Quiz 1(B)

Infinite..
c=4 => x=1,4
c=-6 => x=-1,6
c=-14 => x=-2,7
c=-24 => x=-3,8
c=-36 => x=9,-4
c=-50 => x=-5,10
c=-55 => x=-6,11 and so on...

Additionally, For roots of the equation X^2-5X+C to be
rational integers we need to calculate its
DISCRIMINANT (Dis) which should be greater than or equal to 0.
i.e.

Dis=b^2-4ac
Dis>=0

here b=-5 , a=1 , c=C

hence Dis=(-5)^2 -4*1*C = 25-4C
therefore we can write the inequality as

25-4C>=0

or
25>=4C

there are infinite values of C ranging from -infinity to +6

(ab+bc+ca)/abc=1/(a+b+c)
=> abc=a2b+ab2+ac2+a2c+b2c+bc2+3abc

=> a2b+ab2+ac2+a2c+b2c+bc2+2abc=0

(a+b)(b+c)(c+a)=(ab+ac+b2+bc)(c+a)

=abc+ac2+b2c+bc2+a2c+a2b+b2a+abc
=0

This mathematical problem can be explained well if we use
a basic formula which helps to find out the the No. of Divisors
or Factors of a particular Numbers it has, it generally
goes like this.

If N is a Natural number, Prime Factorization of N = a^x * b^y .
Then the No. of divisors = (x+1)*(y+1)

As the Natural number has exactly 10 divisors & 10 = 5 * 2
So possible value of x = 4 & y = 1 or x = 9 .
--------------------------------------------------------------------------------
Example
if 512 = 2^9 , total No. of factor = (x+1) i.e (9+1) = 10
or
if 48 = 2^4*3^1 , total No. of factor = (x+1)*(y+1) i.e (4+1)*(1+1) = 10
or
if 19683 = 3^9 , total No. of factor = (x+1) i.e (9+1) = 10
or
if 162 = 2^1*3^4, total No. of factor = (x+1)*(y+1) i.e (4+1)*(1+1) = 10
--------------------------------------------------------------------------------
Thus Answer = Either 1 or 2 distinct prime factors.

x= 444444..... / 18
x= 22222..... / 9
the sum of the digits should be divisible by 9.
so 2+2+2....(9times) =18 which is divisible by 9.
(Since 2 is less than 9 therefore, 22 will be the first no. to be divided by 9. Thus deducting one 2 from 9
2s. i.e 9-1=8 )
so there are 8 minimum digits.
and the number will come out to be 24691358.
24691358 * 18 = 444444444.

260*1024+73*512+129*128+90->2
260*128+73*64+129*16+11->3
260*16+73*8+259->32
625->1
78->6
9->1
1->1

reading from down to top 1161332.
so there are two 3\\\'s.

Or

Given,
Z=260*1024+73*512+128*129+81+9(which is in decimal form)

Convert each number in to octal form by placing remainders......
Let\\\'s take for (260)base to the 10 = (404)base to the 8
Similarily 1024= 2000
......

Finally octal form of z is
Z=404*2000+111*1000+200*201+121+11

---111=3*37 && 201=3*67

Z=404*2000+(3*37)*1000+200*(3*67)+121+11

How many times the digit 3 is repeated.......
2 times

99 is smallest value of \\\'a\\\'.

a*47*64*1313 = a*47*8*8*13*101
117 =13*9
88=11*8

so \\\'a \\\' should be 11*9 = 99

x(sqrt(x+5))-sqrt(x+5)=0
(x-1)(sqrt(x+5)=0
so,
x-1=0 or x+5=0;
so,
x=1 or -5

nswer must be both negative (A)
as m+n= -q
and m*n = r

if m and n are of opposite sign, then m*n must be negative
But r is a positive number...

if both m and n are positive, then m+n must be positive..
but m+n is negative here..

which shows that m and n both are negative...

11 will be ans.
99^1=9801
99^2=971299
99^3=96157601
like this......
take only first two digit for n=1 99
n=2 98
n=3 97
n=4 96
......................................................
n=10 90
n=11 89
so for n=11 number start with 8

or

Answer will be 11
because 99^1=99
99^1=99
99^2=9801
99^3=97.....
99^4=96.......
99^5=95......
99^6=94.......
99^7=93.........
99^8=92....
99^9=91.....
99^10=90.......
99^11=89..........
at 99^11=89.......
So answer is n=11

let the digits be x and y. then the number is 10x+y
given:
(10x+y)^2/[(10x+y)/2]=2(10x+y).
[2(10x+y)+36]/2=10y+x
or, 10x+y+18=10y+x
or, 9y=9x+18
or, y=x+2
therefore, 10x+y=13,24,35,46,57,68,79