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

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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.  
 
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>.  
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So the numbers that have this property are <math>40, 41, 42, 43, 44, 45, 46, 47, 48, 49</math>.  
  
Therefore the answer is <math>10\Rightarrow</math> <math>(D)</math>
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Therefore the answer is <math>\boxed{\textbf{(D) }10}</math>
  
 
==See Also==
 
==See Also==

Revision as of 18:31, 13 December 2021

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?

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

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 $\boxed{\textbf{(D) }10}$

See Also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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
All AMC 10 Problems and Solutions

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