Algebra help.Finding four whole numbers that agree with n

Hi guys this is first post on the forum as I am seeking help for this question

The question states:

5. Some statements have the interesting property that they describe themselves. For example:
“This sentence contains thirty six letters”

(i) Find a list of four whole numbers n0, n1, n2, n3 which obeys the following rules

n0 = the number of times 0 occurs in your list
n1 = the number of times 1 occurs in your list
n2 = the number of times 2 occurs in your list
n3 = the number of times 3 occurs in your list

Explain carefully how you obtained your list.

(ii) Find another list of four, different from your answer in part (i)
(iii) Is it possible to form a similar list with only two numbers n0, n1 ? Explain carefully.
(iv) Is it possible to form a similar list with only three numbers n0, n1, n2 ? Explain carefully.
(v) In a list with eight numbers n0, n1, n2, n3, n4, n5, n6, n7 could n4, n5, n6, n7 all be zero? Explain.



Could please help me with at least i)

Hi gazing600000! :)

It's a matter of checking the different cases.

The number of zeroes is either 0,1,2,3, or 4.

Suppose there are 0 zeroes, then your list is (0,x,x,x).
But this list already contains a zero so that is a contradiction.

Suppose there are 4 zeroes, then your list is (4,0,0,0).
But 4 zeroes don't even fit, since the first number (n0) is 4.

And so on...

Can you do a couple of cases?