If m and n are integers such that (m+2n)(2m+n)=27,thenthemaximumpossiblevalueof2m-3n$ is
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Official Correct Answer: 17. First, 27 is a positive number, and since m and n are integers, both (m + 2n) and (2m + n) must be integer factors of 27. So, I should list all the pairs of integers whose product is 27. Let me think, 27 can be factored as:
1 and 27,
3 and 9,
-1 and -27,
-3 and -9.
Wait, but since multiplication is commutative, I should also consider the reverse pairs, like 27 and 1, 9 and 3, etc. So, all possible pairs are:
(1, 27), (27, 1), (3, 9), (9, 3), (-1, -27), (-27, -1), (-3, -9), (-9, -3).
Alright, so for each of these pairs, I can set up equations where (m + 2n) equals the first element and (2m + n) equals the second element. Then, I can solve for m and n in each case and compute 2m - 3n to find the maximum value.
Let me start with the first pair: (1, 27).
So, m + 2n = 1,
and 2m + n = 27.
I can solve this system of equations. Let's use substitution or elimination. Maybe elimination is easier here. If I multiply the first equation by 2, I get:
2m + 4n = 2.
Now, subtract the second equation from this:
(2m + 4n) - (2m + n) = 2 - 27,
Which simplifies to 3n = -25,
So, n = -25/3.
Wait, but n has to be an integer. Hmm, so this pair doesn't give integer solutions. So, I can discard this pair.
Next pair: (27, 1).
So, m + 2n = 27,
and 2m + n = 1.
Again, let's use elimination. Multiply the first equation by 2:
2m + 4n = 54.
Subtract the second equation:
(2m + 4n) - (2m + n) = 54 - 1,
Which gives 3n = 53,
So, n = 53/3. Again, not an integer. So, discard this pair.
Moving on to the next pair: (3, 9).
So, m + 2n = 3,
and 2m + n = 9.
Multiply the first equation by 2:
2m + 4n = 6.
Subtract the second equation:
(2m + 4n) - (2m + n) = 6 - 9,
Which simplifies to 3n = -3,
So, n = -1.
Now, plug n = -1 into the first equation:
m + 2*(-1) = 3,
So, m - 2 = 3,
Thus, m = 5.
So, m = 5 and n = -1. Now, Computing 2m - 3n:
2*5 - 3*(-1) = 10 + 3 = 13.
Okay, that's one value, 13. Let's keep that in mind.
Next pair: (9, 3).
So, m + 2n = 9,
and 2m + n = 3.
Multiply the first equation by 2:
2m + 4n = 18.
Subtract the second equation:
(2m + 4n) - (2m + n) = 18 - 3,
Which gives 3n = 15,
So, n = 5.
Now, plug n = 5 into the first equation:
m + 2*5 = 9,
So, m + 10 = 9,
Thus, m = -1.
Compute 2m - 3n:
2*(-1) - 3*5 = -2 - 15 = -17.
Hmm, that's a negative number, so not the maximum we're looking for.
Next pair: (-1, -27).
So, m + 2n = -1,
and 2m + n = -27.
Multiply the first equation by 2:
2m + 4n = -2.
Subtract the second equation:
(2m + 4n) - (2m + n) = -2 - (-27),
Which simplifies to 3n = 25,
So, n = 25/3. Not an integer. Discard this pair.
Next pair: (-27, -1).
So, m + 2n = -27,
and 2m + n = -1.
Multiply the first equation by 2:
2m + 4n = -54.
Subtract the second equation:
(2m + 4n) - (2m + n) = -54 - (-1),
Which gives 3n = -53,
So, n = -53/3. Not an integer. Discard.
Next pair: (-3, -9).
So, m + 2n = -3,
and 2m + n = -9.
Multiply the first equation by 2:
2m + 4n = -6.
Subtract the second equation:
(2m + 4n) - (2m + n) = -6 - (-9),
Which simplifies to 3n = 3,
So, n = 1.
Plug n = 1 into the first equation:
m + 2*1 = -3,
So, m + 2 = -3,
Thus, m = -5.
Compute 2m - 3n:
2*(-5) - 3*1 = -10 - 3 = -13.
Again, negative, so not the maximum.
Last pair: (-9, -3).
So, m + 2n = -9,
and 2m + n = -3.
Multiply the first equation by 2:
2m + 4n = -18.
Subtract the second equation:
(2m + 4n) - (2m + n) = -18 - (-3),
Which gives 3n = -15,
So, n = -5.
Plug n = -5 into the first equation:
m + 2*(-5) = -9,
So,