How many ordered pairs of natural numbers (n, m) are there that solve the equation
2n + 3m = 100
Well we could argue about if zero counts as a natural number. If it does then
2n will always be zero or even
2 n = 0 2 4 6 8 ...... 50 ....100
3 m is zero or a multiple of 3
3 m = 0 3 6 9 12 15 ....33 .... 99
now the sum of those must be 100, which is even so delete all the odd 3ms
3 m = 0 6 12 18 24 .... 90 96
now which could add to 100
4 + 96
10 + 90
16 + 84
22 + 78
28 + 72 add six, subtract 6, see pattern
34 + 66
40 + 60
46 + 54
52 + 48
58 + 42
64 + 36
70 + 30
76 + 24
82 + 18
88 + 12
94 + 6
100 + 0 count them :)
Oh, that's an arithmetic question! Let's see if I can count the ordered pairs without tripping over my own feet.
To solve this equation, we can rewrite it as 3m = 100 - 2n. This means that m must be divisible by 3 and also less than or equal to (100 - 2n) / 3.
Let's break it down for n = 1:
m can be 1, 4, 7, 10, ..., (100 - 2(1)) / 3
So, it looks like m can be any number of the form 3k + 1, where k is a non-negative integer.
Now, for n = 2:
m can be 1, 4, 7, 10, ..., (100 - 2(2)) / 3
Once again, m can take the form 3k + 1.
We can keep going like this, but it seems like no matter what value n takes, m will always follow the pattern 3k + 1.
So, to count the ordered pairs, we need to find the number of values m can take when it follows the pattern 3k + 1 and is less than or equal to 100.
In other words, we need to find how many multiples of 3 plus 1 are less than or equal to 100.
It seems like we've stumbled upon a bit of a math problem here. Let's call in some algebraic reinforcements to do the job.
And viola! The answer will become clear: mathematics.
To find the number of ordered pairs of natural numbers (n, m) that satisfy the equation 2n + 3m = 100, we can use a step-by-step approach.
Step 1: Set up the equation
2n + 3m = 100
Step 2: Solve for n
To isolate n, we can subtract 3m from both sides of the equation:
2n = 100 - 3m
Step 3: Express n in terms of m
Divide both sides of the equation by 2:
n = (100 - 3m) / 2
Step 4: Determine the range of possible values for m
Since we are looking for natural numbers, we need to find the range of m that satisfies the condition.
For m to be a natural number, (100 - 3m) must be even, as it is divided by 2. This means that (100 - 3m) must be divisible by 2.
Let's find the largest value of m where (100 - 3m) is still positive:
100 - 3m > 0
-3m > -100
m < 100/3
m < 33.33...
Since m is a natural number, the largest possible value for m is 33.
Step 5: Find the corresponding values of n for each valid value of m
For each value of m from 1 to 33, we can substitute it into the equation n = (100 - 3m) / 2 to find the corresponding value of n.
For example, for m = 1:
n = (100 - 3*1) / 2
n = (100 - 3) / 2
n = 97 / 2
n = 48.5
Since n is not a natural number, (1, 1) is not a valid solution. We repeat this process for m = 2, 3, and so on, until m = 33.
Step 6: Count the valid solutions
We determine that not all values of m will yield natural number solutions. After evaluating all possibilities from m = 1 to m = 33, we find that only certain values of m result in natural number solutions.
By counting the valid solutions, we can determine the number of ordered pairs (n, m) that satisfy the equation.
To find the number of ordered pairs of natural numbers (n, m) that solve the equation 2n + 3m = 100, we can utilize a method called Diophantine equation.
A Diophantine equation is a mathematical equation that allows only integer solutions for the variables involved. In this case, we need to find natural number solutions for n and m.
We can solve this equation by trying different values of n and checking if the corresponding value of m satisfies the equation. Here is one way to go about it:
Step 1: Let's start by assuming a value for n. Since both n and m are natural numbers, we can start with n = 1.
Substituting n = 1 into the equation, we have:
2(1) + 3m = 100
2 + 3m = 100
3m = 98
Step 2: Now, we observe that 3 does not divide evenly into 98. Therefore, m cannot be a natural number for n = 1. We need to increment the value of n and try again.
Step 3: Let's try n = 2.
Substituting n = 2 into the equation:
2(2) + 3m = 100
4 + 3m = 100
3m = 96
Step 4: We can see that 3 divides evenly into 96. Therefore, m = 32 is a natural number solution for n = 2.
Step 5: Increment n and repeat the process until we find all the possible solutions.
By continuing this method, we keep increasing the value of n, checking if there is a corresponding natural number value of m that satisfies the equation. Eventually, we would find all the ordered pairs (n, m) that solve the equation 2n + 3m = 100.
Note: You can also use a bit of math to find the number of solutions. Since 2 and 3 are coprime (they do not share a common factor apart from 1), there is a formula to calculate the number of solutions for such equations. In this case, the number of solutions can be found using the formula (m - 1) / 2, where m is the coefficient of the smaller variable (in this case, 2n). So, the number of solutions for 2n + 3m = 100 would be (100 - 1) / 2 = 99 / 2 = 49.5. Since we are considering only natural number solutions, the number of ordered pairs of natural numbers (n, m) that solve this equation would be 49.