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If y is an integer, then the least possible value of \(|23 - 5\mathrm{y}|\) is
Let's think about what this problem is really asking. We have the expression \(|23 - 5\mathrm{y}|\) and we want to make this as small as possible.
The absolute value \(|23 - 5\mathrm{y}|\) represents the distance between 23 and \(5\mathrm{y}\) on a number line. To minimize this distance, we want \(5\mathrm{y}\) to be as close as possible to 23.
Since y must be an integer, we're looking for the integer value of y that makes \(5\mathrm{y}\) land closest to the target value of 23.
Process Skill: TRANSLATE - Converting the absolute value minimization into the intuitive concept of finding the closest distance
If we could choose any value for y (not just integers), what would make \(|23 - 5\mathrm{y}|\) equal to zero?
We'd need: \(23 - 5\mathrm{y} = 0\)
Solving: \(23 = 5\mathrm{y}\)
So: \(\mathrm{y} = 23/5 = 4.6\)
This means if y could be 4.6, then \(5\mathrm{y}\) would equal exactly 23, and our absolute value expression would be zero. But since y must be an integer, we can't use 4.6.
Since our ideal value is \(\mathrm{y} = 4.6\), the two closest integers are:
These are the only two candidates we need to check, since any other integer would be further from 4.6.
Process Skill: CONSIDER ALL CASES - Systematically identifying the relevant integer options
Now let's calculate \(|23 - 5\mathrm{y}|\) for both integer options:
For \(\mathrm{y} = 4\):
\(|23 - 5(4)| = |23 - 20| = |3| = 3\)
For \(\mathrm{y} = 5\):
\(|23 - 5(5)| = |23 - 25| = |-2| = 2\)
Comparing our results:
Since 2 < 3, the minimum value occurs when \(\mathrm{y} = 5\).
The least possible value of \(|23 - 5\mathrm{y}|\) is 2.
This matches answer choice (B) 2.
Students may think they need to find the value of y that makes the expression inside the absolute value bars equal to zero, without recognizing that y must be an integer. They might solve \(23 - 5\mathrm{y} = 0\) to get \(\mathrm{y} = 4.6\) and incorrectly assume this is the final answer, missing that y must be a whole number.
Students may approach this algebraically by trying to remove the absolute value bars through casework (when \(23 - 5\mathrm{y} \geq 0\) vs when \(23 - 5\mathrm{y} < 0\)) instead of recognizing that \(|23 - 5\mathrm{y}|\) represents the distance between 23 and \(5\mathrm{y}\) on a number line. This geometric insight makes finding the minimum much more intuitive.
Some students may overlook that y must be an integer and attempt to minimize the expression using calculus or treating y as a continuous variable, leading them to incorrectly conclude that the minimum value is 0.
Students might only test one integer value (like \(\mathrm{y} = 5\) since it's closest to 4.6) without checking the adjacent integer (\(\mathrm{y} = 4\)) to confirm which gives the smaller result. They need to test both \(\mathrm{y} = 4\) and \(\mathrm{y} = 5\) to compare.
When calculating \(|23 - 5\mathrm{y}|\) for the candidate values, students may make simple computational mistakes such as:
After determining that \(\mathrm{y} = 5\) gives the minimum, students might mistakenly select 5 as their final answer instead of the actual minimum value of the expression \(|23 - 5\mathrm{y}| = 2\). They confuse the input value that produces the minimum with the minimum value itself.