for non negative solution
3y+5z = 30-2x
as x =
3y+5z = 30,28,...,0
besides we know among 0,2,4,..,30 only solution of 3y+5z = 3k1 or 5k2 or 3k3+5k4
so we can filter numbers now... thus giving solution
and for positive solution , we have x =
and 3x+5z = 28,26....,0
same process as stated above

i think it would be 11c4 *7!

i think probability would be 5/12...

find remainder when (6!^7!)^13333 is divided by 13 ?

1
5
8
9
none of these

can u elaborate plz hw share of x = 3000 for nxt 6 month

X, Y and Z start a venture together. Y and Z invest Rs.4000 and Rs.3000 respectively while X invests Rs.1000 initially and after 6 months withdraws his capital and decides to work as a working partner with 30% stake in the profits. If X gets Rs.800 at the end of the year, what was the total profit?
1200
2400
3600
4800
5000

X, Y and Z start a venture together. Y and Z invest Rs.4000 and Rs.3000 respectively while X invests Rs.1000 initially and after 6 months withdraws his capital and decides to work as a working partner with 30% stake in the profits. If X gets Rs.800 at the end of the year, what was the total profit?
1200
2400
3600
4800
5000

The highest power of 2 in 10! + 11! + 12! + 13! + ...+ 1000! is ?
8
9
10
11
12

i think it would be -1 coz z=0

Let x = 100!. How many zeros does the base 6 notation of x end with?

97
75
50
48