Question

Consider the non-decreasing sequence of positive integers:

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5.... in which $n^{th}$ positive number appears n times. Find the remainder when the 2000th term is divided by 4.

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is D

3

Let us see the sequence of the numbers:

Number Last term of the number

$11$
$23$
$36$
$410$
$\stackrel{-}{n}\frac{\stackrel{-}{n}(n+1)}{2}$

We have to find the value of n for the 2000th term.

If n = 62, the last term that ends with n is $(\frac{1}{2}\times 62\times 63)$ = 1953.

Therefore, the next 63 terms are 63. So the 2000th term is 63. So the remainder when divided by 4 is 3.