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Difference between revisions of "2007 iTest Problems/Problem 21"

(Created page with "== Problem == Find the largest integer <math>n</math> such that <math>2007^{1024}-1</math> is divisible by <math>2^n</math> <math>\text{(A) } 1\qquad \text{(B) } 2\qquad \text{...")
 
(Solution to Problem 21 + Fixed Problem 21)
 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
−
Find the largest integer <math>n</math> such that <math>2007^{1024}-1</math> is divisible by <math>2^n</math>
+
James writes down fifteen 1's in a row and randomly writes + or - between each pair of consecutive 1's.
 +
One such example is
 +
<cmath>1+1+1-1-1+1-1+1-1+1-1-1-1+1+1.</cmath>
 +
What is the probability that the value of the expression James wrote down is <math>7</math>?
  
−
<math>\text{(A) } 1\qquad
+
<math>\text{(A) }0\qquad
−
\text{(B) } 2\qquad
+
\text{(B) }\frac{6435 }{2^{14}}\qquad
−
\text{(C) } 3\qquad
+
\text{(C) }\frac{6435 }{2^{13}}\qquad
−
\text{(D) } 4\qquad
+
\text{(D) }\frac{429}{2^{12}}\qquad
−
\text{(E) } 5\qquad
+
\text{(E) }\frac{429}{2^{11}}\qquad
−
\text{(F) } 6\qquad
+
\text{(F) }\frac{429}{2^{10}}\qquad
−
\text{(G) } 7\qquad
+
\text{(G) }\frac{1}{15}\qquad
−
\text{(H) } 8\qquad \\ </math>
+
\text{(H) }\frac{1}{31}\qquad</math>
−
<math>\text{(I) } 9\qquad
 
−
\text{(J) } 10\qquad
 
−
\text{(K) } 11\qquad
 
−
\text{(L) } 12\qquad
 
−
\text{(M) } 13\qquad
 
−
\text{(N) } 14\qquad
 
−
\text{(O) } 15\qquad
 
−
\text{(P) } 16\qquad \\ </math>
 
−
<math>\text{(Q) } 55\qquad
 
−
\text{(R) } 63\qquad
 
−
\text{(S) } 64\qquad
 
−
\text{(T) } 2007\qquad</math>
 
  
 +
<math>\text{(I) }\frac{1}{30}\qquad
 +
\text{(J) }\frac{1}{29}\qquad
 +
\text{(K) }\frac{1001 }{2^{15}}\qquad
 +
\text{(L) }\frac{1001 }{2^{14}}\qquad
 +
\text{(M) }\frac{1001 }{2^{13}}\qquad
 +
\text{(N) }\frac{1}{2^{7}}\qquad
 +
\text{(O) }\frac{1}{2^{14}}\qquad
 +
\text{(P) }\frac{1}{2^{15}}\qquad</math>
 +
 +
<math>\text{(Q) }\frac{2007}{2^{14}}\qquad
 +
\text{(R) }\frac{2007}{2^{15}}\qquad
 +
\text{(S) }\frac{2007}{2^{2007}}\qquad
 +
\text{(T) }\frac{1}{2007}\qquad
 +
\text{(U) }\frac{-2007}{2^{14}}\qquad</math>
  
  
 
== Solution ==
 
== Solution ==
 +
 +
In James’s expression, he is adding or subtracting <math>14</math> ones to one.  Since the wanted result is <math>7</math>, he needs to add a total of <math>6</math> to <math>1</math>.  This is achieved when he writes <math>10</math> plus signs and <math>4</math> minus signs, and there are <math>\binom{14}{10} = 1001</math> possible ways to do that.  Since the total number of ways to put plus signs and minus signs between consecutive ones is <math>2^{14}</math>, the probability of getting a <math>7</math> is <math>\boxed{\textbf{(L) } \frac{1001}{2^{14}}}</math>.
 +
 +
==See Also==
 +
{{iTest box|year=2007|num-b=20|num-a=22}}
 +
 +
[[Category:Intermediate Probability Problems]]
 +
[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 03:34, 14 June 2018

Problem

James writes down fifteen 1's in a row and randomly writes + or - between each pair of consecutive 1's. One such example is \[1+1+1-1-1+1-1+1-1+1-1-1-1+1+1.\] What is the probability that the value of the expression James wrote down is $7$?

$\text{(A) }0\qquad \text{(B) }\frac{6435 }{2^{14}}\qquad \text{(C) }\frac{6435 }{2^{13}}\qquad \text{(D) }\frac{429}{2^{12}}\qquad \text{(E) }\frac{429}{2^{11}}\qquad \text{(F) }\frac{429}{2^{10}}\qquad \text{(G) }\frac{1}{15}\qquad \text{(H) }\frac{1}{31}\qquad$

$\text{(I) }\frac{1}{30}\qquad \text{(J) }\frac{1}{29}\qquad \text{(K) }\frac{1001 }{2^{15}}\qquad \text{(L) }\frac{1001 }{2^{14}}\qquad \text{(M) }\frac{1001 }{2^{13}}\qquad \text{(N) }\frac{1}{2^{7}}\qquad \text{(O) }\frac{1}{2^{14}}\qquad \text{(P) }\frac{1}{2^{15}}\qquad$

$\text{(Q) }\frac{2007}{2^{14}}\qquad \text{(R) }\frac{2007}{2^{15}}\qquad \text{(S) }\frac{2007}{2^{2007}}\qquad \text{(T) }\frac{1}{2007}\qquad \text{(U) }\frac{-2007}{2^{14}}\qquad$


Solution

In James’s expression, he is adding or subtracting $14$ ones to one. Since the wanted result is $7$, he needs to add a total of $6$ to $1$. This is achieved when he writes $10$ plus signs and $4$ minus signs, and there are $\binom{14}{10} = 1001$ possible ways to do that. Since the total number of ways to put plus signs and minus signs between consecutive ones is $2^{14}$, the probability of getting a $7$ is $\boxed{\textbf{(L) } \frac{1001}{2^{14}}}$.

See Also

2007 iTest (Problems, Answer Key)
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Problem 20 		Followed by:
Problem 22
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