Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 1"
(Created page with "== Problem == How many positive <math>3</math>-digit numbers <math>abc</math> are there such that <math>a+b=c</math> For example, <math>202</math> and <math>178</math> have this...") |
Mathisfun04 (talk | contribs) (→Solution) |
||
| (3 intermediate revisions by 2 users not shown) | |||
| Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
| + | |||
| + | Here, we need to find <math>a,b\in \Bbb N_0</math> such that <math>1\le a\le 9</math> , <math>0\le b\le 9</math> and <math>a+b=c</math> where <math>c\in \Bbb N, c\le 9</math>. | ||
| + | |||
| + | If we place <math>a=1</math>, then we can place <math>0,1,2,3,4,5,6,7,8</math> as <math>b</math>, i.e. in <math>9</math> ways. | ||
| + | |||
| + | Similarly, if we place <math>a=2</math>, we can place <math>b=0,1,2,3,4,5,6,7</math> i.e. in <math>8</math> ways. | ||
| + | |||
| + | <cmath>\dots</cmath> | ||
| + | |||
| + | If, we place <math>a=9</math>, we have the only choice <math>b=0</math>, in <math>2</math> ways. | ||
| + | |||
| + | So, in order to get the number of possibilities, we have to add the no. of all the possibilities we got, i.e. the answer is <cmath>\color{red}{1+2+3+4+5+6+7+8+9=\frac {9\times 10}{2}}=\color{blue}{45}</cmath> | ||
== See also == | == See also == | ||
| − | {{ | + | {{UNCO Math Contest box|year=2009|n=II|before=First Question|num-a=2}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 09:45, 26 November 2016
Problem
How many positive 
-digit numbers 
are there such that 
For example, 
and 
have this property but 
and 
do not.
Solution
Here, we need to find 
such that 
, 
and where 
.
If we place 
, then we can place 
as 
, i.e. in 
ways.
Similarly, if we place 
, we can place 
i.e. in 
ways.
![\[\dots\]](http://latex.artofproblemsolving.com/1/6/7/16762e47f22f53e2431935f211fa82570b46b6cb.png)
If, we place 
, we have the only choice 
, in 
ways.
So, in order to get the number of possibilities, we have to add the no. of all the possibilities we got, i.e. the answer is ![\[\color{red}{1+2+3+4+5+6+7+8+9=\frac {9\times 10}{2}}=\color{blue}{45}\]](http://latex.artofproblemsolving.com/d/3/f/d3f93e37c8802607aedddfdcb9905f32e8f74b00.png)
See also
| 2009 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
