Difference between revisions of "2006 iTest Problems/Problem 10"

(Created page with "==Solution== The pattern for <math>64</math> rows of Pascal's Triangle with the multiples of <math>4</math> colored red is here: http://www.catsindrag.co.uk/pascal/?r=64&m=4...")
 
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For the first figure, there are <math>3</math> multiples of <math>4</math> represented by the three red dots.
 
For the first figure, there are <math>3</math> multiples of <math>4</math> represented by the three red dots.
For the second figure, notice the first one of those is on the <math>8</math>th row, meaning there are <math>9</math> total numbers in that row. Then subtract the <math>3</math> black numbers to get <math>6</math> multiples, but that's for both of those lines, so each one is <math>3</math> numbers long. We know from  how these patterns on Pascal's Triangle work that the number of red numbers in the row of the triangle below that one is <math>2</math> numbers long and the last row has <math>1</math> number. Each one of those triangles therefore has <math>6</math> numbers.
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For the second figure, notice the first one of those is on the <math>8</math>th row, meaning there are <math>9</math> total numbers in that row. Then subtract the <math>3</math> black numbers to get <math>6</math> multiples, but that's for both of those lines, so each one is <math>3</math> numbers long. The number of red numbers in the row of the triangle below that one is <math>2</math> numbers long and the last row has <math>1</math> number. Each one of those triangles therefore has <math>3+2+1=6</math> numbers. In each copy of this figure, there are three of these triangles and a single dot adding to <math>19</math> numbers.
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For the third figure, there is one of the smaller triangles from the previous figure and three dots adding to <math>9</math> numbers.
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For the fourth figure, notice the first one of these triangles is on the <math>16</math>th row so there are 17 numbers in that row. Subtract three for <math>14</math> numbers in total for the tops of those two triangles and <math>7</math> for one of them. Once again, that means one triangle has <math>7</math> on the first row, <math>6</math> on the second, until <math>1</math> on the last row. This adds to a total of <math>7+6+5+4+3+2+1=\frac{(7)(8)}{2}=28</math> Since each of these figures are only one triangle, there are <math>28</math> numbers.
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For the fifth figure, we use the same logic to find that each large triangle has <math>15+14+...+1=120</math> numbers

Revision as of 18:24, 19 March 2020

Solution

The pattern for $64$ rows of Pascal's Triangle with the multiples of $4$ colored red is here: http://www.catsindrag.co.uk/pascal/?r=64&m=4 There are five different figures in this triangle.

$1.$ The black triangles with $3$ red dots in them. There are $27$ of these. $2.$ The three small red triangles with a dot in the middle separated by black in between. There are $9$ of these. $3.$ The three red dots with a red triangle in the middle separated by black in between. There are $6$ of these. $4.$ The medium red triangles. There are $10$ of these. $5.$ The large red triangles. There are $3$ of these.

For the first figure, there are $3$ multiples of $4$ represented by the three red dots.

For the second figure, notice the first one of those is on the $8$th row, meaning there are $9$ total numbers in that row. Then subtract the $3$ black numbers to get $6$ multiples, but that's for both of those lines, so each one is $3$ numbers long. The number of red numbers in the row of the triangle below that one is $2$ numbers long and the last row has $1$ number. Each one of those triangles therefore has $3+2+1=6$ numbers. In each copy of this figure, there are three of these triangles and a single dot adding to $19$ numbers.

For the third figure, there is one of the smaller triangles from the previous figure and three dots adding to $9$ numbers.

For the fourth figure, notice the first one of these triangles is on the $16$th row so there are 17 numbers in that row. Subtract three for $14$ numbers in total for the tops of those two triangles and $7$ for one of them. Once again, that means one triangle has $7$ on the first row, $6$ on the second, until $1$ on the last row. This adds to a total of $7+6+5+4+3+2+1=\frac{(7)(8)}{2}=28$ Since each of these figures are only one triangle, there are $28$ numbers.

For the fifth figure, we use the same logic to find that each large triangle has $15+14+...+1=120$ numbers

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