@Logrhythm said:hanji bhai, lattice method ya fir criss cross method for multiplication seekh lo youtube pe link mill jayega aapko.....usko use karke aese expressions ke digits find karna easy ho jata hai
ok...
@Swapno said:@TONYMBA yes.I understand that it would take values form 0 to infinitive.I know basic definations of mod function.But In time materials that answer is given that the inequality lies between :- -infinitiveIs the approach is like this:- |3x+6|.>0 or 3x+6 = 0 or,3x+6>0 or,3x=-6or,x>-2 or,x=-2so the value of x is going to be x>=-2.So the value of x >=-2.how the inequality is between like the given answer.@Estallar12 :- Bro actually I am facing a bit problem with inequalities..
Yeah it will be - infinity to + infinity.
Mod is outside the whole expression and so it will always be greater than - 12.
Yeh approach kya lagayi hai Bro ?? 
:|
:|@missionCAT13 said:Qa-99.94%tiler He is .
you must have me mistaken for someone else, i did not score 99.94 in qa.....kaash kare hote but agle saal kar loonga aap sabke contribution ke sath 
@PURITAN said:@Estallar12 you are back. i am glad.u r a hero..:respect:
everyone is a hero....bas bollywood mein kaam milna chahiye...
Baap re..99.98%..98%..Pata nahi mera kya hoga is baar..Pehli baar mein bhi CAT dunga is saal.. 😞
and please no spamming here puys
@TONYMBA said:everyone is a hero....bas bollywood mein kaam milna chahiye...
@pyashraj said:Baap re..99.98%..98%..Pata nahi mera kya hoga is baar..Pehli baar mein bhi CAT dunga is saal..
You will nail it. Be Confident and Keep working for it :)
Try this one guys :-
Q> What is the 2037th positive number/integer that can be expressed as the sum of two or more consecutive positive integers?
NOTE: The first 3 such numbers are 3=1+2, 5=2+3 and 6=1+2+3
All the Best 
@staaalinnn
All Number except 2 and its power can be..Thus, Till 2037; 2, 4, 8, ....1024 = 10 Numbers cannot be expressed.
Thus, the required number = 2037 + 11 = 2048..However, 2048 = 2^11..
Hence, the actual Number is 2049..
@staaalinnn said:Try this one guys :-Q> What is the 2037th positive number/integer that can be expressed as the sum of two or more consecutive positive integers?NOTE: The first 3 such numbers are 3=1+2, 5=2+3 and 6=1+2+3All the Best
all numbers which have atleast one odd factor can be expressed.... (by odd number here i mean not including 1)...
so we have to exclude 2,4,8,16.... and so on.... before 2037 there are 11 such numbers (including 1).... so we add 11 to 2037 so 2048 however 2048 cannot be expressed hence 2049
@staaalinnn said:Try this one guys :-Q> What is the 2037th positive number/integer that can be expressed as the sum of two or more consecutive positive integers?NOTE: The first 3 such numbers are 3=1+2, 5=2+3 and 6=1+2+3All the Best
2^n form of numbers cannot be expressed as the sum of consecutive numbers
2^10 is 1024, 2^11 is 2048
2^0 till 2^11 are 12 numbers
2048-12 = 2036 numbers till 2048 can be expressed
so 2037th number would be 2048+1 = 2049
@staaalinnn said:Try this one guys :-Q> What is the 2037th positive number/integer that can be expressed as the sum of two or more consecutive positive integers?NOTE: The first 3 such numbers are 3=1+2, 5=2+3 and 6=1+2+3All the Best
Remove the Powers of 2 till 2037. Rest all can be expressed.
So, 2037 + 11 (As 11 powers will be there) = 2048.
2048 is again a power of 2 ( 2^12).
Hence, 2049. :)
A TSD Problem:
26 men named A to Z running at the respective speeds of 'a' to 'z' m/s are participating in a 10 km race on a circular track of length 100 m. Their speeds are in Arithmetic Progression from 'a' to 'z' in that order. If the time taken by Z to meet A for the first time after start is 20 seconds and time taken by M to complete the race is 52 minutes 5 seconds, then find the time taken for all the 26 men to meet for the first time at the starting point.
@staaalinnn said:A TSD Problem:26 men named A to Z running at the respective speeds of 'a' to 'z' m/s are participating in a 10 km race on a circular track of length 100 m. Their speeds are in Arithmetic Progression from 'a' to 'z' in that order. If the time taken by Z to meet A for the first time after start is 20 seconds and time taken by M to complete the race is 52 minutes 5 seconds, then find the time taken for all the 26 men to meet for the first time at the starting point.
A = s
Z = s + 25d
For meeting the first time -
RS = 25d and Distance = 100m.
=> 100/25d = 20
=> d = 1/5
M = 13th => s + 12d = s + 2.4
Find s from here using given time (52 mints 5 sec.)
Now, It can be solved.
P.S. From Tab, so not typing the whole. :splat:
:splat:@staaalinnn said:A TSD Problem:26 men named A to Z running at the respective speeds of 'a' to 'z' m/s are participating in a 10 km race on a circular track of length 100 m. Their speeds are in Arithmetic Progression from 'a' to 'z' in that order. If the time taken by Z to meet A for the first time after start is 20 seconds and time taken by M to complete the race is 52 minutes 5 seconds, then find the time taken for all the 26 men to meet for the first time at the starting point.
Guys if u don't get an exact answer, please PM me..
@staaalinnn
Let A have a speed of x m/sec..B = x+d m/sec..C = x+2d m/sec.....M= x+12d m/sec..Z = x+25d m/sec..
Now, 10,000/3125 = Speed of M..or, x +12d = 16/5...(i)
Again, let after L m from the starting point, do Z meet A for the 1st time..
Hence, L/x = 20 sec, and, (100+L)/(x+25d) = 20..Solving Both, we have, d = 1/5 m..
From Eqn(i), we can say that, x+12/5 = 16/5...Thus, x = 4/5 m/sec..
Thus Speed of A= 4/5 m/sec, B = 1 m/sec....M= 16/5 m/sec..Z= 29/5 m/sec..
They will all meet at LCM[100/(4/5), 100/1, 100/(6/5)......., 100/(29/5)]
Yaha se Option will do.. 
Thus, after 500 secs will all the contestant meet at the starting point for the 1st tym..
