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)

Re: Algebra help.Finding four whole numbers that agree with n

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?