# Difference between revisions of "1995 AHSME Problems/Problem 27"

## Problem

Consider the triangular array of numbers with 0,1,2,3,... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.

$$\begin{tabular}{ccccccccccc} & & & & & 0 & & & & & \\ & & & & 1 & & 1 & & & & \\ & & & 2 & & 2 & & 2 & & & \\ & & 3 & & 4 & & 4 & & 3 & & \\ & 4 & & 7 & & 8 & & 7 & & 4 & \\ 5 & & 11 & & 15 & & 15 & & 11 & & 5 \end{tabular}$$

Let $f(n)$ denote the sum of the numbers in row $n$. What is the remainder when $f(100)$ is divided by 100?

$\mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 30 } \qquad \mathrm{(C) \ 50 } \qquad \mathrm{(D) \ 62 } \qquad \mathrm{(E) \ 74 }$

## Solution 1

Note that if we re-draw the table with an additional diagonal row on each side, the table is actually just two of Pascal's Triangles, except translated and summed. $$\begin{tabular}{ccccccccccccccc} & & & & & 1 & & 0 & & 1 & & & & \\ & & & & 1 & & 1 & & 1 & & 1 & & & \\ & & & 1 & & 2 & & 2 & & 2 & & 1 & & \\ & & 1 & & 3 & & 4 & & 4 & & 3 & & 1 & \\ & 1 & & 4 & & 7 & & 8 & & 7 & & 4 & & 1 \\ 1 & & 5 & & 11 & & 15 & & 15 & & 11 & & 5 & & 1 \end{tabular}$$

The sum of a row of Pascal's triangle is $2^{n-1}$; the sum of two of each of these rows, subtracting away the $2$ ones we included, yields $f(n) = 2^n - 2$. Now, $f(100) = 2^{100} - 2 \equiv 2 \pmod{4}$ and $f(100) = 2^{100} - 2 \equiv 2^{20 \cdot 5} - 2 \equiv -1 \pmod{25}$, and by the Chinese Remainder Theorem, we have $f(100) \equiv 74 \pmod{100} \Longrightarrow \mathrm{(E)}$.

## Solution 2 (induction)

We sum the first few rows: $0, 2, 6, 14, 30, 62$. They are each two less than a power of $2$, so we try to prove it:

Let the sum of row $n$ be $S_n$. To generate the next row, we add consecutive numbers. So we double the row, subtract twice the end numbers, then add twice the end numbers and add two. That makes $S_{n+1}=2S_n-2(n-1)+2(n-1)+2=2S_n+2$. If $S_n$ is two less than a power of 2, then it is in the form $2^x-2$. $S_{n+1}=2^{x+1}-4+2=2^{x+1}-2$.

Since the first row is two less than a power of 2, all the rest are. Since the sum of the elements of row 1 is $2^1-2$, the sum of the numbers in row $n$ is $2^n-2$. Thus, using Modular arithmetic, $f(100)=2^{100}-2 \bmod{100}$. $2^{10}=1024$, so $2^{100}-2\equiv 24^{10}-2\equiv (2^3 \cdot 3)^{10} - 2$ $\equiv 1024^3 \cdot 81 \cdot 81 \cdot 9 - 2 \equiv 24^3 \cdot 19^2 \cdot 9 - 2$ $\equiv 74\bmod{100} \Rightarrow \mathrm{(E)}$.

## Solution 3 (plain recurrence solving)

We derive the recurrence $S_{n+1}=2S_n + 2$ as above. Without guessing the form of the solution at this point we can easily solve this recurrence. Note that one can easily get rid of the "$+2$" as follows: Let $S_n=T_n-2$. Then $S_{n+1}=T_{n+1}-2$ and $2S_n+2 = 2(T_n-2)+2 = 2T_n-2$. Therefore $T_{n+1}=2T_n$. This obviously solves to $T_n=2^{n-1} T_1$. As $S_1=0$, we have $T_1=2$. Therefore $T_n=2^n$ and consecutively $S_n=2^n-2$.