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

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==Problem 10==
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Find  the  number  of  elements  in  the  first  <math>64</math>  rows  of  Pascal's  Triangle  that  are  divisible  by  <math>4</math>.
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<math>\mathrm{(A)}\,256\quad\mathrm{(B)}\,496\quad\mathrm{(C)}\,512\quad\mathrm{(D)}\,640\quad\mathrm{(E)}\,796 \\
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\quad\mathrm{(F)}\,946\quad\mathrm{(G)}\,1024\quad\mathrm{(H)}\,1134\quad\mathrm{(I)}\,1256\quad\mathrm{(J)}\,\text{none of the above}</math>
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==Solution==
 
==Solution==
  
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<math>1.</math> The black triangles with <math>3</math> red dots in them. There are <math>27</math> of these.
 
<math>1.</math> The black triangles with <math>3</math> red dots in them. There are <math>27</math> of these.
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<math>2.</math> The three small red triangles with a dot in the middle separated by black in between. There are <math>9</math> of these.
 
<math>2.</math> The three small red triangles with a dot in the middle separated by black in between. There are <math>9</math> of these.
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<math>3.</math> The three red dots with a red triangle in the middle separated by black in between. There are <math>6</math> of these.
 
<math>3.</math> The three red dots with a red triangle in the middle separated by black in between. There are <math>6</math> of these.
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<math>4.</math> The medium red triangles. There are <math>10</math> of these.
 
<math>4.</math> The medium red triangles. There are <math>10</math> of these.
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<math>5.</math> The large red triangles. There are <math>3</math> of these.
 
<math>5.</math> The large red triangles. There are <math>3</math> of these.
  
 
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.
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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. 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
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Therefore, the total number of red numbers, or multiples of four, are:
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<math>(3)(27)+(19)(9)+(9)(6)+(28)(10)+(120)(3)=\boxed{(F) 946}</math>
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==Sidenote==
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Instead of spitting the problem into five different figures, we could have noted that there are <math>108</math> red dots, <math>33</math> small triangles, <math>10</math> medium triangles, and <math>3</math> big triangles. Then, using similar logic to above, we find there are <math>3+2+1=6</math> dots per small triangle, <math>7+6+5+4+3+2+1=28</math> dots per medium triangle, and <math>15+14+...+1=120</math> dots per large triangle. Then the answer is <math>(108)(1)+(33)(6)+(10)(28)+(3)(120)=946</math>
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~Someonenumber011
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==See Also==
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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]]}}

Latest revision as of 22:33, 3 November 2023

Problem 10

Find the number of elements in the first $64$ rows of Pascal's Triangle that are divisible by $4$.

$\mathrm{(A)}\,256\quad\mathrm{(B)}\,496\quad\mathrm{(C)}\,512\quad\mathrm{(D)}\,640\quad\mathrm{(E)}\,796 \\ \quad\mathrm{(F)}\,946\quad\mathrm{(G)}\,1024\quad\mathrm{(H)}\,1134\quad\mathrm{(I)}\,1256\quad\mathrm{(J)}\,\text{none of the above}$

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

Therefore, the total number of red numbers, or multiples of four, are: $(3)(27)+(19)(9)+(9)(6)+(28)(10)+(120)(3)=\boxed{(F) 946}$

Sidenote

Instead of spitting the problem into five different figures, we could have noted that there are $108$ red dots, $33$ small triangles, $10$ medium triangles, and $3$ big triangles. Then, using similar logic to above, we find there are $3+2+1=6$ dots per small triangle, $7+6+5+4+3+2+1=28$ dots per medium triangle, and $15+14+...+1=120$ dots per large triangle. Then the answer is $(108)(1)+(33)(6)+(10)(28)+(3)(120)=946$

~Someonenumber011

See Also

2006 iTest (Problems, Answer Key)
Preceded by:
Problem 9 		Followed by:
Problem 11
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