likithae

The sum of a two-digit number and the number obtained by reversing the digit is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? explain me how to solve it.........

i think consecutive integers are n,n+1,n+2....... for example 1,2,3,.......

okay.thank you,this only shortcut method or if any..

what is OA?

If the lines 2x-3y=0 and 3x-4y=7 are two diameters of a circle of radius 7, then the equation of the circle is (a)x^2+y^2+2x-4y-47=0 (b)x^2+y^2=49 ©x^2+y^2-2x+2y-47=0 (d)x^2+y^2=17

thank you very much. now i understood how to solve it.

In terms of prime factors 100! can be written as 2^1.3^b.5^c.7^d.......... Now, E_3(100 !)=[100/3]+[100/3^2]+[100/3^3]+[100/3^4] =33+11+3+1=48. Hence the exponent of 3 in 100! is 48 (since 100!=2^a.3^48.5^c.7^d......) please correct me if i am wrong.

Hi, I thought the best ways was to assume values for b in the 2nd equation Let b = -5; The equation becomes 2x^2-5x+2=0 Roots are 1/2 or 2. Roots of 1st equation are 3/2 or 3 try to form the 1st equation (x-3/2) (x-3) = 0 Make sure that the coefficients of x match in both the equation. (as they are equal to b).

Let the heights be 1,2,3,4,5,6 C D 6 1 A B This should be the arrangement. Hence we are left with 4 unknown positions A,B,C,D corresponding to 2,3,4,5 (not in the same order) Now, for D: D>A and D>C Hence D should be greater than at least 2 elements out of 2,3,4,5 Which means D can be 4 or 5 Now, for C, C So C can be 2 or 3 or 4 When D = 4, C can be 2 or 3 -> 2 ways When D = 5, C can take 2 or 3 or 4. -> 3 ways Hence total 5 ways.

thank you.but the OA is 5456. help me to solve it.

thank you very much.great. but how 7 gold coins can be distributed in 3 groups. explain me clearly please.

OA is 52!/(17!)^3 3!. please help me to solve this problem.

The number of ways in which 13 gold coins can be distributed among three persons such that each one gets at least two gold coins is: (a) 6 (b) 12 © 24 (d) 36

thank u....but please explain me the formula little bit clear....

please help me i don't know how to solve it..........