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 x^{2} + bx + c = 0. Let b, c ā {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 : R ā RĀ 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 āthe only way to learn mathematics is to do mathematicsā.
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 x^{2} + bx + c = 0. Condition for real roots is
b^{2} ā 4c ā„ 0 or, b^{2} ā„ 4c.
Now, following cases need to consider:
Case 1. Let b = 1. Then 1 ā„ 4c and c ā {1, 2, 3, 4, 5, 6}.
Therefore, it becomes evident that, b cannot be 1.
Case 2. Let b = 2. Then 4 ā„ 4c and c ā {1, 2, 3, 4, 5, 6}.
Therefore, only possible value of c is 1.
Case 3. Let b = 3. Then 9 ā„ 4c and c ā {1, 2, 3, 4, 5, 6}.
Therefore, only possible values of c are 1 and 2.
Case 4. Let b = 4. Then 14 ā„ 4c and c ā {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 ā„ 4c and c ā {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 ā„ 4c and c ā {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, y ā R 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 = Ļ/2 = y. From the given relation, we get
|f(Ļ/2 + Ļ/2) + sin (Ļ/2) + sin (Ļ/2)| < 2Ā Ā Ā Ā Ā Ā Ā Ā
or, |f(Ļ) + 2| < 2Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā
or, ā2 < f(Ļ) + 2 < 2
or, ā4 < f(Ļ) < 0
or, f(Ļ) < 0Ā Ā Ā Ā Ā (1)
Next, consider x = 3Ļ/2 and y = āĻ/2. From the given relation, we get
|f(3Ļ/2 āĻ/2) + sin (3Ļ/2) + sin (āĻ/2)| < 2Ā Ā Ā
or, |f(Ļ) ā1 ā 1| < 2Ā Ā Ā Ā
or, ā2 < f(Ļ) ā 2 < 2Ā Ā Ā
or, 0 < f(Ļ) < 4Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā
or, f(Ļ) > 0Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā (2)
Clearly, (1) and (2) contradicts each other.
Solutions are Elegant!
LikeLiked by 1 person
Thank you Sir
LikeLiked by 1 person
WONDERFUL Post.thanks for share..more wait .. ā¦
LikeLiked by 1 person
Fantastic blog you have here but I was curious if you knew of any community forums that cover the same topics talked about here? I’d really love to be a part of group where I can get feed-back from other knowledgeable individuals that share the same interest. If you have any recommendations, please let me know. Thank you!
LikeLiked by 1 person
Thank you so much for your support. Please join the Facebook page Math1089 and Facebook group Math1089-Mathematics for All.
LikeLiked by 1 person