Alyssa writes 101 distinct positive integers on a blackboard, in a row, so that the sum of any two consecutive integers is divisible by 5. Let N be the smallest possible sum of all 101 integers. Find N. I started with 1, then 4, then 6.... How do I find the sum? What will be the 101th term? Is there an easier way? Thanks.
Answer
Assume $x$ is the remainder of the first number when divided by $5$, then it follows that the second number must have $5-x$ as the remainder when divided by $5$. Then the third number must have $x$ as the remainder, and the fourth number must have $5-x$ as the remainder when divided by $5$ ...
Hence, the $i$-th numbers all have remainder $x$ when divided by $5$ when $i$ is odd, and the $i$th numbers all have remainder $5-x$ when divided by $5$ when $i$ is even.
So, for a sequence of $101$ numbers, we have to get the smallest $51$ positive integers which have remainder $x$ when divided by $5$, and the smallest $50$ positive integers which have remainder $5-x$ when divided by $5$. Hence $x$ should be $1$ if we want to minimize the sum. This means your $101$ numbers are $1, 6, 11, 16, \dots, ... 251$, and $4, 9, 14, 19, \dots, ... 249$. Adding them together give you the total $= 12751$.
However, there is no distinct answer for your second part of the question "what is the $101$-th term". This is because the question does not specify whether the sequence of numbers must be strictly increasing. So, any one of the smallest $51$ positive integers which have remainder $1$ when divided by $5$, i.e., $1, 6, 11, 16, \dots, 251$, can be the $101$-th term.
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