Difference between revisions of "2005 AMC 10A Problems/Problem 16"

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So <math>(10a+b)-(a+b)=9a</math> must have a units digit of <math>6</math>  
 
So <math>(10a+b)-(a+b)=9a</math> must have a units digit of <math>6</math>  
  
<math>a=4</math> is the only way this can be true.  
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This is only possible if <math>9a=36</math>, so <math>a=4</math> is the only way this can be true.  
  
 
So the numbers that have this property are <math>40</math>, <math>41</math>, <math>42</math>, <math>43</math>, <math>44</math>, <math>45</math>, <math>46</math>, <math>47</math>, <math>48</math>, <math>49</math>.  
 
So the numbers that have this property are <math>40</math>, <math>41</math>, <math>42</math>, <math>43</math>, <math>44</math>, <math>45</math>, <math>46</math>, <math>47</math>, <math>48</math>, <math>49</math>.  
  
Therefore the answer is <math>10\Rightarrow D</math>  
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Therefore the answer is <math>10\Rightarrow D</math>
  
 
==See Also==
 
==See Also==

Revision as of 21:36, 21 May 2013

Problem

The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is $6$. How many two-digit numbers have this property?

$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 19$

Solution

Let the number be $10a+b$ where $a$ and $b$ are the tens and units digits of the number.

So $(10a+b)-(a+b)=9a$ must have a units digit of $6$

This is only possible if $9a=36$, so $a=4$ is the only way this can be true.

So the numbers that have this property are $40$, $41$, $42$, $43$, $44$, $45$, $46$, $47$, $48$, $49$.

Therefore the answer is $10\Rightarrow D$

See Also