Messages posted by sosgf

If the integers from 1 to 222,222,222 are written down in succession, how many

of them have at least one zero?

2^4 = 16 = -1 mod(17 )

(2^4)^64=(-1)^64mod(17) = 1mod(17)

so

remainder =1

Please I want an explanation for a solution of the theory of numbers

What is the 2009th digit in this sequence?

1234567891011121314151617............

Which day of the week will Valentine’s Day be in 2048?

a_n = ( 3^(n+1) - 2 ) /3^n= 3- 2(1/3)^n

sum of GM from 1 to 2n ......... esy

no of 2,s in 1000!

= [1000/2] + [1000/4] + ...+ [1000/2^9]+ 0= 994

no of 5,s in 1000!

= [1000/5] + [1000/25]+ [1000/125]+ [1000/625]+ 0= 249

so

1000!= 2^994 * 5^249* k = 2^745* 10^249 * k

so

n = 249

no. of 2,s in 1000! = [ 1000/2] + [1000/4]+ [1000/8]+ ...+[1000/512]

= 500+ 250 + 125+ 62+ 31 + 15+ 7+ 3 + 1 = 994

so

1000!=( 2^994)*I ----> n = 994

x= 2sexu

I=4int(sec^3u ) -4 int(secu)

d(p^7) = (7+1) = 8

d(8) = d(2^3) = (3+1) =4

d(4) = d(2^2) = (2+1)=3

so

d(d(d(p^7)))=3

For any positive constant

c, find the maximum value of xy(c − x − y) over all positivex and y.

numbers

use AM-GM x/y + 3y/z + 9z/x >= 3 (27)^(1/3 ) = 9

All choices are true !!!!Options missing???

Find the least value of :

x/y + 3y/z + 9z/x

over all positive real numbers

x, y,and z.

MR.Yagyadutt Mishra

You created a total of the border indefinitely, while the required total N limit

Please do correct my answer

T1= a

T2 = a+ d

T7 = a+ 6d -----> d= ( 9-a )/6

T1 T2 T3 = a( a+ d ) ( a+ 6d) = a(a + (9 - a )/6) (9) = (3( 10a + 9)^2 )/40 - 243/40 >= - 243/40

so least value = -243/40

and

(3( 10a + 9)^2 )/40 = 0 ----> a = - 9/10 and d = 33/20

D) 64

-729/10

number of words are formed from word SUCCESS such that two C and two S are together.=5!

number of words are formed from word SUCCESS = 7!/(2!*2!)

number of words are formed from word SUCCESS such that no two C and no two S are together = 7!/(2!*2!) - 5!

=1140

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