Question

Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55,$ then his sequence contains $5$ numbers: $\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array}$ Let $N$ be the smallest number for which Jim’s sequence has $8$ numbers. What is the units digit of $N?$

$1$

$3$

$5$

$7$

$9$

Answer: B

Concepts:perfect square recursion work backwards

Difficulty rating: 2440

Solution:

We can work backwards starting with $0.$ From this, we can add on $1^2$ to get $1.$

We can again add on $1^2$ to get $2.$ Again, adding on $1^2$ gives us $3.$

If we add on $1^2$ now, we get $4,$ but then $1^2$ is not the greatest square less than or equal to $4.$

Then adding on $2^2$ gives us $7.$ Continuing until there are eight terms in the forward sequence gives $7223.$

$\begin{array}{ccccc} {}&{}&{}&{}&7223\\ 7223&-&84^2&=&167\\ 167&-&12^2&=&23\\ 23&-&4^2&=&7\\ 7&-&2^2&=&3\\ 3&-&1^2&=&2\\2&-&1^2&=&1\\1&-&1^2&=&0\end{array}$

Thus, B is the correct answer.

2010 AMC 10A Problem 25

Answer: B

Difficulty rating: 2440

Solution:

Problem 25 in Other Years