@mailtoankit said:

haha..same here!!basically to find out the last three digits..you have to divide the given number by 1000now 3^1994 = 3^2*3^1992 = 9*(3^4)^498 = 9*(81)^498....now In (81)^498 last digit will always be 1....to find the second last digit...multiply the digit at tens place in bracket i.e. 8 , to the last digit of the exponent i.e. again 8 = 8*8 = 64(this method is used only in the case where last digit is 1)...so this will give us the two digits of original no. as 41....now this 9*(81)^498 will become = 9*41 = 369....which will the final last three digitsPS - you can check more about finding the last one,two,three digits at totalgadha.com..

@mailtoankit said:

369 ?3^2*(3^4)^498 = 9*41 = 369...

@jain4444 said:

How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20?

@bodhi_vriksha said:

I would like to say that your estimate is in the ball park range Let me create a few nice options for you a) 8 b)9 c>10 d>12 e>15 But the question is about finding those numbers and not just the number of such numbers!

@Faruq said:

Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?

@Faruq said:

Thirty six numbers are filled in the cells of a matrix as shown in the figure given below. Six numbers are chosen from the matrix such that no two numbers belong to the same row or the same column. In how many ways can the numbers be chosen?1 2 3 4 5 67 8 9 10 11 1213 14 15 16 17 1819 20 21 22 23 2425 26 27 28 29 3031 32 33 34 35 36a. (6^2 × 5^2 × 4^2 × 3^2 × 2^2 × 1^2) b. (6 × 6 × 6 × 6 × 6 × 6)c. 720 d. None of these