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A-probability of getting red
B- probability of getting 6 or 7 or 8 or 9
P(A union B)=P(A)+P(B)-P(A intersect B)
P(A)=1/2
P(B)=16/52=4/13
Intersecct=P(A/B)*P(B)
1/2*16/52=2/13
so answer = 1/2+4/13-2/13
=17/26

are u sure the answer is 19/26 or is it 17/26

elastic and jagery are exactly correct the answer is 225

never worry ..enjoy reading

actually there is no restriction on the number of digits
reason:
divisibility by 11 :abcde(a,b,c,d,e are digits)
condition is (a+c+e)-(b+d)=11k where k is whole number
that is it can be equal to 0 or 11 or 22 etc
now consider palindromes
2 digit number (11 itself obviously is divisible)
3 digits 121
4 digits 1001
5 digits 11011
....
so it is so easy to generate a palindrome divisible by 11
technique is
even no of digits means put 1st and last digit as 1 rest as 0
odd no of digits means 1st 2 digits and last 2 digits as 1 and rest as 0
this ofcourse can include few of the possible palindromes
rest can be derived similarly
so there is no restriction on the no of digits

well done elastic,nudge if useful rate if useless

sin(a+45)=-pi/4
expand using formula
sina+cosa=-pi/4*2^1/2
squaring
sin2a=pi^2/8 - 1
now writing sin2a as 2 sina cosa and then cosa as root of 1-sin^2a
we get fourth degree eqn in sina
solving which we get sina =root of (8+ or - pi root of 16-pi^2)/16

I had given a detailed answer when u asked with respect to 92! question

19^92 = 19*19*19.....92 times
which is an odd number

any multiple of 92 is even because it contains a factor of 2 in it
so obviously 19^92 not divisible by 92

consider this way
1111(4 1's) divided by 270 Rem = 270
11111(5 1's) Rem = 0
111111(6 1's ) Rem= 1
(7 1's) Rem = 11
for 8 Rem = 111
for 9 Rem = 1111
for10Rem = 0
so similarly ....
for 15 Rem = 0
for 20 Rem = 0
.....
for 1110 Rem =0
so by pattern for 1111 Rem =1
that is 1111......(1111 times)
divided by 271 leaves 1 as remainder

I didnt read the first line my friends

so its my miss

anyway as far as i know mr feynmann is right

f(1)<0

f(2)>0

and by checking up we find only one root (real)

so i think there is only 1 real root

pls correct if wrong

a^2-b^2 (a-b)(a+b)is divisible by a-b

a^3-b^3 (a-b)(a^2+ab+b^2)is divisible by a-b

similarly a^4-b^4 is also divisible by a-b

it can be thus shown that a^n-b^n is always divisble by a-b for natural number n

thus a^1992-b^1992/a-b is an integer

similarly the other two

so int+int+int which is also an integer

sorry for being hasty
i didnt notice the square in the question
next time type the word square if u dont have special editor
anyway using same technique
answer is
-1/6*(1+1)
=-1/3

lim(x-0) [1/x-1/sinx]
=(sinx-x)/xsinx
consider sinx-x/x cubed
by applying l'hospital's rule we get value to b -1/6(when x tends to 0)
so given = (sinx-x)/x cubed * x square /sinx
so answer is 0

Could u retype the last question u have posted
there seem to be something missing in that