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Difference between revisions of "2007 iTest Problems/Problem 25"

(Created page with "== Problem == Ted's favorite number is equal to <cmath>1\cdot{2007\choose 1}+2\cdot {2007\choose 2}+3\cdot {2007\choose 3} + \cdots + 2007\cdot {2007 \choose 2007}</cmath> Find...")
 
(Solution to Problem 25)
 
Line 1: Line 1:
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== Problem ==
+
==Problem==
  
 
Ted's favorite number is equal to
 
Ted's favorite number is equal to
 
<cmath>1\cdot{2007\choose 1}+2\cdot {2007\choose 2}+3\cdot {2007\choose 3} + \cdots + 2007\cdot {2007 \choose 2007}</cmath>
 
<cmath>1\cdot{2007\choose 1}+2\cdot {2007\choose 2}+3\cdot {2007\choose 3} + \cdots + 2007\cdot {2007 \choose 2007}</cmath>
  
−
Find the remainder when Ted's favorite number is divided by 25.
+
Find the remainder when Ted’s favorite number is divided by 25.
  
 
<math>\text{(A) } 0\qquad
 
<math>\text{(A) } 0\qquad
Line 14: Line 14:
 
\text{(G) } 6\qquad
 
\text{(G) } 6\qquad
 
\text{(H) } 7\qquad
 
\text{(H) } 7\qquad
−
\text{(I) } 8\qquad \\ </math>
+
\text{(I) } 8\qquad</math>
  
 
<math>\text{(J) } 9\qquad
 
<math>\text{(J) } 9\qquad
Line 23: Line 23:
 
\text{(O) } 14\qquad
 
\text{(O) } 14\qquad
 
\text{(P) } 15\qquad
 
\text{(P) } 15\qquad
−
\text{(Q) } 16\qquad \\ </math>
+
\text{(Q) } 16\qquad</math>
  
 
<math>\text{(R) } 17\qquad
 
<math>\text{(R) } 17\qquad
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\text{(U) } 20\qquad
 
\text{(U) } 20\qquad
 
\text{(V) } 21\qquad
 
\text{(V) } 21\qquad
−
\text{(A) } 22\qquad
+
\text{(W) } 22\qquad
 
\text{(X) } 23\qquad
 
\text{(X) } 23\qquad
 
\text{(Y) } 24</math>
 
\text{(Y) } 24</math>
  
−
== Solution ==
+
==Solution==
 +
 
 +
Let <math>T = \sum_{i=1}^{2007} i \cdot \binom{2007}{i}</math>.  That means
 +
<cmath>2T = 2 \cdot \sum_{i=1}^{2007} (i \cdot \binom{2007}{i}</cmath>
 +
We know that <math>\binom{2007}{i} = \binom{2007}{2007-i}</math>, so
 +
<cmath>2T = \sum_{i=1}^{2007} (i \cdot \binom{2007}{i}) + \sum_{j=1}^{2007} (j \cdot \binom{2007}{2007-j})</cmath>
 +
If <math>2007-j= i</math>, then <math>i+j = 2007</math>, so
 +
<cmath>2T = 2007 \cdot \sum_{i=0}^{2007} \binom{2007}{i}</cmath>
 +
<cmath>2T = 2007 \cdot 2^{2007}</cmath>
 +
<cmath>T = 2007 \cdot 2^{2006}</cmath>
 +
Note that <math>2^{10} \equiv -1 \pmod{25}</math>.  Using the properties of [[modular arithmetic]], we can find the remainder when <math>T</math> is divided by <math>25</math>.
 +
<cmath>T \equiv 2007 \cdot 2^{2000} \cdot 2^6 \pmod{25}</cmath>
 +
<cmath>T \equiv 2007 \cdot (2^{10})^{200} \cdot 64 \pmod{25}</cmath>
 +
<cmath>T \equiv 7 \cdot 1 \cdot 14 \pmod{25}</cmath>
 +
<cmath>T \equiv 98 \pmod{25}</cmath>
 +
<cmath>T \equiv 23 \pmod{25}</cmath>
 +
The remainder when Ted’s favorite number is divided by 25 is <math>\boxed{\textbf{(X) } 23}</math>.
 +
 
 +
==See Also==
 +
{{iTest box|year=2007|num-b=24|num-a=26}}
 +
 
 +
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 13:35, 29 July 2018

Problem

Ted's favorite number is equal to \[1\cdot{2007\choose 1}+2\cdot {2007\choose 2}+3\cdot {2007\choose 3} + \cdots + 2007\cdot {2007 \choose 2007}\]

Find the remainder when Ted’s favorite number is divided by 25.

$\text{(A) } 0\qquad \text{(B) } 1\qquad \text{(C) } 2\qquad \text{(D) } 3\qquad \text{(E) } 4\qquad \text{(F) } 5\qquad \text{(G) } 6\qquad \text{(H) } 7\qquad \text{(I) } 8\qquad$

$\text{(J) } 9\qquad \text{(K) } 10\qquad \text{(L) } 11\qquad \text{(M) } 12\qquad \text{(N) } 13\qquad \text{(O) } 14\qquad \text{(P) } 15\qquad \text{(Q) } 16\qquad$

$\text{(R) } 17\qquad \text{(S) } 18\qquad \text{(T) } 19\qquad \text{(U) } 20\qquad \text{(V) } 21\qquad \text{(W) } 22\qquad \text{(X) } 23\qquad \text{(Y) } 24$

Solution

Let $T = \sum_{i=1}^{2007} i \cdot \binom{2007}{i}$. That means \[2T = 2 \cdot \sum_{i=1}^{2007} (i \cdot \binom{2007}{i}\] We know that $\binom{2007}{i} = \binom{2007}{2007-i}$, so \[2T = \sum_{i=1}^{2007} (i \cdot \binom{2007}{i}) + \sum_{j=1}^{2007} (j \cdot \binom{2007}{2007-j})\] If $2007-j= i$, then $i+j = 2007$, so \[2T = 2007 \cdot \sum_{i=0}^{2007} \binom{2007}{i}\] \[2T = 2007 \cdot 2^{2007}\] \[T = 2007 \cdot 2^{2006}\] Note that $2^{10} \equiv -1 \pmod{25}$. Using the properties of modular arithmetic, we can find the remainder when $T$ is divided by $25$. \[T \equiv 2007 \cdot 2^{2000} \cdot 2^6 \pmod{25}\] \[T \equiv 2007 \cdot (2^{10})^{200} \cdot 64 \pmod{25}\] \[T \equiv 7 \cdot 1 \cdot 14 \pmod{25}\] \[T \equiv 98 \pmod{25}\] \[T \equiv 23 \pmod{25}\] The remainder when Ted’s favorite number is divided by 25 is $\boxed{\textbf{(X) } 23}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 24 		Followed by:
Problem 26
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
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