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? | ||
− | <math> \ | + | <math> \textbf{(A) } 5\qquad \textbf{(B) } 7\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 19 </math> |
− | ==Solution== | + | ==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. | ||
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. | + | 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 | + | 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\ | + | Therefore the answer is <math>\boxed{\textbf{(D) }10}</math> |
+ | |||
+ | ==Solution 2== | ||
+ | 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. | ||
+ | |||
+ | From the problem, we have <math>10a+b-(a+b)=9a</math> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | So the answer is <math>\boxed{\textbf{(D) }10}</math>, since there are <math>10</math> numbers that have <math>a=4</math>. | ||
+ | |||
+ | ~BurpSuite | ||
==See Also== | ==See Also== | ||
− | |||
− | + | {{AMC10 box|year=2005|ab=A|num-b=15|num-a=17}} | |
− | + | [[Category:Introductory Number Theory Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 10:57, 17 July 2023
Contents
Problem
The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is . How many two-digit numbers have this property?
Solution 1
Let the number be where and are the tens and units digits of the number.
So must have a units digit of
This is only possible if , so is the only way this can be true.
So the numbers that have this property are .
Therefore the answer is
Solution 2
Let a two-digit number equal , where and are the tens and units digits of the number.
From the problem, we have
Now let , where and are the tens and units digits of the number. Then it must be that as stated in the problem.
Note that ends in , but ends in , so . We need not to care about , since it cancels out in the calculation.
So the answer is , since there are numbers that have .
~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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