Food for a squirrel

Let number of pines be P and number of acorns be A

Thus by calculating the total days we get: 3P + 2A = 150

Here as 150 is even number and 2 multiplied by something is also even thus;

P has TO BE an even number

In that case it can be easily seen that p can take values from 0 to 50 in steps of 2

That is {0,2,4,6,8,10................,48,50}

The number of numbers in the set is the answer

There are 26 numbers in the set

[50/2] even numbers starting from 2 =25 numbers

As 0 is included thus answer is 25+1

Let $x$ and $y$ are the amount of pine and acorn respectively. It's easy to generate a formula

$3x+2y=150$

It's easy to make a conclusion that $x={0,2,...,50}$ and $y={0,3,...,75}$ , but it will be easier if we make $x=2k$ and $y=3l$ . In other words, we make new variables.

Proof: By modulating $3x+2y=150$

$2y \equiv 0 \pmod 3$

$y \equiv 0 \pmod 3$

$y=3l$

and

$3x+2y=150$

$3x \equiv 0 \pmod 2$

$x \equiv 0 \pmod 2$

$x=2k$

We put all $x=2k$ and $y-3l$ into the first equation, $3x+2y=150$

$3x+2y=150$

$3(2k)+2(3l)=150$

$6k+6l=150$

$6(k+l)=150$

$k+l=25$

By using stars and banners or combinations with repetition, we get

${25+2-1 \choose 25} = {26 \choose 25} = 26$

3p+2a=150, solving for p allows for us to see that the squirrel must collect an even number of pine cones,p, so, 0,2,4,6,8..... 50 would be the possible solutions of p, thus there are 26 combinations

Devemos verificar quantas soluções inteiras existem na equação 3x+2y=150 em que x representa o número de pinhas e y, o número de bolotas. Vamos separar em casos em que x assume os valores de 1,2,3,...,50 (50 é o limite já que teríamos valores negativos de y para x>50). É fácil ver que, para valores ímpares (1,3,5,..), não há valores inteiros para y que satisfaça a equação. Logo, x pode apenas assumir valores pares até 50 (0,2,4...,50)=50/2+1 (conta-se o zero) = 26

This is a rather non-elementary solution that I include for fun. The desired number is the coefficient on the $x^{150}$ term of the series expansion of $\frac{1}{1-x^2}\frac{1}{1-x^3}$ since summing $n+m$ is isomorphic to multiplying $x^nx^m$ . Using Taylor's formula or expanding with some computer algebra, we get 26.

We can set up an equation $3p + 2a = 150$

Variables can't be negative so Max $p = 50$ and Max $a = 75$ as this is where the other variable becomes zero

$p$ is an integer $0 \leq p \leq 50$ and $a$ is an integer $0 \leq a \leq 75$

$p$ must always be even as $3p = 150 - 2a$ $\Rightarrow$ $3p = 2(75 - a)$ which is of the form of an even number $2n$ where n is $75-a$ this means $3p$ must be divisible by $2$ so $p$ is therefore even for integer solutions

Rearranging formed equation $3p + 2a = 150$

$a = \frac {150 - 3p} {2}$

$a = \frac {3} {2} (50 - p)$

So as $p$ is even $50 - p$ is even

This means for all even $p$ such that $0 \leq p \leq 50$ you will get an integer value for $a$

So number of different combinations is $\frac {50} {2} + 1$ (adding one for when $p = 0$ )

Giving the final solution of $26$

A simple way to solve it would be to assume that 150 days can be initially divided into the proportion of 75 acorns (at the maximum), and that if we put pine 1 we get (150-3 = 147) so 147 does not have zero rest in the division by 2 (what are the Acorns), then to divide correctly we have to use : 75 (Acorns) (1 Possibility) or 2 (Pines) + 72 (Acorns) (2 Possibility) or 4 (Pines) + 69 (Acorns) (3 Possibility) or 6 (Pines) + 66 (Acorns) (4 Possibility) or ... 50 (Pines) (26 Possibility).

The Min and Max of x and y

Denote x=pine corn and y=acorn

The number of food the squirrel get will be

3x + 2y = 150

If x=0 y=75

If y=0 x=50

Diophantine Equation:

let x=50 and y=0 thus (If n=0)

x = 50+2n and

y= 0-3n

Since the maximum of y = 75

-3n = 75 or n=-25

The statement is true for n is (0,-1,-2,-3,...,-25) Thus there are 26 different combinations in the statement

(2 x a)+(3 x y)=150 (a=no of acorn && y=no of pine cone) (2 x a)=3(50-y) therefore (50-y)=2 x k (where k is an integer) as a canot be -ve (50-y) also cannot be -ve therefore y ranges from 0 to 50 even numbers therefore k ranges from 0 to 25. number of integers from 0 to 25 is 26

• all acorn - 75 acorns - 1
• subtitute 3 acorn with 2 pine cones - 72 acorns - 2
• subtitute 6 acorn with 4 pine cones - 69 acorns - 3
• ...
• ....
• subtitute 75 acorns with 50 pine cones - 0 acorns - 26

We have $3p+2a=150$ , where $p$ is the number of pine cones and $a$ is the number of acorns. Solving for $p$ , we have $p=50-\dfrac{2a}{3}$ . Since $p,a$ are integers, then $3|a$ because $\gcd(2,3)=1$ . Note that since $3|a$ , we can write $a=3x$ where $x$ is an integer. Therefore, $p=50-2x\Rightarrow p=2(25-x)$ . Since $p,a$ are both non-negative, $x$ can be any integer in the set $\{0,1,\cdots,24,25\}$ . Therefore the answer is $\boxed{26}$

Avoid double-counting; (pines, acorns) - (0, 75). 75/3 = 25, add one for (50,0) It works vice versa, 50/2 = 25, add one for (0, 75)

First, we know that a possible combination is 50 pine cones and 0 acorns. Next, we can see that every combination in the form of $2n$ pine cones and $3n - 3$ acorns. Since the maximum value of $n$ is 25, and the minimum is 0, there are 26 different combinations.

length (filter even [0,3..150])

Let (p, a) denote p pine cones and a acorns. For example, (50, 0) means he collects 50 pine cones and no acorns and this is a valid combination because 50 x 3 = 150. I found the following. (50, 0) (48, 3) (46, 6) (44, 9) ... (2, 72) (0, 75) That's 26 combinations.

2x+3y=150

x {0}=30 y {0}=30

x=30-2t>0

y=30+3t>0

30>2t

t<15

30>-3t

3t>-30

t>-10

Logo, como existem 26 números inteiros de -10 a 15, então existem 26 pares distintos de x e y que satisfazem a Equação Diofantina Linear dada (2x+3y=150) de modo que x e y sejam não-negativos. Então há 26 possíveis combinações.

I found the end possibilities: 50 pinecones 0 acorns, or 0 pinecones 75 acorns.

In terms of amounts of food, 3 acorns = 2 pinecones. Thus there are 24 ways to make different combinations between the end possibilities; 24 + 2 = 26.

2a+3p=150 P must be between 0 & 50. Consider the values generated by P values. Only even values of P give a solution that works with A. There are 25 solutions involving P, plus the one solution with A. Hence 26 combinations.

Let (p, a) denote p pine cones and a acorns. For example, (50, 0) means he collects 50 pine cones and no acorns and this is a valid combination because 50 x 3 = 150. I found the following. (50, 0) (48, 3) (46, 6) (44, 9) ... (2, 72) (0, 75) That's 26 combinations.

Number of Pines = p ,

Number of Acorns = a

Where a and p are integers

Now we can have our simple equation ,

3p + 2a = 150

$\Rightarrow$ 2a = 150 - 3p

$\Rightarrow$ 2a = 3 (50 - p)

$\Rightarrow$ a = $\frac{3}{2}$ (50 - p)

From the Equation we can see that a will be an integer only when (50-p) is not an odd number.

Values of p can be from 0 to 50 . From 0 to 50 there are 51 numbers and 25 of them are odd .

so, answer = (51 - 25) = 26

We have: $3x + 2y = 150$ . So $x = 0, 2, 4, 6, ... , 50$ . There are $26$ combinations.

you have to use the 2 days in every par chance, if you divide 150/3.2 you have 25 plus 1 because 0 is par

First of all, let us note the following facts:

Fact 1

The product of $2$ odd integers will always be an odd integer.

Proof:

$(2k+1)(2k+1)=4k^2+4k+1=2(2k^2+2k)+1$

Since the integer $2(2k^2+2k)$ is divisible by $2$ , we are free to call it even. If we add $1$ to any even integer we will get an odd one. Therefore, the integer $2(2k^2+2k)+1$ is odd, which means that the product of any $2$ odd integers is an odd integer.

Fact 2

The product of an odd and an even integer will always be an even integer.

Proof:

$2k(2k+1)=4k^2+2k=2(2k^2+k)$

Since the number $2(2k^2+k)$ is divisible by $2$ , we are free to call if even. Consequently, this proves that the product of any even and any odd integer will be an even integer.

With these things in mind, let us turn our attention to the actual problem. The minimum number of pine cones the squirrel can use is $0$ , whereas the maximum is $50$ . However, following the facts given above we can notice that every odd number of pine cones used (( $2k+1)\cdot3$ , which is an odd integer) is not a possible solution of the problem, since if we subtract an odd integer from an even one we would get an odd one

$2k-(2k+1)=2k-2k-1=-1$ , which is an odd integer.

Therefore, we cannot evenly divide this integer with $2$ in order for the squirrel to last exactly $150$ days. That leaves us only with the even number of pine cones (possible because of the even number of days left that way), which is $\frac{50}{2}=25$ . There is also one more case not counted in this equation, which is the one with $0$ pine cones used. Therefore, the total number of combinations our squirrel can use is $26$ .

The situation could be modeled by an equation 3x+2y=150, where x is the number of pine cones and y is the number of acorns. The ordered pairs are (50,0);(48,3);(46,6)... until it ends at (0,75). Looking at x-values, they form even positive integers up to 50 (including 0). 50/2=25, plus the ordered pair (0,75) is 26 .

We would like to determine how many non-negative integer pairs $(a,p)$ satisfy $3p + 2a = 150$ . Since $a,p$ must be integral, it must be the case that $2 | 150 - 3p$ ; thus $2 | p$ . Combine this result with the constraint that $0 \leq p \leq \frac{150}{3} = 50$ gives 26 possible pairs.

Let us focus on the squirrel collecting pine cones: For the amount of pine cones he collects that will last him, 150 subtracted from that number must be even, in order to have a whole number of acorns. Therefore our number of pine cones must be an even number, with the max being 50 because 150/3=50. So at first we think the answer is just all even numbers from 1-50, giving us 25, but we forget to take into account that the squirrel could have 0 pine cones and all acorns. Therefore our total is 25+1=26

Suppose he collects $x$ pine cones and $y$ acorns, then $$3x + 2y = 150$$ that means $3 \equiv 0 \pmod{2}$ , i.e. $x = 2n$ , hence $y = 75 - 3n$ , we need $x \geq 0$ and $y \geq 0$ , so $$0 \leq n \leq 25$$ So there are $$\boxed{26}$$ different combinations.

The simple way to solve that problem is using Diophantine Equation, which from equation with x = pines and y - acorn, so we can make it become

$3x + 2y = 150$ ----- > $3x = 150 - 2y$ ------> $x = 50 - \frac{2}{3}y$

since x and y must be non-negative integers, so the positive value of y is spreaded from 0, 3, 6, .... , until 75, which the value of x will depends from the value o y, which we finally found all of the pairs of (x,y), that is 26 pair of solutions.

QED

First we must make a working equation The working equation must be 3x+2y=150. using the formula for Diophantine equations and (50,1) as value of (x,y), x= 50-2k and y=0+3k using basic counting, the only possible values of k such that x and y are positive integers are k=[0,25] which gives 26 possible combinations

When the squirrel has only acorns, he needs 75 of them. When he has only pine corns, he needs 50 of them. We can always exchange 2 pine corns for 3 acorns. So when starting from only pine corns, and every time exchanging them once, we get 50/2=25 exchanges. After adding the starting position (50 pine corns) we get 26 combinations.

Let x and y be the number of pine and acorn collected by the squirrel respectively. Then,

                                3x+2y=150

                           i.e  y=(150-3x)/2
                                  =75-3x/2

For y to be an integer x should be a multiple of 2 and the maximum value of x=50 where y becomes 0. Therefore, the answer is the no. of even numbers from 0-50 i.e 26.

Let p=number of pines the squirrel collects and a=number of pines the squirrel collects

3p+2a=150; p=50-(2/3)a. For p to be a positive integer, a must be a non negative integer and a multiple of 3 and (2/3)a<=50. Therefore 0<=a<=75, a=0,3,6,9,...,75. Therefore, there are 26 combinations.

Well, let us say that the number of pines collected is x and the number of acorns is y. We have the following equation, 3x+2y = 150. following from which y=(150-3x)/2. Now, basically, we need 150-3x to be even(and positive), which is possible only when x=0,2,4,6...........50. 0 and 50 are inclusive because it is well possible that the poor squirrel lands up with no pines or no acorns. So we have 26 possible values of x, and along with each x, 26 values of y, so in a way we have 26 ordered pairs (x,y), resulting in 26 combinations.

3x+2y=150;

x=0,y=75;(0,75)

x=2,y=72;(2,72)

...

x=50, y=0;(50,0)

3(25,24,23,...,2,1,0)

n=25-0+1=26.

Let $x$ and $y$ be the number of pine cone and acorn, respectively.

Thus we can model the problem as the number of integral nonnegative solutions $(x,y)$ of $3x+2y=150$ .

Suppose the squirrel didn't get any pine cone. Thus, to last for $150$ days, the squirrel must get $75$ acorns. That is the maximum number of acorns the squirrel can have. We all know that $3|150$ and $3|3x$ . thus it is clear to say that $3|2y$ where $gcd(3,2)=1$ . Hence, $3|y$ . But $y$ ranges from $0$ to $75$ . Thus $y$ can have $26$ nonnegative integral values.

One can systematically count all combinations:

75 acorns

72 acorns and 2 pine cones

69 acorns and 4 pine cones

...

0 acorns and 50 pine cones

Notice that the number of acorns decreases by 3 in each term.

Thus there are (75/ 3 ) + 1 = 26 possibilities. (The final plus one includes the last term with 0 acorns)