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

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==Solution 1==
 
==Solution 1==
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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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Let the number be <math>10a+b</math>, where <math>a</math> is its tens digit and <math>b</math> is its units digit. Then <math>(10a+b)-(a+b) = 9a</math> must have a units digit of <math>6</math>, and as <math>a</math> is the tens digit, we can only have <math>1 \leq a \leq 9</math>, so <math>9 \leq 9a \leq 81</math>.
  
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So <math>(10a+b)-(a+b)=9a</math> must have a units digit of <math>6</math>  
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Accordingly, <math>9a</math> has units digit <math>6</math> only if <math>9a = 36 \iff a=4</math>. Thus the numbers that have the required property are all those with tens digit <math>4</math>, from <math>40</math> to <math>49</math>, so the answer is <math>49-40+1 = \boxed{\textbf{(D) } 10}</math>.
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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.  
 
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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>\boxed{\textbf{(D) }10}</math>
 
  
 
==Solution 2==
 
==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.
+
As in Solution 1, suppose that <math>a</math> and <math>b</math> are the tens and units digits of the number respectively, so the result of the subtraction is <math>10a+b-(a+b)=9a</math>. Thus <math>9a</math> must have units digit <math>6</math>.
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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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We now observe that <math>10a</math> is a multiple of 10, so has units digit <math>0</math>, and hence <math>10a-9a = a</math> will have units digit <math>10-6 = 4</math> (where the <math>0</math> will become <math>10</math> in the subtraction by 'borrowing' <math>1</math> from the tens digit). Since <math>a</math> is a single digit, this simply means <math>a = 4</math>, while <math>b</math> can be any digit from <math>0</math> to <math>9</math> (since it cancelled out in the subtraction above).
  
−
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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Thus, as in Solution 1, there are <math>10</math> possible choices for <math>b</math> with <math>a = 4</math>, so the answer is <math>\boxed{\textbf{(D) } 10}</math>.
  
 
~BurpSuite
 
~BurpSuite

Latest revision as of 18:00, 1 July 2025

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 $a$ is its tens digit and $b$ is its units digit. Then $(10a+b)-(a+b) = 9a$ must have a units digit of $6$, and as $a$ is the tens digit, we can only have $1 \leq a \leq 9$, so $9 \leq 9a \leq 81$.

Accordingly, $9a$ has units digit $6$ only if $9a = 36 \iff a=4$. Thus the numbers that have the required property are all those with tens digit $4$, from $40$ to $49$, so the answer is $49-40+1 = \boxed{\textbf{(D) } 10}$.

Solution 2

As in Solution 1, suppose that $a$ and $b$ are the tens and units digits of the number respectively, so the result of the subtraction is $10a+b-(a+b)=9a$. Thus $9a$ must have units digit $6$.

We now observe that $10a$ is a multiple of 10, so has units digit $0$, and hence $10a-9a = a$ will have units digit $10-6 = 4$ (where the $0$ will become $10$ in the subtraction by 'borrowing' $1$ from the tens digit). Since $a$ is a single digit, this simply means $a = 4$, while $b$ can be any digit from $0$ to $9$ (since it cancelled out in the subtraction above).

Thus, as in Solution 1, there are $10$ possible choices for $b$ with $a = 4$, so the answer is $\boxed{\textbf{(D) } 10}$.

~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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