddus|mathkiddus]] | |
== Solution 3 == | == Solution 3 == | ||
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~Thesmartgreekmathdude | ~Thesmartgreekmathdude | ||
| + | |||
| + | ==Video Solution by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=_o5zagJVe1U | ||
==See also== | ==See also== | ||
Latest revision as of 21:44, 13 November 2024
- The following problem is from both the 2024 AMC 10A #7 and 2024 AMC 12A #6, so both problems redirect to this page.
Contents
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 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. 
