Difference between revisions of "2005 AMC 10A Problems/Problem 16"
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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> | <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. | ||
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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 | + | 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== | ||
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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}} | ||
| − | [[Category:Introductory | + | [[Category:Introductory Number Theory Problems]] |
| − | |||
{{MAA Notice}} | {{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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
