Diophantine Solutions

We first see that $b$ must be even, so we set $b=2c$ . Dividing by $2$ we obtain $a+3c=50$ , that is $a=50-3c$ Since $a\ge0$ and $c\ge0$ , $c$ can assume all the values between $0$ and $16$ , i.e. $17$ different values

When b is odd, there are no solutions as this would require 2a to be odd. For each even b (i.e. $b = 2c$ ), there exists a unique integer solution for a, that is $a = 50 - 3c$
Therefore the total number of solutions is equal to the number of multiples of 3 below 50. As $16\cdot3 = 48$ there are 17 possible non-negative values of c and therefore 17 unique pairs.

$100 \equiv 1 \pmod{3}$ , $3b \equiv 0 \pmod{3}$ . Hence $2a \equiv 1 \equiv -2 \pmod{3}$ , $a \equiv -2/2 \equiv -1 \equiv2 \pmod{3}$

Smallest possible $a = 2$ since $a \geq 0$ . Possible values of $a$ are in the form of an arithmetic progression $(3a +2)$ where $a \geq 0$ . Greatest value of $a$ is $50$ since $2a = 100$

No. of values of $a = (50-2)/3 + 1) = 17$ .Since the equation is linear, it is a one-to-one function. Hence the number of ordered pairs of non-negative integers $(a,b)= 17$

Rearranging, a = (100-3b)/2 Since we want a to be a non-negative integer, 100-3b must be divisible by 2 And we get b to be even numbers from 0 to 32 which will give 17 pairs.

2 a + 3 * b = 100; a = (100-3 b)/2; a and b non negative integers a, b >= 0;

3*b has to be even so that a is not a fraction;

b is even;

maximum 3b = 100; b <= 100/3 = 33.3

values for b even numbers between 0 to 33;

1+ 32/2 = 17;

0,2,4,6, 8, 10,12,14,16,18 20,22,24,26,28

30,32

2 and 3 are coprime numbers and therefore the equation has at least 1 solution.We can easily verify that (50,0) is a solution to the Diophantine equation and build the full set of its solutions.

$a=50-3t$

$b=2t$ where t is an integer.

In order to count all pairs we just need to count how many t's produce nonnegative integer solutions for a and b. Setting

$a\geq{0}$
and $b\geq{0}$

we get that $t\in{[0,16]}$ and since t can assume only integer values, it is obvious that there are only 17 values of t that produce nonnegative integer solutions.The answer is 17 pairs.

In order to solve this problem, first a order pair that solves the above equation has to be found. Through basic trial and error it is easy to see that $a_0=50,b_0=0$ is a possible solution-pair. From there, it is possible to find all the other solution-pairs.

First it is important to note that $2(3)-3(2)=0$ . This means that if $a=a_0$ and $b=b_0$ are solutions to $2a+3b=100$ , then so are $a=a_0-3x$ and $b=b_0+2y$

Since $a_0=50$ and $b_0=0$ , all that remains is to find the maximum value of $x$ such that $50-3x$ is non-negative. This turns out to be $\lfloor \frac {50}{3}=16 \rfloor$ . Along with the original solution set, this makes a total of $16+1=17$ ordered-pairs.

I solved this by choosing integers a for which there could exist an integer b to satisfy the conditions. For this to happen, 2a would have to be less than or equal to 100 (as 2a added to some non-negative integer equals 100), and 2a must be 1 mod 3 (as 2a added to a multiple of 3 equals 100, which is 1 mod 3). From these two requirements, we get that $0 \leq a \leq 50$ and $a \equiv 2 \pmod{3}$ . One out of every three consecutive integers between 0 and 50 inclusive will satisfy these conditions, and since there are 51 such integers, there are 51/3 = 17 solutions.

We know that $2a \equiv 0,2,4 \pmod{6}$ and $3b \equiv 0,3 \pmod{6}$ Since $100 \equiv 4 \pmod{6}$ , therefore $2a \equiv 4 \pmod{6}, 3b \equiv 0 \pmod{6}$ is the only solution. This implies $a \equiv 2 \pmod{3}, b \equiv 0 \pmod{2}$ . The minimum value of a is 2. When a is 2, b is 32. Every time we add 3 to a(still congruent to 2 modulo 3), there must be a integer solution for b(by subtracting 2 from b). So, there are 17 "a"s which is congruent to 2 modulo 3 and not more than $\frac{100}{2}=50$ . Done.

2a+3b=100 b=100-2a/3 so the values of a for which 50-a is divisible by 3 50-a is divisible by 3 16 times as 48 is divisble by three 16 times so 16 values of a there are 16 values of b satisfying the given eqn and also b can be zero so totally 17 values

As it is 2a+3b=100, and multiple of 2 is even, so the other term added must also be a positive even number, not exceeding 100. so, 3a= even not exceeding hundred. therefore, 3a = {3 2, 3 4, 3 6.........3 30, 3 32} 3 32 is 96 and 3*34 is 102 so the number 34 from the set of b is excluded. therefore b has 16 numbers in its set and for each number in b there is a ordered pair in a. So there are 16 ordered pairs.

$2a+3b$ must be even if it is to equal 100. $2a$ will always be even, so 3b must be even too. If 3b must be even, then b must be even. Therefore, the number of ordered pairs is the cardinality of the set of all multiples of six such that $0 \le 6x <100$ , which is $\boxed{17}$ .

manipulating factor is b because the coefficient of b is 3, which is an odd number, so we take it as the manipulating factor. then we take a number which is less than 100 that can be divided by 3 but the answer must be an even number so the number is 96. 96/3 is 32. and with the factor b=0 so we have 33 solutions.

a=2,b=32; a=5,b=30, a=8,b=28 which proceed for a up to 50 maintaining A.P and for b=32, 30, 28, ....up to 0 maintaining A.P with common difference 2 in descending order. As a,b are non-negative integer we take smallest value of b is zero(0) and that of a=2

Consider parity . Since $2a$ and 100 are both even, $3b$ has to be even too. Since $3$ is odd, this implies that $b$ has to be even (Worked example 2). Set $b=2b^*$ , where $b^*$ is a non-negative integer. Then $2a + 3 (2b^*) = 100 \Rightarrow a + 3b^* = 50$ . Since $a \geq 0$ , we get $a = 50 - 3b^* \geq 0 \Rightarrow b^* \leq \frac {50} {3}$ . Since $b^*$ is a non-negative integer, it can take on integer values from $0$ to $16$ , for a total of $16 - 0 + 1 = 17$ values.

For each integer value of $b^*$ from 0 to 16, we can set $(a,b) = ( 50 - 3 b^*, 2b^*)$ , which would give a pair of non-negative integers that satisfy $2a + 3b = 100$ . To see this, we substitute the pair into the original equation: $2\times(50 - 3b^*) + 3\times(2b^*) = 100 - 6b^* + 6b^* = 100$ . Hence, there are 17 possible solutions in all.

Rearranging the equation gives us 2a=100-3b. From this we know that 100-3b has to be even for it needs to be divisible by 2. It also follows that 3 times an even number will be even. List out all the even values of b (be sure to top after 32). There are 17 possible values of b that will satisfy the equation.

3b must be even since 2a=3b is even and 2a for all integers a.

looking at the sequence 3 \times 0, 3x2, ... , 3x30, 3x32 we can show there are accompanying even numbers that will sum to 100. We can then use y(n)= a+(n-1)*d where d=2 a=0 and y(n)=32 on the sequence 0,2,4,6...,30,32 to find the number of terms n.

n=17