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Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 1"

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== Solution ==
 
== Solution ==
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Here, it is a slight trick.
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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>.
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 +
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.
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 +
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.
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<cmath>\dots</cmath>
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If, we place <math>a=9</math>, we have the only choice <math>b=0</math>, in <math>2</math> ways.
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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 ==

Revision as of 17:24, 5 October 2015

Problem

How many positive $3$-digit numbers $abc$ are there such that $a+b=c$ For example, $202$ and $178$ have this property but $245$ and $317$ do not.


Solution

Here, it is a slight trick.

Here, we need to find $a,b\in \Bbb N_0$ such that $1\le a\le 9$ , $0\le b\le 9$ and $a+b=c$ where $c\in \Bbb N, c\le 9$.

If we place $a=1$, then we can place $0,1,2,3,4,5,6,7,8$ as $b$, i.e. in $9$ ways.

Similarly, if we place $a=2$, we can place $b=0,1,2,3,4,5,6,7$ i.e. in $8$ ways.

\[\dots\]

If, we place $a=9$, we have the only choice $b=0$, in $2$ 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 \[\color{red}{1+2+3+4+5+6+7+8+9=\frac {9\times 10}{2}}=\color{blue}{45}\]

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
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