2010 AMC 10A Problem 25

Problem 25 of 25HarderAlgebra Number Theory

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?

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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.

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