If each term in the sum a_1+a_2+a_3+...+a_n is either 7 or 77 and the sum equals 350, which of the...

1. Translate the problem requirements: We need to find how many terms total (n) when each term is either 7 or 77, and all terms sum to 350. We need to determine which of the given options could work.
2. Set up the equation structure: If we have x terms equal to 7 and y terms equal to 77, then \(7\mathrm{x} + 77\mathrm{y} = 350\) and \(\mathrm{x} + \mathrm{y} = \mathrm{n}\). We can solve for one variable in terms of the other.
3. Simplify and find constraints: Divide the sum equation by 7 to get \(\mathrm{x} + 11\mathrm{y} = 50\), which means \(\mathrm{x} = 50 - 11\mathrm{y}\). Since x must be non-negative, y can be at most 4 (since \(11 \times 5 = 55 > 50\)).
4. Test feasible values and match with answer choices: For each possible value of y (0 through 4), calculate the corresponding x and total n = x + y, then check which matches our answer options.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have and what we need to find.

We have a sum where each term is either 7 or 77. Think of it like having a collection of coins where each coin is worth either 7 points or 77 points. The total value of all our coins must equal exactly 350 points.

We need to find how many coins (terms) we could have in total. This total number of coins is what we call 'n'.

So we're looking for: How many total terms could we have when some terms equal 7, some equal 77, and they all add up to 350?

Process Skill: TRANSLATE - Converting the word problem into mathematical understanding

2. Set up the equation structure

Let's use simple variables to represent our unknowns:

- Let x = number of terms that equal 7
- Let y = number of terms that equal 77

Now we can write two important relationships:

1. The sum condition: All the 7's plus all the 77's must equal 350
So: \(7\mathrm{x} + 77\mathrm{y} = 350\)

2. The total count: The number of terms that equal 7 plus the number that equal 77 gives us our total n
So: \(\mathrm{x} + \mathrm{y} = \mathrm{n}\)

This gives us our system: \(7\mathrm{x} + 77\mathrm{y} = 350\) and \(\mathrm{x} + \mathrm{y} = \mathrm{n}\)

3. Simplify and find constraints

Let's make the first equation easier to work with. Since both 7 and 77 are multiples of 7, we can divide the entire equation by 7:

\(7\mathrm{x} + 77\mathrm{y} = 350\)
Dividing by 7: \(\mathrm{x} + 11\mathrm{y} = 50\)

This is much simpler! Now we can solve for x:
\(\mathrm{x} = 50 - 11\mathrm{y}\)

But wait - x must be a non-negative number (we can't have a negative number of terms). So:

\(50 - 11\mathrm{y} \geq 0\)
\(50 \geq 11\mathrm{y}\)
\(\mathrm{y} \leq 50/11\)
\(\mathrm{y} \leq 4.54...\)

Since y must be a whole number, y can be at most 4.
Also, y must be non-negative, so y can be 0, 1, 2, 3, or 4.

Process Skill: APPLY CONSTRAINTS - Recognizing that variables must be non-negative integers

4. Test feasible values and match with answer choices

Let's check each possible value of y and see what total n we get:

When y = 0 (no 77's):
\(\mathrm{x} = 50 - 11(0) = 50\)
\(\mathrm{n} = \mathrm{x} + \mathrm{y} = 50 + 0 = 50\)

When y = 1 (one 77):
\(\mathrm{x} = 50 - 11(1) = 39\)
\(\mathrm{n} = \mathrm{x} + \mathrm{y} = 39 + 1 = 40\)

When y = 2 (two 77's):
\(\mathrm{x} = 50 - 11(2) = 28\)
\(\mathrm{n} = \mathrm{x} + \mathrm{y} = 28 + 2 = 30\)

When y = 3 (three 77's):
\(\mathrm{x} = 50 - 11(3) = 17\)
\(\mathrm{n} = \mathrm{x} + \mathrm{y} = 17 + 3 = 20\)

When y = 4 (four 77's):
\(\mathrm{x} = 50 - 11(4) = 6\)
\(\mathrm{n} = \mathrm{x} + \mathrm{y} = 6 + 4 = 10\)

So the possible values of n are: 10, 20, 30, 40, 50

Looking at our answer choices (38, 39, 40, 41, 42), only 40 appears in our list!

Process Skill: CONSIDER ALL CASES - Testing each valid possibility systematically

Final Answer

Let's verify: When n = 40, we have y = 1 and x = 39.
Check: 39 terms of 7 plus 1 term of 77 = \(39(7) + 1(77) = 273 + 77 = 350\) ✓

The answer is C. 40

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the problem setup
Students often confuse what 'n' represents. They might think n is the number of 7's or the number of 77's, rather than understanding that n is the total number of terms in the sum. This leads to setting up incorrect equations from the start.

2. Incorrect variable assignment
Some students struggle with creating two variables (x and y) and instead try to solve with just one variable. They might attempt to express everything in terms of just the number of 7's or just the number of 77's, missing the relationship between the two types of terms.

3. Missing the constraint recognition
Students may set up the equation \(7\mathrm{x} + 77\mathrm{y} = 350\) correctly but fail to recognize that both x and y must be non-negative integers. This constraint is crucial for limiting the possible solutions, and without recognizing it, students get lost in the solution process.

Errors while executing the approach

1. Arithmetic errors in simplification
When dividing \(7\mathrm{x} + 77\mathrm{y} = 350\) by 7 to get \(\mathrm{x} + 11\mathrm{y} = 50\), students commonly make calculation mistakes. They might incorrectly divide 77 by 7 (getting something other than 11) or divide 350 by 7 incorrectly (getting something other than 50).

2. Incorrect constraint application
When finding the maximum value of y from \(\mathrm{x} = 50 - 11\mathrm{y} \geq 0\), students often make errors. They might incorrectly solve \(50 - 11\mathrm{y} \geq 0\), getting the wrong upper bound for y, or forget that y must be a whole number, leading to testing non-integer values.

3. Incomplete case testing
Students may correctly identify that y can be 0, 1, 2, 3, or 4, but then fail to systematically test all cases. They might test only a few values of y and miss calculating n for all possible scenarios, potentially missing the correct answer.

Errors while selecting the answer

1. Confusing calculated values
After finding all possible values of n (10, 20, 30, 40, 50), students sometimes confuse these with the intermediate values they calculated (like x = 39 when y = 1). They might incorrectly select 39 instead of 40, mixing up the number of 7's with the total number of terms.

2. Verification errors
Even when students identify that n = 40 is the answer, they may make mistakes in the final verification step. When checking \(39(7) + 1(77) = 350\), they might miscalculate \(39 \times 7 = 273\) or the final sum, leading them to doubt their correct answer and potentially change it.