Sneaky snakes

How many combinations of iguanas ( $i$ ) and humans ( $h$ ) can you have so that there are 200 legs? We can test this by varying $i$ .

We can write an equation to represent the number of legs:

$4i + 2h = 200$

And rewrite it so that we can find $h$ if we know $i$ :

$h = 100 - 2i$

Now, let's cycle through the different possibilities for $i$ .

• $i = 1,$ $h = 98$
• $i = 2,$ $h = 96$
• $i = 3,$ $h = 94$

...

• $i = 48,$ $h = 4$
• $i = 49,$ $h = 2$

It stops there because the next iguana would make it so that no more humans are in the room, and of course we need to think about our good friend Calvin!

In all of these 49 possible numbers of iguanas, the sum of the iguanas and humans (all of the heads) were less than 103, leaving room for a varying amount of snakes.

Therefore, there are $49$ possibilities for the number of snakes that our little reptile fan got to see.

We have s+i+p=103 and p+2i=100. Subsituting p+i=100-i gives us s-i=3. Now i can range from 1 to 49, so s ranges from 4 to 52, or 49 possible numbers.

From the problem above, we can use here algebra to find the possible number of combinations of the snakes, iguanas and people. Let x be the number of snakes, y be the number of iguanas and z be the number of people. Then we can have, x + y + z = 103 and 4y + 2z = 200. From the two equations, we can x - y = 3. We can conlude that the maximum number of iguana in the zoo is 50. But this is impossible so 49 is the maximum number of it. Then if there are 49 iguanas, there are 52 snakes. The minimum number of snakes is 4 so there are 49 different possibilities for the number of snakes.

Denote the respective numbers of animals/persons by first letter in the name, then $4i + 2p = 200 \\ i + s + p = 103$ yielding $p = 100 - 2i \\ s = 3 + i$ Hence $s \geq 4$ and $i \leq 49$ , i.e. $s \leq 52$ , hence there are $52 - 4 + 1 = \boxed{49}$ possible numbers of snakes.

Let n be the number of people, k be the number of iguanas and m be the number of snakes. Then we have two equations:

$2n+4k=200$

$n+k+m=103$

They will give $m=k+3$ and $n+2k=100$ . We can see that k can have 49 different values (k can only be a whole number not 0 nor 50 (because that means n=0)) which gives n 49 distinct values and also m 49 distinct possible values. Hence the answer 49.

Let s be the number of snakes, p be the number of people and i be the number of iguanas.

We immediately see that we can form a system of linear equations using the number of heads and number of legs respectively:

$s+p+i=103$ - heads

$2p + 4i = 200$ - legs

Now, since both people and iguanas have bilateral symmetry, we can simplify by only considering the left leg.

$p + 2i = 100$ - left leg

From the left leg equation, we can see that p must be even, and create a set of tuples T containing all 49 solutions to this equation of the form.

$T_n = (p_n,i_n)$

$T = {(2,49),(4,48),(6,47),(8,46), ... ,(98,1)}$

$|T| = 49$

Observe that the sum of the elements of each tuple is unique. Each of the 49 tuple element sums $p_n+i_n$ gives a unique value of s. Hence, $s = 49$

Since there is at least one each of iguanas, snakes, and people, we subtract $3$ heads from our total to get $100$ and $6$ legs from out total to get $194$ . We consider the maximum and minimum numbers of extra snakes (ones beyond our original value of $1$ snake): the minimum is obtained when when we have the maximum number of humans, as then we have the maximum number of heads per leg and thus the fewest needed snakes. To account for all the legs in this way, we need $194/2=97$ people, which means we need $3$ snakes. For the maximum number of extra snakes, we need the minimum number of heads per leg, which we get by using only iguanas. However, we cannot only have iguanas, as the number of legs ( $194$ ) is not divisible by $4$ . The maximum number of iguanas is $\lfloor 194/4 \rfloor=48$ , which means we have $194-192=2$ legs left over for humans, or $1$ human. This is a total of $49$ heads, so we have $51$ extra snakes. So, our maximum and minimum numbers of extra snakes, respectively, are $3$ and $51$ . The total number of different possibilities is thus $51-2=\boxed{49}$ , as long as each solution is valid. We prove this as follows:

Say we have a $s$ number of extra snakes, thus $100-s$ heads and $194$ legs remaining. Let us first say we have $100-s$ humans, so we have $200-2s$ legs. This takes into account the case where $194=200-2s$ , or $s=3$ . Now we consider when $194 \neq 200-2s$ . Our number of legs $200-2s \le 194$ because $s \ge 3$ , so we only need to consider the cases when $200-2s < 194$ . Let $200-2s+x=194$ . It is obvious that $x$ is even. For each human we replace with an iguana, we add $2$ to our leg total. So if we replace $\frac{x}{2}$ humans with iguanas, we add $2\cdot\frac{x}{2}=x$ legs to our total, meaning our total is equal to $194$ . This means every possible number of snakes in our previously stated parameters is valid, so we have $\boxed{49}$ as our answer.

Let the number of snake ,iguana and men are S,I,M.

So, we can make two equations:

S+I+M=103--------------------------------------------------------------(1)

2M+4I=200---------------------------------------------------------------(2)

By doing (2)-(1) we get,

S-I=3

the least number of iguana is 1 and so least value of I is 1.

So, * least number of snake is 3+1=4. *

Now we will put I=2,3,4,5,6,7,8,9,----------------------------------------------49.

We will get S=5,6,7,8,9,10,11,12,---------------------------------------------------52.

When I=50 then S=53.

Now** in this case total head of snake and iguana is 53+53=103 which is the total number of heads.

In this case there exists no man.**

But there must be at least a man.

So ,I=50 is not acceptable.

So,we can get total 49 possibilities (4,5,6,7,8,9,10,11,----------------------------------------------52.)

So, we get * 49. *