2005 AMC 10A Problems/Problem 14

Revision as of 16:51, 5 November 2006 by Xantos C. Guin (talk | contribs) (added category)

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 $9$, possible numbers range from $999$ to $999$. So there are $1$ numbers in this case.

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

See Also

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