Difference between revisions of "1995 AHSME Problems/Problem 27"
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Revision as of 13:12, 17 June 2008
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.
\[\begin{tabular}{ccccccccccc} & & & & & 0 & & & & & \\ & & & & 1 & & 1 & & & & \\ & & & 2 & & 2 & & 2 & & & \\ & & 3 & & 4 & & 4 & & 3 & & \\ & 4 & & 7 & & 8 & & 7 & & 4 & \\ 5 & & 11 & & 15 & & 15 & & 11 & & 5 & \end{tabular}\] (Error compiling LaTeX. Unknown error_msg)
Let denote the sum of the numbers in row
. What is the remainder when
is divided by 100?
Solution
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 be
. 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
. If
is two less than a power of 2, then it is in the form
.
. 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
, the sum of the numbers in row
is
. Thus, using Modular arithmetic,
.
, so
.
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 |