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?