Understanding the Difference of Squares
The problem asks us to find the value of the product \(m \times n\). We are given the equation \(m^2 - n^2 = 199\), where \(m\) and \(n\) must be natural numbers (positive integers, \(m, n \in \{1, 2, 3, ...\}\)) and it's specified that \(m > n\).
Using the Difference of Squares Formula
The expression \(m^2 - n^2\) is a standard algebraic identity known as the difference of squares, which can be factored as:
\( m^2 - n^2 = (m - n)(m + n) \)
Using this factorization, our equation becomes:
\( 199 = (m - n)(m + n) \)
Identifying Factors of 199
Let's define two terms based on the factors: \(a = m - n\) and \(b = m + n\). The equation is now \(199 = a \times b\).
Based on the conditions given (\(m, n\) are natural numbers and \(m > n\)), we can deduce the properties of \(a\) and \(b\):
- Since \(m > n \ge 1\), \(a = m - n\) must be a positive integer (\(a \ge 1\)).
- Similarly, \(b = m + n\) must be a positive integer (\(b \ge 1+1=2\)).
- Since \(m+n\) is always greater than \(m-n\) for positive \(n\), we must have \(b > a\).
We need to find pairs of positive integer factors \((a, b)\) of 199 such that \(a \times b = 199\) and \(b > a\).
Solving for \(m\) and \(n\)
We have a system of two linear equations:
- \(m - n = a\)
- \(m + n = b\)
Adding these two equations gives \(2m = a + b\), which means \(m = \frac{a+b}{2}\).
Subtracting the first equation from the second gives \(2n = b - a\), which means \(n = \frac{b-a}{2}\).
For \(m\) and \(n\) to be integers, the sum \(a+b\) and the difference \(b-a\) must both be even. This occurs only when \(a\) and \(b\) have the same parity (i.e., both are even or both are odd).
Analyzing the Factors of 199
The number 199 is a prime number. This means its only positive integer factors are 1 and 199.
Considering the condition \(b > a\), the only possible pair of factors \((a, b)\) for 199 is \((1, 199)\).
Checking the Solution for \(m\) and \(n\)
Let's use the factor pair \((a, b) = (1, 199)\):
- Check parity: \(a=1\) is odd, and \(b=199\) is odd. They have the same parity, so \(m\) and \(n\) will be integers.
- Calculate \(m\): \(m = \frac{1+199}{2} = \frac{200}{2} = 100\).
- Calculate \(n\): \(n = \frac{199-1}{2} = \frac{198}{2} = 99\).
We verify if these values meet the problem's conditions: \(m=100\) and \(n=99\) are indeed natural numbers, and \(m > n\) (\(100 > 99\)).
Calculating the Product \(mn\)
Using the derived values \(m=100\) and \(n=99\), we calculate the product \(mn\):
\( mn = 100 \times 99 = 9900 \)
Considering Uniqueness
The mathematical steps show that for the number 199, there is only one pair of natural numbers \((m, n)\) such that \(m^2 - n^2 = 199\) and \(m > n\). This pair is \((100, 99)\), which leads to a single, unique value for the product \(mn\), namely 9900.
However, the question provides 'Cannot be uniquely determined' as an option. This typically happens when a number can be represented as a difference of squares in multiple ways. For example, composite odd numbers (like 15, 105, etc.) have multiple factor pairs with the same parity, leading to different values for \(m\), \(n\), and consequently \(mn\).
While the specific case of 199 yields a unique result, the structure of the question and the presence of the option 'Cannot be uniquely determined' might suggest considering the general principle or potential interpretations.
Based on the options provided, the conclusion is that the value cannot be uniquely determined.
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How many zeroes are there at the end of the following product?
1 x 5 x 10 x 15 x 20 x 25 x 30 x 35 x 40 x 45 x 50 x 55 x 60
Let XYZ be a three-digit number, where (x + y + Z) is not a multiple of 3. Then (XYZ + YZX + ZXY) is not divisible by
