Let m and n be natural numbers. What is the minimum value of (m + n) such that 33m + 22n is divisible by 121 ?
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Finding the Minimum Value of m+n for Divisibility
The problem asks for the minimum value of the sum of two natural numbers, m and n, such that the expression \(33m + 22n\) is perfectly divisible by 121.
Natural numbers are positive integers, meaning \(m \ge 1\) and \(n \ge 1\).
Setting up the Divisibility Condition
When we say that \(33m + 22n\) is divisible by 121, it means that \(33m + 22n\) can be written as \(121\) multiplied by some integer, let's call it \(k\). Since m and n are positive, \(33m + 22n\) will be positive, so \(k\) must be a positive integer.
We can write this as an equation:
\(33m + 22n = 121k\)
Simplifying the Equation
Let's look at the coefficients: 33, 22, and 121. We can see that 11 is a common factor for all these numbers.
- \(33 = 11 \times 3\)
- \(22 = 11 \times 2\)
- \(121 = 11 \times 11\)
We can factor out 11 from the left side of the equation:
\(11(3m + 2n) = 121k\)
Now, we can divide both sides of the equation by 11:
\(3m + 2n = \frac{121k}{11}\)
\(3m + 2n = 11k\)
This simplified equation tells us that the expression \(3m + 2n\) must be a multiple of 11 for some positive integer \(k\).
Finding the Smallest Possible Value for 3m + 2n
Since m and n are natural numbers, the smallest possible value for m is 1 and the smallest possible value for n is 1. Let's find the minimum value of \(3m + 2n\) with \(m \ge 1\) and \(n \ge 1\):
Minimum \(3m + 2n = 3(1) + 2(1) = 3 + 2 = 5\).
So, \(3m + 2n\) must be a multiple of 11 and must be greater than or equal to 5. The smallest multiple of 11 that is greater than or equal to 5 is 11 itself (when \(k=1\)).
Therefore, we need to find natural numbers m and n such that \(3m + 2n = 11\).
Solving 3m + 2n = 11 for Natural Numbers
We are looking for integer solutions \((m, n)\) for the equation \(3m + 2n = 11\), with the conditions that \(m \ge 1\) and \(n \ge 1\).
Let's test possible values for m (starting from 1) and see if we get a natural number for n:
- If \(m = 1\): \(3(1) + 2n = 11\) \(3 + 2n = 11\) \(2n = 11 - 3\) \(2n = 8\) \(n = \frac{8}{2}\) \(n = 4\) Since \(n=4\) is a natural number, \((m, n) = (1, 4)\) is a valid solution.
- If \(m = 2\): \(3(2) + 2n = 11\) \(6 + 2n = 11\) \(2n = 11 - 6\) \(2n = 5\) \(n = \frac{5}{2}\) \(n = 2.5\) Since \(n=2.5\) is not a natural number, \((m, n) = (2, 2.5)\) is not a valid solution in natural numbers.
- If \(m = 3\): \(3(3) + 2n = 11\) \(9 + 2n = 11\) \(2n = 11 - 9\) \(2n = 2\) \(n = \frac{2}{2}\) \(n = 1\) Since \(n=1\) is a natural number, \((m, n) = (3, 1)\) is a valid solution.
If we try \(m = 4\), \(3(4) = 12\), which is already greater than 11. For \(n \ge 1\), \(3m + 2n\) would be even larger. So, there are no more solutions for \(m \ge 4\).
Calculating m+n for Valid Solutions
We found two pairs of natural numbers \((m, n)\) that satisfy \(3m + 2n = 11\):
- For \((m, n) = (1, 4)\): \(m + n = 1 + 4 = 5\).
- For \((m, n) = (3, 1)\): \(m + n = 3 + 1 = 4\).
Considering Larger Multiples of 11
We found the minimum \(m+n\) for the smallest possible multiple of 11 for \(3m+2n\), which was 11. What if we considered the next multiple, 22 (when \(k=2\))? The equation would be \(3m + 2n = 22\). Let's find some solutions:
- If \(m = 2\), \(6 + 2n = 22\), \(2n = 16\), \(n = 8\). \((2, 8)\) is a solution, \(m+n = 10\).
- If \(m = 4\), \(12 + 2n = 22\), \(2n = 10\), \(n = 5\). \((4, 5)\) is a solution, \(m+n = 9\).
- If \(m = 6\), \(18 + 2n = 22\), \(2n = 4\), \(n = 2\). \((6, 2)\) is a solution, \(m+n = 8\).
For \(3m+2n=22\), the minimum value of \(m+n\) we found is 8. This is larger than the minimum value 4 found when \(3m+2n=11\).
In general, as \(11k\) increases, the required values for \(m\) and \(n\) in \(3m + 2n = 11k\) will tend to increase, leading to larger values for \(m+n\). Therefore, the minimum value of \(m+n\) will occur for the smallest possible valid multiple of 11 for \(3m+2n\), which is 11.
Conclusion
Comparing the sums \(m+n\) for the valid natural number solutions when \(3m + 2n = 11\), we found sums of 5 and 4. The minimum of these values is 4.
Thus, the minimum value of \((m + n)\) such that \(33m + 22n\) is divisible by 121 is 4, which occurs when \(m=3\) and \(n=1\).
| Equation | Valid \((m, n)\) Natural Number Pairs | Sum \((m+n)\) |
|---|---|---|
| \(3m + 2n = 11\) | \((1, 4)\) | 5 |
| \(3m + 2n = 11\) | \((3, 1)\) | 4 |
| \(3m + 2n = 22\) | \((2, 8)\), \((4, 5)\), \((6, 2)\) ... | Minimum is 8 |
Revision Table: Understanding Divisibility and Natural Numbers
| Concept | Explanation | Example in Problem |
|---|---|---|
| Divisibility | An integer 'a' is divisible by an integer 'b' if 'a' can be written as 'b' multiplied by another integer 'k' (\(a = bk\)). | \(33m + 22n\) is divisible by 121 means \(33m + 22n = 121k\). |
| Natural Numbers | Positive integers: 1, 2, 3, ... | m and n must be from the set \{1, 2, 3, ...\}. This is important for finding valid solutions. |
| Linear Diophantine Equation | An equation of the form \(ax + by = c\) where we seek integer solutions for x and y. Our simplified equation \(3m + 2n = 11\) is a type of Diophantine equation, and we seek positive integer solutions. | We solved \(3m + 2n = 11\) for \(m \ge 1, n \ge 1\). |
Additional Information: Co-prime Coefficients
The equation \(3m + 2n = 11k\) involves coefficients 3 and 2, which are co-prime (their greatest common divisor is 1). For any integer value of \(11k\), the equation \(3m + 2n = 11k\) is guaranteed to have integer solutions. However, we specifically need natural number (positive integer) solutions.
The general integer solution for \(3m + 2n = C\) (where \(C\) is a multiple of gcd(3,2)=1) can be found, but finding the natural number solutions requires checking or using inequalities based on the specific value of C (like 11, 22, 33, etc.). As demonstrated, for \(C=11\), there are two natural number solutions; for \(C=22\), there are several; and so on.
The process involved finding the smallest multiple of 11 for which natural number solutions \((m, n)\) exist for \(3m + 2n = 11k\), and then finding the minimum sum \(m+n\) among those solutions.
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