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Difference between revisions of "2005 AMC 10A Problems/Problem 14"

(→‎Solution 3)
(→‎Solution 3)
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==Solution 3==
 
==Solution 3==
−
By listing the first two hundred numbers you'll see a pattern. The (-) is where the middle number should go.
+
By listing the first two hundred numbers you'll see a pattern. The <math>(-)</math> is where the middle number should go.
  
−
1-0 Can't do (1+0)/2=0.5
+
<math>1-0</math> Can't do <math>(1+0)/2=0.5</math>
  
−
1-1 Can do (1+1)/2=1
+
<math>1-1</math> Can do <math>(1+1)/2=1</math>
  
−
1-2 Can't do (1+2)=3/2
+
<math>1-2</math> Can't do <math>(1+2)=3/2</math>
  
−
1-3 Can do (1+3)/2=2
+
<math>1-3</math> Can do <math>(1+3)/2=2</math>
  
−
1-4 Can't do (1+4)=5/2
+
<math>1-4</math> Can't do <math>(1+4)=5/2</math>
  
−
1-5 Can do (1+5)/2=2
+
<math>1-5</math> Can do <math>(1+5)/2=3</math>
  
−
1-6 Can't do (1+6)=7/2
+
<math>1-6</math> Can't do <math>(1+6)=7/2</math>
  
−
1-7 Can do (1+7)/2=2
+
<math>1-7</math> Can do <math>(1+7)/2=2</math>
  
−
1-8 Can't do (1+8)=9/2
+
<math>1-8</math> Can't do <math>(1+8)=9/2</math>
  
−
1-9 Can do (1+9)/2=2
+
<math>1-9</math> Can do <math>(1+9)/2=2</math>
  
−
Total can do: 5
+
Total can do: <math>5</math>
  
−
2-0 Can't do (2+0)/2=0.5
+
<math>2-0</math> Can't do <math>(2+0)/2=0.5</math>
  
−
2-1 Can do (2+1)/2=1
+
<math>2-1</math> Can do <math>(2+1)/2=1</math>
  
−
2-2 Can't do (2+2)=3/2
+
<math>2-2</math> Can't do <math>(2+2)=3/2</math>
  
−
2-3 Can do (2+3)/2=2
+
<math>2-3</math> Can do <math>(2+3)/2=2</math>
  
−
2-4 Can't do (2+4)=5/2
+
<math>2-4</math> Can't do <math>(2+4)=5/2</math>
  
−
2-5 Can do (2+5)/2=2
+
<math>2-5</math> Can do <math>(2+5)/2=2</math>
  
−
2-6 Can't do (2+6)=7/2
+
<math>2-6</math> Can't do <math>(2+6)=7/2</math>
  
−
2-7 Can do (2+7)/2=2
+
<math>2-7</math> Can do <math>(2+7)/2=2</math>
  
−
2-8 Can't do (2+8)=9/2
+
<math>2-8</math> Can't do <math>(2+8)=9/2</math>
  
−
2-9 Can do (2+9)/2=2
+
<math>2-9</math> Can do <math>(2+9)/2=2</math>
  
−
Total can do: 5
+
Total can do: <math>5</math>
  
 
SUMMARY
 
SUMMARY
  
−
You can now see the pattern. There are 5 can do's each hundred. So then 9*5=<math>45</math>.  
+
You can now see the pattern. There are <math>5</math> can do's each hundred. So then <math>9*5= \boxed{45} \Rightarrow \mathrm{(E)}</math>.
 +
 
 
==See Also==
 
==See Also==
 
*[[2005 AMC 10A Problems]]
 
*[[2005 AMC 10A Problems]]

Revision as of 22:53, 3 January 2016

Problem

How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

$\mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45$

Solution

If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.

Doing some casework:

If the middle digit is $1$, possible numbers range from $111$ to $210$. So there are $2$ numbers in this case.

If the middle digit is $2$, possible numbers range from $123$ to $420$. So there are $4$ numbers in this case.

If the middle digit is $3$, possible numbers range from $135$ to $630$. So there are $6$ numbers in this case.

If the middle digit is $4$, possible numbers range from $147$ to $840$. So there are $8$ numbers in this case.

If the middle digit is $5$, possible numbers range from $159$ to $951$. So there are $9$ numbers in this case.

If the middle digit is $6$, possible numbers range from $369$ to $963$. So there are $7$ numbers in this case.

If the middle digit is $7$, possible numbers range from $579$ to $975$. So there are $5$ numbers in this case.

If the middle digit is $8$, possible numbers range from $789$ to $987$. So there are $3$ numbers in this case.

If the middle digit is $9$, the only possible number is $999$. So there is $1$ number in this case.

So the total number of three-digit numbers that satisfy the property is $2+4+6+8+9+7+5+3+1=45\Rightarrow E$

Solution 2 (much faster and slicker)

Alternatively, we could note that the middle digit is uniquely defined by the first and third digits, since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:

If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are $5\cdot5=25$ numbers in this case.

If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are $4\cdot5=20$ numbers here.

The total number, then, is $20+25=45\Rightarrow E$

Solution 3

By listing the first two hundred numbers you'll see a pattern. The $(-)$ is where the middle number should go.

$1-0$ Can't do $(1+0)/2=0.5$

$1-1$ Can do $(1+1)/2=1$

$1-2$ Can't do $(1+2)=3/2$

$1-3$ Can do $(1+3)/2=2$

$1-4$ Can't do $(1+4)=5/2$

$1-5$ Can do $(1+5)/2=3$

$1-6$ Can't do $(1+6)=7/2$

$1-7$ Can do $(1+7)/2=2$

$1-8$ Can't do $(1+8)=9/2$

$1-9$ Can do $(1+9)/2=2$

Total can do: $5$

$2-0$ Can't do $(2+0)/2=0.5$

$2-1$ Can do $(2+1)/2=1$

$2-2$ Can't do $(2+2)=3/2$

$2-3$ Can do $(2+3)/2=2$

$2-4$ Can't do $(2+4)=5/2$

$2-5$ Can do $(2+5)/2=2$

$2-6$ Can't do $(2+6)=7/2$

$2-7$ Can do $(2+7)/2=2$

$2-8$ Can't do $(2+8)=9/2$

$2-9$ Can do $(2+9)/2=2$

Total can do: $5$

SUMMARY

You can now see the pattern. There are $5$ can do's each hundred. So then $9*5= \boxed{45} \Rightarrow \mathrm{(E)}$.

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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