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# Godly math problems

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6)Prove that 10002-999+9982-9972+...+22-12 is equal to the sum of the first thousand positive integers

7) suppose that p>q>0. sketch a square of side p and another square of side q, and give a pictorial demonstration of the identity p2-q2= (p+q)(p-q)

8) verify that 13! = 789122 -2882 without actually calculating the value of 13!

9) verify that 13! = 1122962-798962 without actually calculating the value of 13!

10) write 17! as the difference of two integer perfect squares

11) it is possible to arrive at the factorization of x2-y2 by the technique of adding in an extra term and subtracting it out again:

x2-y2=x2-xy+xy-y2

=x(x-y)+y(x-y)

=(x+y)(x-y)

apply this technique to determine a factorization (with integer exponents) of x-y3

Jan 1, 2019

#1
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6)  This forms an arithemetic series as follows:
First term =3994, Number of term =[10,002 - 12 / 10 + 1] =1,000 / 4 =250, Common difference =-16
Sum = N/2[2F + D(N-1)
Sum =250/2[2*3,994 + (-16*(249)
Sum = 125 [7,988 - 3,984]
Sum =500,500
Sum of 1 to 1,000 =[1,000 x 1001] / 2 =500,500

Jan 1, 2019
#2
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11.    x^3  -  y^3

x^3 - x^2 y + x^2y - xy^2 + xy^2 - y^3   =

x^2 ( x - y)  + xy (x - y) + y^2 (x - y) =

(x - y) ( x^2 + xy + y^2)

Jan 2, 2019
#3
+111325
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8) verify that 13! = 78912^2 -288^2 without actually calculating the value of 13!

78912^2 - 288^2  =

(78912 + 288) (78912 - 288)  =

(79200) (78624)

(2^5 * 3^2  * 5^2* 11 )  ( 2^5 * 3^3 * 7 * 13)

( 2  * 3 *  2^2 * 5 * ( 2  * 3)  *  2 * 5 * 11)  ( 2 * 2 * 2^3 * 3^2 * 3  * 7 * 13)  =

( 2 * 3 * 4 * 5 * 6 )  ( 2 *5* 11) ( 7 *  2^3 * 3^2 ) ( 2 * 2 * 3 * 13 )  =

(2 * 3 * 4 * 5 * 6 * 7 * 8 * 9)  (  (2 * 5) * 11 * ( 2 * 2 * 3) * 13 )  =

1 * (2 * 3 * 4 * 5 * 6 * 7 *8 * 9) ( 10 * 11 * 12 * 13 )   =   13!

Jan 2, 2019
#4
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I'm not the question asker, but I'm confused as to how you did question 8?

Jan 2, 2019