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# Powers of 11

Alignments to Content Standards: A-APR.C.5 A-APR.A.1

Felicia notices what appears to be an interesting pattern between powers of 11 and powers of $x + 1$:

\begin{align} 11^0 &= 1 & (x+1)^0 &= 1 \\ 11^1 &= 11 & (x+1)^1 & = x+1 \\ 11^2 &= 121 & (x+1)^2 &= x^2 + 2x +1 \end{align}
The digits of the number $11^n$ are the same as the coefficients of the polynomial $(x+1)^n$. Is this always true?
1. Does this pattern continue for $n = 3$ and $n = 4$?
2. What is the answer to Felicia's question?

## IM Commentary

This task has students combine polynomial arithmetic with pattern-matching. Students can expand powers of $x+1$ using either repeated multiplication (A-APR.1) or by the binomial theorem (A-APR.5), and then are asked to analyze the question of whether the similarity of coefficients with the digits of powers of 11 is a coincidence. Identifying patterns, as Felicia has done, is an important part of mathematics. In this case, there is a deep relationship between the numbers and polynomials that Felicia is investigating; on the other hand, further consideration shows that the pattern does not continue. It is important for students not only to identify patterns but also to look more deeply to understand whether or not the patterns are ''generalizable'' or true because of some essential mathematical structure. For these reasons, the task provides a good illustration of Standard for Mathematical Practice 7: Look for and make use of structure.

This task might be used as either practice with polynomial arithmetic or as an introduction to the binomial theorem, providing a process for raising binomials to powers without dredging through many repetitive applications of the distributive law.

## Solution

The pattern continues! We have $11^3=1331$ and $(x+1)^3=x^3+3x^2+3x+1$, and $11^4=14641$ and $(x+1)^4=x^4+4x^3+6x^2+4x+1$. To see why this is happening, let's start with the simple case of $n = 2$. Here we have $11^2 = 121$ and $(x+1)^2 = x^2 + 2x + 1$. The hundreds digit of $11^2$ is the same as the coefficient of $x^2$ in $(x+1)^2$ and similarly for the tens digit and ones digit. In this case, however, we can start to see what is happening by substituting $x = 10$ in the expression $(x+1)^2$. When $x = 10$ we have $x + 1 = 11$. Moreover, $x^2 = 100$ and $x = 10$ so the coefficient of $x^2$ in $(x+1)^2$ tells us how many hundreds we have (1), the coefficient of $x$ tells us how many tens (2), and the constant term is the number of ones (1).

Now we see why the pattern continues: the binomial theorem tells us how to expand $(x+1)^n$, and the digits of $11^n$ are what we get when we substitute $x=10$. When $n = 3$, we have $$(x+1)^3 = x^3 + 3x^2 + 3x + 1.$$ Substituting $x = 10$ we find

\begin{align} 11^3 &= 1000 + 3 \times 100 + 3 \times 10 + 1\\ &= 1331 \end{align}

When $n = 4$ we have (using the binomial theorem or successively multiplying polynomials) $$(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x+ 1.$$ Substituting $x = 10$ we find

\begin{align} 11^4 &= 10,000 + 4 \times 1,000 + 6 \times 100 + 4 \times 10 + 1\\ &= 14,641 \end{align}

So Felicia is right that there is a very close relationship between the coefficients of the polynomial $(x+1)^n$ and the digits of the number $11^n$. The two are not always the same, however, because the coefficients of $(x+1)^n$ depend on $n$ and become larger as $n$ grows. The digits of $11^n$, however, can only range from 0 to 9 because of our decimal system for writing numbers. So when $n = 5$, for example, $$(x+1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1.$$ Substituting $x = 10$ we find

\begin{align} 11^5 &= 100,000 + 5 \times 10,000 + 10 \times 1,000 + 10 \times 100 + 5 \times 10 + 1\\ &= 161,051 \end{align}

In this case, both the hundreds and the thousands get regrouped when we write this number in base 10. If we were allowed to put any whole number we like in each place value, then the coefficients of $(x+1)^n$ would give us directly an expression for $11^n$ for all values of $n$.