2024 AMC 10A Problems/Problem 23

The following problem is from both the 2024 AMC 10A #23 and 2024 AMC 12A #17, so both problems redirect to this page.

Problem

Integers , , and satisfy , , and . What is ?

Solution 1

Subtracting the first two equations yields . Notice that both factors are integers, so could equal one of and . We consider each case separately:

For , from the second equation, we see that . Then , which is not possible as is an integer, so this case is invalid.

For , we have and , which by experimentation on the factors of has no solution, so this is also invalid.

For , we have and , which by experimentation on the factors of has no solution, so this is also invalid.

Thus, we must have , so and . Thus , so . We can simply trial and error this to find that so then . The answer is then .

~eevee9406

minor edits by Lord_Erty09

Solution 2

Adding up first two equations:

Subtracting equation 1 from equation 2:

Which implies that from

Giving us that

Therefore,

~lptoggled

Solution 3 (Guess and check)

The idea is that you could guess values for , since then and are factors of . The important thing to realize is that , , and are all negative. Then, this can be solved in a few minutes, giving the solution , which gives the answer ~andliu766

Solution 4

Note that , and the only possible pair of results that yields this is and , so .

Therefore,

~luckuso, yuvag, Technodoggo (LaTeX credits to the latter two and editing to the latter)

Solution 5

\begin{align*} (1) - (2) \implies ab + c -bc - a &=(a-c)(b-1)=13 \\ (2) - (3) \implies bc + a -ca - b &=(b-a)(c-1)=27 \\ (3) - (1) \implies ca + b -ab - c &=(c-b)(a-1)=-40 \end{align*}

There are ordered pairs of : , , .

However, only the last ordered pair meets all three equations.

Therefore,

~luckuso, megaboy6679 (formatting), Technodoggo (LaTeX optimization/clarity adjustments)

Solution 6 (Elimination)

Before we start, keep in mind that the problem is asking for the sum \(ab+bc+ac\). This is nothing but \(100+87+60-a-b-c\), or \(247-(a+b+c\)).

To solve the problem, we systematically test the options using elimination:

Let's first check options A and B, since they only happen when a,b, and c sum to 35 or 0. We begin by testing three positive values, but none satisfy the equation when there is a plus sign. For example, \( (12, 8, 4) \) satisfies \( ab + c = 100 \), but does not satisfy \( bc + a = 87\), or \( ac + b = 60\). If \(a+b+c=0\), then not all of the numbers can be positive or negative, so this would not work. From this observation, we conclude that the answer cannot be \( \textbf{A} \) or \( \textbf{B} \).

Now let's test the next option, option C. Option \( \textbf{C} \) states \( ab + bc + ca = 258 \). If true, then:

\(a + b + c = -11\)

This sum is too large. Furthermore, if all three numbers are negative, the solution still fails. For example, testing \( (-4, -5, -2) \) confirms the equation is not satisfied, as we get results that are too small. Thus, we eliminate option \( \textbf{C} \).

Finally, let's test the last two options: D and E. For option \( \textbf{E} \), the sum \( a + b + c \) would be:

\(247 - 284 = -37\)

Testing values such as \( (-11, -12, -14) \), the resulting sums \( ab + c \), \( bc + a \), and \( ac + b \) are far too large to satisfy the equation. Therefore, \( \textbf{E} \) is also eliminated.

Once we have this answer, we still need to verify it by testing out numbers: Finally, we test option \( \textbf{D} \). Using \( ab + bc + ca = 276 \), we get that \(a+b+c = -29\). Also note that a, b, and c all have to be different, because the sums from the three equations are all different. We want to get the three closest values of a, b, and c such that they are all different, and the sum \(a+b+c = -29\). The values \( (-9, -12, -8) \) are the closest three numbers. When we try them, they satisfy all three equations. So, the correct answer is:

~pimathmonkey

Video Solution by Power Solve

https://www.youtube.com/watch?v=LNYzBhf3Ke0

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.