In the following sequence, consisting of positive integers, the first term is a one digit number.

?, ? , 65, 61, 37, 58, 89, 145, ...

A) What is the first term?

B) What is the next term that is a one digit number?

Lt S(n) = nth term of the sequence.

By the problem, since S(n) can contain atmost three digits, without any loss of generality:

S(i) = 100*A(i) + 10*B(i) + C(i) for i=1,2,...........,n

with B(k)=C(k) =0 whenever S(k) is a one digit number and A(k)=B(k) =0 whenever S(k) is a one digit number and C(k) =0 whenever S(k) is a two digit number.

Now, by conditions of the problem;

S(i+1) = A(i)^2 +B(i)^2 + C(i)^2 so that S(3) = 65 gives

A(2)^2 +B(2)^2 + C(2)^2 =65; or, (A(2),B(2),C(2)) =(0,8,1) since, the three choices (A(2),B(2),C(2)) = (0,7,4),(0,4,7) and (0,1,8) does not yield A(1)=B(1) =0 which is a contradiction.

Accordingly, S(2) = 81 giving, A(1)^2 +B(1)^2 + C(1)^2 = 81 or, (A(1),B(1),C(1)) = (0,0,9) ( since, S(1) is a one digit number). This gives, S(1) =9.

Also, S(8) = 145 or, S(9) = 1^2 +4^2 + 5^2 = 42;

or, S(10) = 4^2 + 2^2 = 20;S(11) = 2^2 = 4.

HENCE;

(i) The first term is 9.

(ii) The eleventh term of the sequence is a one digit number and the numerical magnitude of the said number is 4.