We have k identical mugs. In an n-storey building, we have to determine the highest floor from which, when a mug is dropped, it still does not break. The experiment we are allowed to do is to drop a mug from a floor of our choice. How many experiments are necessary to solve the problem in any case, for sure? P.S. I just know the answer..so i won't give any options
for larger values of 'k', expression will be a bit complicated
for k = 2, minimum no of experiments will be 'a', where a is the smallest integer such that
a(a + 1)/2 ≥ n
for k = 3, minimum no of experiments will be 'a', where a is the smallest integer such that summation a = 1 to a {a(a + 1)/2} ≥ n
=> a(a + 1)(a + 2)/6 ≥ n
In general for k = r, minimum no of experiments will be 'a', where a is the smallest integer such that (summation a) from a = 1 to a {(r - 1) times} ≥ n
OR, for k = r, it will be a, where C{(a + r - 1), r} ≥ n
In a locality, the most read newspapers are " Hindu" , "TOI" and "IE" . 60% of the residents read at most one of these newspapers. 30% of the residents read exactly two of these newspapers. The number of residents who read all these papers is 5/6th of those who read none of the papers. 10% read "Hindu" and "TOI" but not "IE" . 40% read Hindu or "TOI" but not "IE" . 44% of residents read "Hindu".. 12% read only "TOI" and "IE". 2400 residents read only "Hindu" and "IE"Q) How many residents read TOIa) 2800 b)4200 c)3500 d) 4900Q) How many residents don't read any of the 3 newspapersa) 2400 b)3600 c) 3000 d) 4800Please provide answers with detailed solutions
1b and 2b?
H = a HT = b T = c TI = d I = e IH = f THI = g none = h
Given : h + a + c + e = 60%______________(1) b + d + f = 30%_________(2) g = 5h/6 or h = 6g/5 b = 10% a +b + c = 40%____________(3) a + b + f + g = 44%___________(4) d = 12% f = 2400
from (2), f = 8% from (3), a + c = 30% or c = 30% - a from (4), a + g = 26% or g = 26% - a from (1), 6/5 (26% - a) + a + 30% - a + e = 60% 156% - 6a + 5e = 150% e = 6/5*(a -1)
a + b + c + d + e + f + g + h = 100% a + 10 + 30 - a + 12 + 6/5*(a-1) + 8 + 11/5*(26 - a) = 100 6a - 6 + 286 - 11a = 200 a = 16% c = 14% e = 18% g = 10% h = 12% ------------------------------------------------ Q1 answer: c = 14% = 14 * 2400/8 = 4200
guys i have found in this thread the use of fermat theorem and eular theorem for finding out remainder of complex number. guys please explain both the theorem in detail, so that application used by u guys can be understood easily.
In a locality, the most read newspapers are " Hindu" , "TOI" and "IE" . 60% of the residents read at most one of these newspapers. 30% of the residents read exactly two of these newspapers. The number of residents who read all these papers is 5/6th of those who read none of the papers. 10% read "Hindu" and "TOI" but not "IE" . 40% read Hindu or "TOI" but not "IE" . 44% of residents read "Hindu".. 12% read only "TOI" and "IE". 2400 residents read only "Hindu" and "IE"Q) How many residents read TOIa) 2800 b)4200 c)3500 d) 4900Q) How many residents don't read any of the 3 newspapersa) 2400 b)3600 c) 3000 d) 4800Please provide answers with detailed solutions
Q) Determine the remainder when the number 1+22+333+4444+...+(sss...s times) written in base s+1, is divided by s-1.Don't know the answer. Will the answer vary from base to base?
@Torque024 said:Q) Determine the remainder when the number 1+22+33+4444+...+(sss...s times) written in base s+1, is divided by s-1.Don't know the answer. Will the answer vary from base to base? for binary 1 for base 3 (1+22)base 3 = 1+ 8=9/2 = 1 for base 4 1+22+333 base 4, 1+10+3+12+48 = 64/3 =1 one in all cases.
guys i have found in this thread the use of fermat theorem and eular theorem for finding out remainder of complex number. guys please explain both the theorem in detail, so that application used by u guys can be understood easily.
Q) Determine the remainder when the number 1+22+33+4444+...+(sss...s times) written in base s+1, is divided by s-1.Don't know the answer. Will the answer vary from base to base?
@Torque024 said:Q) Determine the remainder when the number 1+22+33+4444+...+(sss...s times) written in base s+1, is divided by s-1.Don't know the answer. Will the answer vary from base to base? for binary1for base 3(1+22)base 3 = 1+ 8=9/2 = 1for base 41+22+333 base 4,1+10+3+12+48 = 64/3 =1one in all cases.
Let s=9 1+22+333+4444+...+99999999=1097393685 in base 10 divide by s-1=8 mod[685]=5 Where am I wrong?
In a locality, the most read newspapers are " Hindu" , "TOI" and "IE" . 60% of the residents read at most one of these newspapers. 30% of the residents read exactly two of these newspapers. The number of residents who read all these papers is 5/6th of those who read none of the papers. 10% read "Hindu" and "TOI" but not "IE" . 40% read Hindu or "TOI" but not "IE" . 44% of residents read "Hindu".. 12% read only "TOI" and "IE". 2400 residents read only "Hindu" and "IE"Q) How many residents read TOIa) 2800 b)4200 c)3500 d) 4900Q) How many residents don't read any of the 3 newspapersa) 2400 b)3600 c) 3000 d) 4800Please provide answers with detailed solutions
We are given 1997 distinct positive integers, any 10 of which have the same least common multiple. Find the maximum possible number of pairwise coprime numbers among them.
guys i have found in this thread the use of fermat theorem and eular theorem for finding out remainder of complex number. guys please explain both the theorem in detail, so that application used by u guys can be understood easily.
sahi kaha sirji