Difference between revisions of "2006 iTest Problems/Problem 10"
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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. | + | |
| + | 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. | ||
| + | |||
| + | For the third figure, there is one of the smaller triangles from the previous figure and three dots adding to <math>9</math> numbers. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | 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 17:24, 19 March 2020
Solution
The pattern for 
rows of Pascal's Triangle with the multiples of 
colored red is here: http://www.catsindrag.co.uk/pascal/?r=64&m=4
There are five different figures in this triangle.
The black triangles with 
red dots in them. There are 
of these.
The three small red triangles with a dot in the middle separated by black in between. There are 
of these.
The three red dots with a red triangle in the middle separated by black in between. There are 
of these.
The medium red triangles. There are 
of these.
The large red triangles. There are of these.
For the first figure, there are multiples of represented by the three red dots.
For the second figure, notice the first one of those is on the 
th row, meaning there are total numbers in that row. Then subtract the black numbers to get multiples, but that's for both of those lines, so each one is numbers long. The number of red numbers in the row of the triangle below that one is 
numbers long and the last row has 
number. Each one of those triangles therefore has 
numbers. In each copy of this figure, there are three of these triangles and a single dot adding to 
numbers.
For the third figure, there is one of the smaller triangles from the previous figure and three dots adding to numbers.
For the fourth figure, notice the first one of these triangles is on the 
th row so there are 17 numbers in that row. Subtract three for 
numbers in total for the tops of those two triangles and 
for one of them. Once again, that means one triangle has on the first row, on the second, until on the last row. This adds to a total of 
Since each of these figures are only one triangle, there are 
numbers.
For the fifth figure, we use the same logic to find that each large triangle has 
numbers
