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

Revision as of 20:49, 1 August 2006

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

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

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 $2+4+6+8+9+7+5+3+1=45\Rightarrow E$

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