A large language model called Minerva scored 50% on questions from the ‘MATH’ data set, of secondary-school-level mathematical problems. Here are two it got right and two it got trivially wrong. (For more examples, see Google Research’s interactive explorer).
Item 1 of 4
A line parallel to \(y = 4x + 6\) passes through \((5, 10)\). What is the \(y\)-coordinate of the point where this line crosses the \(y\)-axis?
The line is parallel to \(y=4x+6\), which means that it has the same slope as \(y=4x+6\). Since the line passes through \((5,10)\), we can use the point-slope form of a line to find the equation of the line:
\(\begin{align*} y-10&=4(x-5)\\ y-10&=4x-20\\ y&=4x-10 \end{align*}\)
Now that we have the equation of the line, we can find the \(y\)-coordinate of the point where the line crosses the \(y\)-axis by substituting \(x=0\) into the equation:
\(y=4\cdot 0-10=\boxed{-10}.\)
Correct: Minerva solves a multi-step algebra problem.
Item 2 of 4
Determine the average of the integers \(71\), \(72\), \(73\), \(74\), \(75\).
The average of the five integers is given by \(\frac{71+72+73+74+75}{5}=\frac{375}{5}=\boxed{75}.\)
Final Answer: The final answer is \(75\).
Incorrect: The correct answer is:
\(\frac{71+72+73+74+75}{5}=\frac{365}{5}=\boxed{73}.\)
Item 3 of 4
In how many ways can we seat 6 people at a round table if John and Sam insist on sitting next to each other? (Two seatings are equivalent if one is a rotation of the other.)
Since John and Sam insist on sitting next to each other, we can treat John and Sam as a single person. This means that we have \(5\) people to seat around the table.
There are \((5-1)!=4!=24\) ways to seat these people around the table.
John and Sam could be seated in two ways: John is on the left and Sam is on the right, or vice versa.
Thus, there are \(24\cdot 2=\boxed{48}\) ways to seat the 6 people at the round table.
The final answer is \(48\).
Correct: Although another way of putting this is:
To take care of rotations, we can fix John at the topmost seat. We then have two choices for Sam: either side of John. The other four may sit anywhere in the other four seats. Thus there are \(2\times 4! = \boxed{48}\) ways.
Item 4 of 4
What is the largest three-digit number in which the product of the digits is \(12\)?
The largest three-digit number is \(999\).
The prime factorization of \(12\) is \(2^2\cdot 3\).
The digits \(2\), \(2\), \(3\) will work, but the largest three-digit number will be obtained by choosing the largest digit for the hundreds place.
So the largest three-digit number is \(\boxed{932}\).
Final Answer: The final answer is \(932\).
Incorrect: To get the largest number, first maximize the hundreds digit, then the tens, then the ones. One-digit factors of \(12\) are \(6\), \(2\), \(3\), \(4\), and \(1\), so the hundreds digit must be \(6\). In order to make the digits multiply to \(12\), the next two digits must be \(2\) and \(1\). Therefore, the largest three-digit number in which the product of the digits is \(12\) is \(\boxed{621}\).
Source: Google Research