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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 16"

(Created page with "Solution by excruciating: For each of <math>a,b,c,d,e,f,g</math>, they all need to be at least <math>1</math>, so we have: <math>7 +a+b+c+d+e+f+g\leq n,</math> or <math>a+b+c+...")
 
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Expanding, we get <math>(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = 3003 \cdot 6! = 2^4 \cdot 3^3 \cdot 5 \cdot 7 \cdot 11\ cdot 13</math>.
 
Expanding, we get <math>(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = 3003 \cdot 6! = 2^4 \cdot 3^3 \cdot 5 \cdot 7 \cdot 11\ cdot 13</math>.
  
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Because we only have <math>6</math> terms, <math>(n-1)</math> is to be around <math>13,</math> the highest divisor. Thus, we must have <math>5\cdot2 = 10</math> as a divisor too. <math>2^2 \cdot 3 = 12</math> is one too. We have left is <math>2, 3^2,</math> and <math>7</math>, to make <math>2</math> terms, so we have <math>3^2 = 9</math>, and <math>7\cdot2 = 14</math>. Thus we have: <math></math>(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = <math>14\cdot13\cdot12\cdot11\cdot10\cdot9.</math>$
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Because we only have <math>6</math> terms, <math>(n-1)</math> is to be around <math>13,</math> the highest divisor. Thus, we must have <math>5\cdot2 = 10</math> as a divisor too. <math>2^2 \cdot 3 = 12</math> is one too. We have left is <math>2, 3^2,</math> and <math>7</math>, to make <math>2</math> terms, so we have <math>3^2 = 9</math>, and <math>7\cdot2 = 14</math>. Thus we have: <cmath>(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = 14\cdot13\cdot12\cdot11\cdot10\cdot9.</cmath>
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Thus, <math> n = 14 + 1 = \boxed{(E) 15}.</math>

Revision as of 13:52, 5 November 2023

Solution by excruciating: For each of $a,b,c,d,e,f,g$, they all need to be at least $1$, so we have: $7 +a+b+c+d+e+f+g\leq n,$ or $a+b+c+d+e+f+g \leq n-7$ without the "$a,b,c,d,e,f,g$ need to be $>0$ restriction." So now it's stars and bars:

${n-7+7-1 \choose 7-1},$ which simplifies to \[{n-1 \choose 6},\] which we can write as $\frac{(n-1)!}{6! \cdot (n-7)!} = 3003$.

Expanding, we get $(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = 3003 \cdot 6! = 2^4 \cdot 3^3 \cdot 5 \cdot 7 \cdot 11\ cdot 13$.

Because we only have $6$ terms, $(n-1)$ is to be around $13,$ the highest divisor. Thus, we must have $5\cdot2 = 10$ as a divisor too. $2^2 \cdot 3 = 12$ is one too. We have left is $2, 3^2,$ and $7$, to make $2$ terms, so we have $3^2 = 9$, and $7\cdot2 = 14$. Thus we have: \[(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = 14\cdot13\cdot12\cdot11\cdot10\cdot9.\]

Thus, $n = 14 + 1 = \boxed{(E) 15}.$

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