Difference between revisions of "2006 iTest Problems/Problem 10"
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{{iTest box|year=2006|num-b=9|num-a=11|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}} | {{iTest box|year=2006|num-b=9|num-a=11|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}} | ||
Latest revision as of 22:33, 3 November 2023
Contents
Problem 10
Find the number of elements in the first 
rows of Pascal's Triangle that are divisible by 
.
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
Therefore, the total number of red numbers, or multiples of four, are:
Sidenote
Instead of spitting the problem into five different figures, we could have noted that there are 
red dots, 
small triangles, medium triangles, and big triangles. Then, using similar logic to above, we find there are dots per small triangle, 
dots per medium triangle, and dots per large triangle. Then the answer is 
~Someonenumber011
See Also
| 2006 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 9 |
Followed by: Problem 11 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10 | ||
