Mistakes give rise to Problems- 4

Algebra Level 5

If you distribute the modulus \textbf{modulus} sign over the content in it, it's a mistake, i.e. a + b = a + b \left| a+b \right| = \left| a \right| + \left| b \right| This is a big mistake \color{#D61F06}{\textbf{big mistake}} , it's a false property \color{#3D99F6}{\textbf{false property}} !


But for how many pairs of integers ( a , b ) (a,b) such that 10 a 10 -10 \leq a \leq 10 and 10 b 10 -10 \leq b \leq 10 , is the above said "false" property seen to be "true" ?


Note :- a \left| a \right| means the absolute value of a a .


The answer is 241.

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6 solutions

Daniel Liu
Jul 24, 2014

The property is false when arg ( x ) arg ( y ) \text{arg}(x)\ne \text{arg}(y) .

Since we are working in the reals, this happens when either x x is positive and y y is negative or x x is negative and y y is positive.

There are 10 10 negative numbers to choose from and 10 10 positive ones to choose from; thus, there is 100 100 ways to make the property false in each case giving 200 200 ways to make the property false.

There are 21 21 numbers total so there are 2 1 2 = 441 21^2=441 total different ways to pick x x and y y .

Thus, our answer is 441 200 = 241 441-200=\boxed{241} .

I wrote 242 because I did (11^2*2) and forgot that that way 0,0 appears twice...

Shoham Grunblat - 6 years, 10 months ago

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

Carlos David Nexans - 6 years, 10 months ago

Not sure what you mean by arg ( x ) \text{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 ) (0,5) and ( 0 , 5 ) (0,-5) , which I think contradicts your statement arg ( x ) = arg ( y ) \text{arg}(x)=\text{arg}(y) , but I am not exactly sure what you mean by it, probably the sign of the number.

mathh mathh - 6 years, 10 months ago

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Your assumption is right. Also, you're right, I forgot to mention the special case arg ( 0 ) \text{arg}(0) , which to be honest I'm not entirely sure what it equals.

Daniel Liu - 5 years, 11 months ago

200 ways because order matters....

Yuliya Skripchenko - 6 years, 10 months ago

lol..so, the question is saying that you can be right in a total of 241 ways while being incorrect!

Yoogottam Khandelwal - 5 years, 11 months ago

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.


Lukas Leibfried
Dec 13, 2015

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.

11 × 11 = 121 11 \times 11 = 121

Next, we count the pairs including negative numbers or 0.

11 × 11 = 121 11 \times 11 = 121

Now we sum the two.

121 + 121 = 242 121 + 121 = 242

However, we must recognize that we have counted the ordered pair ( 0 , 0 ) (0,0) twice, so we must subtract 1 1 .

242 1 = 241 242 - 1 = \boxed{241}

Lu Chee Ket
Sep 27, 2015

2 (11)^2 - 1 = 241

Andrea Palma
Jul 19, 2015

2 1 2 1 0 2 1 0 2 = 241 21^2 - 10^2 - 10^2 = 241

Aayush Patni
Jan 7, 2015

|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

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