@Brooklyn said:how did u get b=r-2 ??
it is said prob of red is 1/2
form an equation with it taking red as r and blue as b
r-b=2
b= r-2
@Brooklyn said:how did u get b=r-2 ??
getting red is 1/2 ... so half of total balls must b red
so n =2r
2r = r+b+y
r = b+2
b = r -2
@maddy2807 said:it is said prob of red is 1/2form an equation with it taking red as r and blue as br-b=2b= r-2
r-b=2 whyy??? 😲
dekh let it be r ,b , y
r=b+y !! naa ???
@Hypertexter said:Q. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p isA. b/n 2 and 10B. b/n 10 and 20C. b/n 20 and 30D. b/n 30 and 40E. Greater than 40p.s. b/n is between
h(n) is product of all even numbers from 2 to n
So, h(100) is divisible by all prime numbers from 2 to 47
Hence, we can say that h(100) + 1 will not be divisible by any of the prime numbers from 2 to 47
So, least prime number that will divide h(100) + 1 will be 53 or more than that.
Option (4)
@Brooklyn said:r-b=2 whyy??? dekh let it be r ,b , yr=b+y !! naa ???
but y is 2 given in the question.
@Hypertexter said:Q. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p isA. b/n 2 and 10B. b/n 10 and 20C. b/n 20 and 30D. b/n 30 and 40E. Greater than 40p.s. b/n is between
4) greater than 40
@Hypertexter said:
It's right.. Approach?
pata nai aaj sab ans sahi to aa rahe bai but apprch galat ho rahi hai
fir bhi....
h(100) = 2*4*6*....*100 = 2^50*1*2*...*50 = 2^50*50!
h(100)+1 = 2^50*50! + 1
now all no.s till 50 can divide 2^50*50! .. so on dividing h(100)+1 we get remainder 1
so clearly that prime no is greater than 50
@soumitrabengeri said:A man standing at the bottom of a staircase starts tossing a coin. Every time it shows Heads, he climbs two steps, while every time it shows Tails he climbs one step. After a while, he finds that he has climbed 8 steps. How many possible sequences of Heads and Tails could he have thrown?1) 256 2) 283) 36 4) 34Please explain with approach
2 2 2 2 - 1
2 2 2 1 1 - 4c2+4=10
2 2 1 1 1 1 - 5c2+5=15
2 1 1 1 1 1 1 - 7c1=7
1 1 1 1 1 1 1 1 - 1
34
2 2 2 1 1 - 4c2+4=10
2 2 1 1 1 1 - 5c2+5=15
2 1 1 1 1 1 1 - 7c1=7
1 1 1 1 1 1 1 1 - 1
34
Q) Each number 0 to 9, is written on a separate slip and the 10 slips are dropped into a box. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?
@Torque024 said:Q) Each number 0 to 9, is written on a separate slip and the 10 slips are dropped into a box. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?
7
@Torque024 said:Q) Each number 0 to 9, is written on a separate slip and the 10 slips are dropped into a box. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?
7?
@Torque024 said:Q) Each number 0 to 9, is written on a separate slip and the 10 slips are dropped into a box. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?
7?
@Torque024 said:Q) Each number 0 to 9, is written on a separate slip and the 10 slips are dropped into a box. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?
7?
@Torque024 said:Q) Each number 0 to 9, is written on a separate slip and the 10 slips are dropped into a box. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?
7????
consider worst possible case...
0,1,2,3,4,5 .... these are 6 consecutive draws... after this any no (6,7,8,9)can give us sum on 2 slip 10
@Torque024
@Torque024 said:Q) Each number 0 to 9, is written on a separate slip and the 10 slips are dropped into a box. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?
9,8,7,6,5,0 after this any of the nos 1,2,3,4 should give a sum of 10. Hence 7?
Let a, b, c be positive integers, with c > 1.
How many solutions are there to the equation a^c ˆ' b^c = 2^100?
How many solutions are there to the equation a^c ˆ' b^c = 2^100?