Difference between revisions of "2019 Mock AMC 10B Problems/Problem 16"
Excruciating (talk | contribs) (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>. | ||
− | 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: < | + | 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> |
+ | |||
+ | Thus, <math> n = 14 + 1 = \boxed{(E) 15}.</math> |
Latest revision as of 14:52, 5 November 2023
Solution by excruciating:
For each of , they all need to be at least
, so we have:
or
without the "
need to be
restriction." So now it's stars and bars:
which simplifies to
which we can write as
.
Expanding, we get .
Because we only have terms,
is to be around
the highest divisor. Thus, we must have
as a divisor too.
is one too. We have left is
and
, to make
terms, so we have
, and
. Thus we have:
Thus,