Scrat the squirrel has a few nuts. When he put them in groups of 5 nuts, two nuts were left, and when he laid them out in groups of 3 nuts, there were no nuts left. How many nuts among the
options provided below could belong to Scrat?

The options are not shown, so i will solve this in a general way.

Let's suppose that there are N nuts

If we divide the N nuts into groups of 5, there will be a remainder of 2 nuts.

If we divide the N nuts into groups of 3, there is no remainder.

This means taht N is a multiple of 3. then we can write:

N = 3*k

where k is a random integer.

Now, we know that when we divide N by 5, there is a remainder of 2 nuts, this means that N is equal to a multiple of 5 plus 2, then we can write:

N = 5*j + 2

where j is a random integer.

Now we only need to find a pair of k and j (both positive integers) such that:

3*k = 5*j + 2

We can rewrite this as:

k = (5/3)*j + (2/3)

Now we could just input different values of j, and see if k is also an integer:

j = 2

k = (5/3)*2 + 2/3 = 10/3 + 2/3 = 12/3 = 4

Then the pair j = 2, k = 4 is a possible solution.

N = 3*k = 3*4 = 12

N = 5*j + 2 = 5*2 + 2 = 12

From this we can conclude that N = 12, so Scrat has 12 nuts.

Now with the equation

k = (5/3)*j + (2/3)

We can find other possible combinations of j and k, that will give different values for N.

for example, if j = 5:

k = (5/3)*5 + 2/3 = 25/3 + 2/3 = 27/3 = 9

Then the pair j = 5, k = 9 is also a possible solution:

N = 3*k = 3*9 =  27

N = 5*j + 2 = 5*5 + 2 = 27

In this case, Scrat has 27 nuts.

step-by-step explanation:

descending order means to order the polynomials from the term with the largest exponent to the term with the smallest exponent.  

therefore, the correct answer is d.