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

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The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is <math>6</math>. How many two-digit numbers have this property?  
 
The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is <math>6</math>. How many two-digit numbers have this property?  
  
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<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 19 </math>
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<math> \textbf{(A) } 5\qquad \textbf{(B) } 7\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 19 </math>
  
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==Solution==
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==Solution 1==
 
Let the number be <math>10a+b</math> where <math>a</math> and <math>b</math> are the tens and units digits of the number.  
 
Let the number be <math>10a+b</math> where <math>a</math> and <math>b</math> are the tens and units digits of the number.  
  
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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.  
  
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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>.  
  
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Therefore the answer is <math>10\Rightarrow D</math>
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Therefore the answer is <math>\boxed{\textbf{(D) }10}</math>
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==Solution 2==
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Let a two-digit number equal <math>10a+b</math>, where <math>a</math> and <math>b</math> are the tens and units digits of the number.
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From the problem, we have <math>10a+b-(a+b)=9a</math>
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Now let <math>9a=10x+y</math>, where <math>x</math> and <math>y</math> are the tens and units digits of the number. Then it must be that <math>y=6</math> as stated in the problem.
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Note that <math>10a</math> ends in <math>0</math>, but <math>9a</math> ends in <math>6</math>, so <math>a=4</math>. We need not to care about <math>b</math>, since it cancels out in the calculation.
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So the answer is <math>\boxed{\textbf{(D) }10}</math>, since there are <math>10</math> numbers that have <math>a=4</math>.
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~BurpSuite
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2005|ab=A|num-b=15|num-a=17}}
 
{{AMC10 box|year=2005|ab=A|num-b=15|num-a=17}}
  
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[[Category:Introductory Geometry Problems]]
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[[Category:Introductory Number Theory Problems]]
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[[Category:Area Ratio Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:57, 17 July 2023

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 1

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

Solution 2

Let a two-digit number equal $10a+b$, where $a$ and $b$ are the tens and units digits of the number.

From the problem, we have $10a+b-(a+b)=9a$

Now let $9a=10x+y$, where $x$ and $y$ are the tens and units digits of the number. Then it must be that $y=6$ as stated in the problem.

Note that $10a$ ends in $0$, but $9a$ ends in $6$, so $a=4$. We need not to care about $b$, since it cancels out in the calculation.

So the answer is $\boxed{\textbf{(D) }10}$, since there are $10$ numbers that have $a=4$.

~BurpSuite

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
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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