Monthly Archives: May 2012

Problem Of the Day-2


Here is Yesterday Problem

Yesterday Winner was Akash Agarwal

You must be knowing that If you win the Problem of the day for 4 times in a week – then I will send you a Really interesting book on any topic you want 🙂

Today’s Problem  :

Find all possible values of x satisfying :

[x]/[x-2] – [x-2]/[x] = (8{x} + 12)/([x-2][x])

(.) = Normal bracket
{.} = Fractional part function
[.] = GIF/Floor function

Modular Arithmetic


Modular arithmetic is a notation and set of mathematics that were first introduced by Carl Friedrich Gauss. 🙂

The major insight is that equations can fruitfully be analyzed from the perspective of remainders. Standard equations use the ‘=’ sign. Modular arithmetic uses the ‘‘ sign. Two values that are ‘≡’ to each other are said to be congruent relative the modulus. In the case below, the modulus is 3.  🙂

Here’s an example of a modular equation:

7 ≡ 1 (mod 3).

By definition, this means that 3 divides 7 – 1.

Definition 1: a ≡ b (mod c) if and only if c divides a – b.

This definition tells us the following is true:

7 ≡ 1 ≡ 10 ≡ -2 (mod 3).

Now, one of the most interesting things about ‘≡’ is that it follows many of the same relations as ‘=’ .  🙂 🙂

Notice 1: For any value a,b,c,d,n where a ≡ b (mod n) and c ≡ d (mod n):  🙂

(a) a + c ≡ b + d (mod n)
 Proof :  We know that n divides (a + c) – (b + d) since this is equal to: (a -b) + (c – d).

(b) a – c ≡ b – d (mod n)
Proof : We know that n divides (a – c) – (b – d) since this is equal to: (a – b) – (c – d).

(c) ac ≡ bd (mod n)
Proof : We know that n divides ac – bd since this is equal to : c(a – b) + b(c – d).

(Q.E.D) 

Notice  2: If a ≡ b (mod n) then:

(a) a + c ≡ b + c (mod n)
Proof : We know (a) since n divides a + c – (b + c) = a – b.

(b) a – c ≡ b – c (mod n)
 Proof : We know (b) since n divides a – c – (b – c) = a – b.

(c) ac ≡ bc (mod n)
Proof: We know (c) since n divides ac – bc = c(a – b)

(Q.E.D)

Corrolary 2.1: a ≡ d (mod n), b ≡ e (mod n), c ≡ f (mod n), then:
a + b + c ≡ d + e + f  mod n 

(1) We know that a + b ≡ d + e from above.
(2) We therefore know that (a + b) + c ≡ (d + e) + f.

(Q.E.D)

Notice  3: a + b + c ≡ 0, a ≡ 0 (mod p), then b + c ≡ 0 (mod p).

(1) a + b + c ≡ 0 (mod p) [Definition of ≡ ]

(2) b ≡ c (mod p) → a + b ≡ a + c (mod p) [See above]

(3) So, 0 ≡ a + b + c ≡ 0 + b + c ≡ b + c (mod p).

(Q.E.D)

I will cover modular arithmetic in depth 🙂 🙂
Today you must we wondering about its application ..
But after whole course You will surely love this topic
Thank you 🙂 🙂

If any doubts , then dont hesitate in asking Your doubts 🙂 🙂
I will surely love to solve your doubts

Problem of the day !


 We have started one more interesting feature of the blog 
I will be posting problem of the day , and the best answer will be highly appreciated 

If you win problem of the day for more than 4 times in a  week then I will send you a mathematics very interesting E-Book 

So here is Question 1 )

QUESTION 1 )

How many digits does the number 21000 contain?

NOTE : 

You have to give solution of your answer too 

Important theorems of Geometry(Triangles) – 1


Today we will be discussing about  2 important Triangle theorems which have very High applications in field of Geometry

1) Angle Bisector theorem
2) Stewart’s Theorem

Angle bisector theorem :

Introduction :

The Angle Bisector Theorem states that given triangle \triangle ABC and angle bisector AD, where D is on side BC, then c/m= b/n . Likewise, the converse of this theorem holds as well.

size(200);defaultpen(fontsize(10));real a,b,c,d;pair A=(1,4), B=(-5,0), C=(3,0), D;b = abs(C-A);c = abs(B-A);D = (b*B+c*C)/(b...

Proof :

Because of the ratios and equal angles in the theorem, we think of similar triangles. There are not any similar triangles in the figure as it now stands, however. So, we think to draw in a carefully chosen line or two. Extending AD until it hits the line through C parallel to AB does just the trick:

size(200);defaultpen(fontsize(10));real a,b,c,d,m,n;pair A=(1,4), B=(-5,0), C=(3,0), D, E;b = abs(C-A);c = abs(B-A);D = (b*B+...

Since AB and CE are parallel, we know that \angle BAE=\angle CEA and \angle BCE=\angle ABC. Triangle ACE is isosceles, with AC = CE.

By AA similarity\triangle DAB \cong \triangle DEC. By the properties of similar triangles, we arrive at our desired result:

c/m = b/n

Stewarts Theorem :

Introduction :

Given a triangle \triangle ABC with sides of length a, b, c opposite vertices ABC, respectively. If cevian AD is drawn so that BD = mDC = n and AD = d, we have that b^2m + c^2n = amn + d^2a. (This is also often written man + dad = bmb + cnc, a form which invites mnemonic memorization, e.g. “A man and his dad put a bomb in the sink.”)

Stewart's theorem.png

Proof :

Applying the Law of Cosines in triangle \triangle ABD at angle \angle ADB and in triangle \triangle ACD at angle \angle CDA, we get the equations

  • n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2}
  • m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2}

Because angles \angle ADB and \angle CDA are supplementarym\angle ADB = 180^\circ - m\angle CDA. We can therefore solve both equations for the cosine term. Using the trigonometric identity  \cos{\theta} = -\cos{(180^\circ - \theta)} gives us

  • \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}
  • \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n. However, m+n = a som^2n + n^2m = (m + n)mn and we can rewrite this as man + dad= bmb + cnc (A man and his dad put a bomb in the sink).
Thank you

Source :
Wikepedia
Art of problem solving
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