If you distribute the modulus sign over the content in it, it's a mistake, i.e. ∣ a + b ∣ = ∣ a ∣ + ∣ b ∣ This is a big mistake , it's a false property !
But for how many pairs of integers ( a , b ) such that − 1 0 ≤ a ≤ 1 0 and − 1 0 ≤ b ≤ 1 0 , is the above said "false" property seen to be "true" ?
Note :- ∣ a ∣ means the absolute value of a .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
I wrote 242 because I did (11^2*2) and forgot that that way 0,0 appears twice...
Nice solution! Hahaha I tried without taking notes and my answer was 11 11 2=242 I included a pair of (0,0) xD I loved this problem
Not sure what you mean by arg ( x ) , it appears to be the angle of a complex number, but these numbers are not complex. The property is true when the pairs are ( 0 , 5 ) and ( 0 , − 5 ) , which I think contradicts your statement arg ( x ) = arg ( y ) , but I am not exactly sure what you mean by it, probably the sign of the number.
Log in to reply
Your assumption is right. Also, you're right, I forgot to mention the special case arg ( 0 ) , which to be honest I'm not entirely sure what it equals.
200 ways because order matters....
lol..so, the question is saying that you can be right in a total of 241 ways while being incorrect!
When a>0, b >= 0.........10 * 11 = 110 choices.
When a<0, b <= 0.........10 * 11 = 110 choices.
When a = 0, b has all.........21 = 21 choices....Total 241.
Daniel Liu probably has better shorter approach.
In order for the false property to be true, a number will have to be paired with a number of the same sign or with 0. First, we count the pairs including positive numbers or 0.
1 1 × 1 1 = 1 2 1
Next, we count the pairs including negative numbers or 0.
1 1 × 1 1 = 1 2 1
Now we sum the two.
1 2 1 + 1 2 1 = 2 4 2
However, we must recognize that we have counted the ordered pair ( 0 , 0 ) twice, so we must subtract 1 .
2 4 2 − 1 = 2 4 1
2 1 2 − 1 0 2 − 1 0 2 = 2 4 1
|a|+|b|=|a+b|
This is only possible if
1) Both a and b are positive
For a=0,1,2......10
b=0,1,2....10
therefore 121 possibilities
2) Both a and b are negative
Therefore possibilities=121 but since we have already counted a=0,b=0 once
Thus the total number of possibilities= 121+121-1=241
Problem Loading...
Note Loading...
Set Loading...
The property is false when arg ( x ) = arg ( y ) .
Since we are working in the reals, this happens when either x is positive and y is negative or x is negative and y is positive.
There are 1 0 negative numbers to choose from and 1 0 positive ones to choose from; thus, there is 1 0 0 ways to make the property false in each case giving 2 0 0 ways to make the property false.
There are 2 1 numbers total so there are 2 1 2 = 4 4 1 total different ways to pick x and y .
Thus, our answer is 4 4 1 − 2 0 0 = 2 4 1 .