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1995 AHSME Problems/Problem 27

Revision as of 11:05, 7 January 2022 by Isabelchen (talk | contribs) (→‎Side Note (Chinese Remainder Theorem))

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 $f(100) \equiv 74 \pmod{100} \Longrightarrow \mathrm{(E)}$.

Side Note (Chinese Remainder Theorem)

Solution 1 needs to calculate the last $2$ digits of $2^{100}$, meaning to solve $2^{100} (\bmod{ 100})$. $100 = 4 \cdot 25$, we are going to get $n \bmod{ 100}$ from $n \bmod{ 4}$ and $n \bmod{ 25}$.

By Chinese Remainder Theorem, the general solution of the system of $2$ linear congruences is: 

$n \equiv r_1 (\bmod { \quad m_1})$, $n \equiv r_2 (\bmod { \quad m_2})$, $(m_1, m_2) = 1$
Find $k_1$ and $k_2$ such that $k_1 m_1 \equiv 1 (\bmod{ \quad m_2})$, $k_2 m_2 \equiv 1 (\bmod{ \quad m_1})$
Then $n \equiv k_2 m_2 r_1 + k_1 m_1 r_2 (\bmod{ \quad m_1 m_2})$

In this problem, $2^{100} \equiv (2^{10})^{10} \equiv 1024^{10} \equiv 24^{10} \equiv (-1)^{10} \equiv 1 (\bmod{ \quad 25})$, $2^{100} \equiv 0 (\bmod{ \quad 4})$: 

$n \equiv 1 (\bmod{ \quad 25})$, $n \equiv 0 (\bmod{ \quad 4})$, $(25, 4) = 1$
$1 \cdot 25 \equiv 1 (\bmod{ \quad 4})$, $19 \cdot 4 \equiv 1 (\bmod{ \quad 25})$
Then $n \equiv 19 \cdot 4 \cdot 1 + 1 \cdot 25 \cdot 0 \equiv 76 (\bmod{ \quad 100})$

~isabelchen

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 $S_n=2^n-2$.

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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