Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 1"
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== Solution == | == Solution == | ||
+ | Here, it is a slight trick. | ||
+ | |||
+ | 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 == |
Revision as of 18:24, 5 October 2015
Problem
How many positive -digit numbers
are there such that
For example,
and
have this property but
and
do not.
Solution
Here, it is a slight trick.
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.
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
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 |