**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* ∈ {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*.

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 *x*^{2} + *bx* + *c* = 0. Condition for real roots is

*b*^{2} – 4*c* ≥ 0 or, *b*^{2} ≥ 4*c*.

Now, following cases need to consider:

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

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

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

Therefore, only possible value of *c* is 1.

**Case 3**. Let *b* = 3. Then 9 ≥ 4*c* 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 ≥ 4*c* 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 ≥ 4*c* 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 ≥ 4*c* 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.

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

Solutions are Elegant!

Thank you Sir

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