Week 19

Example 1. Doing only one multiplication, prove that

(666)(222) + (1)(333) + (333)(222) + (666)(333) + (1)(445) + (333)(333) +(666)(445) + (333)(445) + (1)(222) = 1000000.

Example 2. Consider the equation of the form x2 + bx + c = 0. Let b, c ∈ {1, 2, 3, 4, 5, 6}, where b may be equal to c. Find the number of such equations that have real roots and have coefficients b and c.

Example 3. How many functions f : RR  are there satisfying the inequality |f(x + y) + sin x + sin y| < 2?

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Solution 1. Though the given problem looks wired, we can solve it by doing only one multiplication. The idea actually depends on the rearrangement of the terms.

(666)(222) + (1)(333) + (333)(222) + (666)(333) + (1)(445) + (333)(333) +(666)(445) + (333)(445) + (1)(222)

= {(666)(222) + (666)(333) + (666)(445)} + {(333)(222) + (333)(333) + (333)(445)} + {(1)(222) + (1)(333) + (1)(445)}

= 666{222 + 333 + 445} + 333{222 + 333 + 445} + 1{222 + 333 + 445}

= (666 + 333 + 1)(222 + 333 + 445)

= (1000)(1000)

=1000000.

Solution 2. Given equation is x2 + bx + c = 0. Condition for real roots is

b2 – 4c ≥ 0       or, b2 ≥ 4c.

Now, following cases need to consider:

Case 1. Let b = 1. Then 1 ≥ 4c and c ∈ {1, 2, 3, 4, 5, 6}.

Therefore, it becomes evident that, b cannot be 1.

Case 2. Let b = 2. Then 4 ≥ 4c and c ∈ {1, 2, 3, 4, 5, 6}.

Therefore, only possible value of c is 1.

Case 3. Let b = 3. Then 9 ≥ 4c and c ∈ {1, 2, 3, 4, 5, 6}.

Therefore, only possible values of c are 1 and 2.

Case 4. Let b = 4. Then 14 ≥ 4c and c ∈ {1, 2, 3, 4, 5, 6}.

Therefore, only possible values of c are 1, 2, 3 and 4.

Case 5. Let b = 5. Then 25 ≥ 4c and c ∈ {1, 2, 3, 4, 5, 6}.

Therefore, only possible values of c are 1, 2, 3, 4, 5 and 6.

Case 6. Let b = 6. Then 36 ≥ 4c and c ∈ {1, 2, 3, 4, 5, 6}.

Therefore, only possible values of c are 1, 2, 3, 4, 5 and 6.

Therefore, total number of equations = 1 + 2 + 4 + 6 + 6 = 19.

Solution 3. We will show that, there is no such function f for all x, yR exists satisfying the given conditions.

To show this, we need to consider a set of values of x, y for which the given condition fail to happen.

First, consider x = π/2 = y. From the given relation, we get

|f(π/2 + π/2) + sin (π/2) + sin (π/2)| < 2        

or, |f(π) + 2| < 2                      

or, −2 < f(π) + 2 < 2

or, −4 < f(π) < 0

or, f(π) < 0      (1)

Next, consider x = 3π/2 and y = −π/2. From the given relation, we get

|f(3π/2 −π/2) + sin (3π/2) + sin (−π/2)| < 2   

or, |f(π) −1 − 1| < 2    

or, −2 < f(π) – 2 < 2   

or, 0 < f(π) < 4           

or, f(π) > 0                  (2)

Clearly, (1) and (2) contradicts each other.

5 comments

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    Liked by 1 person

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