find the sum of the series 1.2+ 2. 2^2 + 3. 2^2 +...............+ 100. 2^100
- 100. 2^101 + 2
- 99 . 2^100 + 2
- 99 . 2^101 + 2
- 100. 2^100 + 2
- none of these
if an be the nth term of an ap and if a7=15, then the value of the common difference that would make a2a7a12 greatest is....
- 3
- 3/2
- 7
- 0
- none of these
Find the sum of the series 1.2+ 2. 2^2 + 3. 2^2 +...............+ 100. 2^100
- 100. 2^101 + 2
- 99 . 2^100 + 2
- 99 . 2^101 + 2
- 100. 2^100 + 2
- none of these
S = 1.2 + 2.2^2 +3.2^3.............100.2^100
2S = 1.2^2+2.2^3+...........+99.2^100 + 100.2^101
subtracting eqn 1 from 2=>
S = -(1.2 + 1.2^2 + ..............+2^100) + 100.2^101
= -1. 2(2^100 -1) + 100. 2^101 (G.P.)
= 99.2^101 + 2
Can anybody help in this question
what are the last 2 digits in the multiplication of 35.34.33.32.31.30.29.28.27.26 ???
i tried the approach 35!/26! but not getting the answer
Thanx in advance
Also can somebody tell me the method of finding the last 2 digits of a factorial??
Can anybody help in this question
what are the last 2 digits in the multiplication of 35.34.33.32.31.30.29.28.27.26 ???
i tried the approach 35!/26! but not getting the answer
Thanx in advance
My take on this is 60.
soln.
divide the given series by 100(since last two digits have been asked)
=>(35.34.33.32.31.30.29.28.27.26)/100
=>(35.34.33.32.31.6.29.28.27.26)/20 (30/100 = 6/20, we can further duduce it to get the denominator 10, we divide the denominator seeing the no.s in numerator, since the no.s in numerator are greater then 10 as well as 20, we can take either, I have deduced the numerator to 20)
=>(15.14.13.12.11.6.9.8.7.6)/20 (numerator are the remainders obtained after dividing each numerator by 20 )
=>(90.98.104.108.66)/20 (multiply any two of numerator here i multiplied 15*6, 14*7, 13*8, 12*9, 11*6)
=> (10*18*4*8*6)/20
=>(180*32*6)/20
=>12*6/20
=>72/20= 12(remainder)
again since we had deduced the denominator to 20 we multiply 5 to the obtained remainder.
Hence the answer is 12*5=60.
My take on this is 60.
soln.
divide the given series by 100(since last two digits have been asked)
=>(35.34.33.32.31.30.29.28.27.26)/100
=>(35.34.33.32.31.6.29.28.27.26)/20 (30/100 = 6/20, we can further duduce it to get the denominator 10, we divide the denominator seeing the no.s in numerator, since the no.s in numerator are greater then 10 as well as 20, we can take either, I have deduced the numerator to 20)
=>(15.14.13.12.11.6.9.8.7.6)/20 (numerator are the remainders obtained after dividing each numerator by 20 )
=>(90.98.104.108.66)/20 (multiply any two of numerator here i multiplied 15*6, 14*7, 13*8, 12*9, 11*6)
=> (10*18*4*8*6)/20
=>(180*32*6)/20
=>12*6/20
=>72/20= 12(remainder)
again since we had deduced the denominator to 20 we multiply 5 to the obtained remainder.
Hence the answer is 12*5=60.
Hello everyone , one of my friends a TIME student told me to solve this type of question
Q)Find the remainder when 123456789.....40 is divided by 36
He(my friend) told me that 36=9x4 therefore first find remainder when 123456...40 is divided by 9 which is found by diving sum of digits 244/9 => remainder = 1
Then find remainder when 123456...40 is divided by 4 which is remainder of dividing last 2 digits by 4 ; 12/4=> remainder= 0
Now to find the final remainder we find the lowest multiple of 4 which will give remainder 1 with 9 and such a no. can be 28 ; Hence final remainder answer is 28
But now if this approach is applied to one of the ARUN SHARMA question then the answer is not coming correct
The question is as follows
Q) Find the remainder when 123456789...4950 is divided by 16
Now my approach is 16=2x8 , Dividing 1234567...4950 by 2 will yield remainder=0
Dividing 123456...4950 by 8 which is found by dividing last 3 digits i.e 950/8 => remainder = 6
Now to find the final remainder we need to find lowest multiple of 2 that gives remainder 6 with 8 which will be 2x7 =14 as 14/8 => remainder = 6 ; the final answer by TIME's approach should be 14 BUT THE ANSWER IS GIVEN AS 6 IN THE ARUN SHARMA BOOK .
CAN SOMEBODY HELP??????????????????
But now if this approach is applied to one of the ARUN SHARMA question then the answer is not coming correct
The question is as follows
Q) Find the remainder when 123456789...4950 is divided by 16
When you need to check reminder by 16...
just check last 4 digits are divisible by 2 or not....
(abcdef...)/2^n
then just divide last n digits of the number by 2...and the reminder which you will get will be the reminder of the whole number...
also that approach(T.I.M.E) is known as Chinese reminder theorem...it is applied in a bit different manner
Can you explain it in some detail like what should be done according to u "then just divide last n digits of the number by "
whether rem (4950/2) or rem(50/2) ?????????????????????
OK....
for ex-
1.)find reminder when 52687456987456321456322/2048,1024,512,248,128,64,32,16,8,4,2(by any of these numbers)
then for 2=we check whether last digit of that number is divisible by 2 or not
for 4=we will check 22 is divisible by 4 or not because 4=2^2 so we will check last 2 digits
for 8=check 322/8 or not because 8=2^3..
for 16=check 6322/16 or not because 16=2^4...
similarly you can proceed for all the powers of 2...
i hope this will help...
:cheers:
hello puys..wud b appearing for cat for the first time...solving arun sharma of quant...but i usually get stuck sometimes...just wanted to know that is solving arun sharma enuf for cat or shud i go for some other buks 2....:)
vipulthapa91 Sayshello puys..wud b appearing for cat for the first time...solving arun sharma of quant...but i usually get stuck sometimes...just wanted to know that is solving arun sharma enuf for cat or shud i go for some other buks 2....:)
thnks fr d reply...just wnted 2 knw 1 more thing....taking a look at previous year papers i realized that profit and loss,interest are not the imp. topics fr cat( from quant point of view) should i hone my skills in these topics or shud i focus on imp. topics of cat.....
Hey Guys please help me out in solving the following question
Which one of the following is true
a) log 275 to the base 17 = log 375 to base 19
b) log 275 to the base 17 c)log 275 to the base 17 > log 375 to base 19
d)cannot be determined
e)none of these
Hey Guys please help me out in solving the following question
Which one of the following is true
a) log 275 to the base 17 = log 375 to base 19
b) log 275 to the base 17 c)log 275 to the base 17 > log 375 to base 19
d)cannot be determined
e)none of these
Hey Guys please help me out in solving the following question
Which one of the following is true
a) log 275 to the base 17 = log 375 to base 19
b) log 275 to the base 17 c)log 275 to the base 17 > log 375 to base 19
d)cannot be determined
e)none of these
i got stuck...plz help me out....
Q-two pipes A and B can fill up half full tank in 1.2 hours.the tank was initially empty.pipe B was kept open for half the time required by pipe A to fill the tank by itself.then pipe A was kept open for as much required by pipe B to fill up 1/3 of the tank itself.it was found that the tank was 5/6 full.the least time in which any of the pipe can fill the tank fully is
a) 4.8 h
b) 4 hours
c) 3.6 hours
d) 6 hours