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Number System
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If log 2, log (2x -1) and log (2x + 3) are in A.P, then x is equal to ?
Bhars.
Bhars.
Hello people,
As the number system thread has been around for about 10000 posts,it need to be closed down for server issues.But a new one has been opened here:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/43728-number-system-ii.html#post1642745
Please continue with yo...
As the number system thread has been around for about 10000 posts,it need to be closed down for server issues.But a new one has been opened here:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/43728-number-system-ii.html#post1642745
Please continue with yo...
11
I too found it divisible by 2
the question must have a not as it is divisible by all three other than 11.. i forgot to post that
1
this question is wrong....its should be not divisible..then answer will be 11.
2
i'm getting weird 3 answers for these 2,3,37 not 11 though----naga how did u get 11 as the ans dude ...plz explain (i used fermat's little theorem for this one)
1
It is divisible by 2,3,37 but not by 11...
so probably question was wrong...it should be "NOT divisible by... "
then answer would be 11
so probably question was wrong...it should be "NOT divisible by... "
then answer would be 11
4
yes, divisible by 3 & not by 11...
1
2. 333^555+555^333 is divisible by
a.2 b.3 c.37 d.11
i got it 3 but answer given is 11 plz confirm...
333 and 555 is divisible by 3 so the ans should be 3 by reminder thrm
am i wrong ?
a.2 b.3 c.37 d.11
i got it 3 but answer given is 11 plz confirm...
333 and 555 is divisible by 3 so the ans should be 3 by reminder thrm
am i wrong ?
for the first one,
3^20 is divisible by 50
so now we r left with 3^12
3^12 - last two digit is 41
so decimal part is 2(41) = 82
so 0.82
3^20 is divisible by 50
so now we r left with 3^12
3^12 - last two digit is 41
so decimal part is 2(41) = 82
so 0.82
2
Can u elaborate please....
the 2nd one i feel its 3
the 2nd one i feel its 3
A few questions from arun sharma.
1. 3^32/50 give remainder and {.}denotes the fractional part.the fractional part is of the form(0.bx) value of x is??
2. 333^555+555^333 is divisible by
a.2 b.3 c.37 d.11
i got it 3 but answer given is 11 plz confirm...
for the first one it should be 0.82
second one its 11
You are not rich until you own your mistakes - Linda Poindexter
A few questions from arun sharma.
1. 3^32/50 give remainder and {.}denotes the fractional part.the fractional part is of the form(0.bx) value of x is??
2. 333^555+555^333 is divisible by
a.2 b.3 c.37 d.11
i got it 3 but answer given is 11 plz confirm...
1. 3^32/50 give remainder and {.}denotes the fractional part.the fractional part is of the form(0.bx) value of x is??
2. 333^555+555^333 is divisible by
a.2 b.3 c.37 d.11
i got it 3 but answer given is 11 plz confirm...
(41*40)%43=6
(39*38 )%43=20
multiplying the remainders u have 120p
its clear that 120p%43=1
so 120p-1 should be divisible by 43
therefore 120p-1=43k
(39*38 )%43=20
multiplying the remainders u have 120p
its clear that 120p%43=1
so 120p-1 should be divisible by 43
therefore 120p-1=43k
1
plz explain these areas
Its like 42!/43 will leave a remainder of 42
42!=41!*42
so for 42! to leave a remainder of 42 41! must leave a remainder of 1 only
therefore 41!%43=1
42!=41!*42
so for 42! to leave a remainder of 42 41! must leave a remainder of 1 only
therefore 41!%43=1
By wilson theorem ....
(p-1)! + 1 mod p = 0 when p is prime>5..
so
42! mod 43 = 42
=> 41! mod 43 = 1
(p-1)! + 1 mod p = 0 when p is prime>5..
so
42! mod 43 = 42
=> 41! mod 43 = 1
it comes from Wilson's theorem
(p-1)! +1 mod p = 0, where p is prime > 5
(p-1)! +1 mod p = 0, where p is prime > 5
how do you get
41! mod 43 = 1
could you explain pls?
41! mod 43 = 1
could you explain pls?
Like ankur said... Different problems diff approaches... try to find the easiest way out
for ex in this case
65x29x37x63x71x87x62
You know u have one 5 and one 2.... multiply it first
65*62 = 4030%100 = 30
Now ur effort is to find last digit of rest n multiply it by 3 to get second...
for ex in this case
65x29x37x63x71x87x62
You know u have one 5 and one 2.... multiply it first
65*62 = 4030%100 = 30
Now ur effort is to find last digit of rest n multiply it by 3 to get second...
4
Shouldnt it be 120p = 43k +1
=> 34p = 43k + 1
=> 34p = 43*15 + 1
=> p=19
So remainder should be 19
P.S: Mere Dimag par thoda jung laga gaya hai...so please do verify it
=> 34p = 43k + 1
=> 34p = 43*15 + 1
=> p=19
So remainder should be 19
P.S: Mere Dimag par thoda jung laga gaya hai...so please do verify it
5
how did u do that???:shocked: i mean getting 85,31,94.. cud u lil' explain the process?
TIA
TIA
not every question can be done that way
65x29x37x63x71x87x62 ==> 85*31*94 ==>90*31==>90
65x29x37x63x71x87x62 ==> 85*31*94 ==>90*31==>90
41! mod 43 = 1
37! mod 43 = p
41*40*39*38p mod 43 = 1
120p-1 = 43k
p = 19
remainder = 19
37! mod 43 = p
41*40*39*38p mod 43 = 1
120p-1 = 43k
p = 19
remainder = 19
Thanks Ankurb... probably the soln comes quick cos of the symmetric nature of nos. Wat if the expression is: 65x29x37x63x71x87x62 or something as asymmetric as this? hw to go about...
can u explain how?
Pls help me out wth this question.
what is the remainder when 37!/43?
is the answer 0 ?
what is the remainder when 37!/43?
is the answer 0 ?
thanks a lot
clear now
clear now
it wyd b -4 * -14 * -2
8 * -14
or 8 * 86 = 688 = rem 8
8 * -14
or 8 * 86 = 688 = rem 8
Pls help me out wth this question.
what is the remainder when 37!/43?
what is the remainder when 37!/43?
196*186*198 last two digits
-4*-14*-2=-112/100=-12/100=88 last 2 digits......
remainders when 196,186,198 are divided by 100,we get-4,-14,-2
-4*-14*-2=-112/100=-12/100=88 last 2 digits......
remainders when 196,186,198 are divided by 100,we get-4,-14,-2
what about if it is in the form of 196*186*198?
its nothing but remainder obtained when each term is divided by 100...
hope this helps
hope this helps
2
Ankur,
Could you please explain this method with an example.
Could you please explain this method with an example.
1152 = 2^7*3*2
(4!)^3 and above will contain that.
(1!)^3 + (2!)^3 + 3!)^3 =
1+8+216 = 225
(4!)^3 and above will contain that.
(1!)^3 + (2!)^3 + 3!)^3 =
1+8+216 = 225
5
last 2 digits = 1*2*3*-3*-2*-1 = -36 = 64
2
For the 2nd prob..2 find the last 2 digits..u nid2 consider d last2digits of ech term..in dis case..its simple..becoz..d 1st 3 terms hav zero in d tens place..(01*02*03)=06...now considering (97*98*99)..do d multiplcation only upto the last 2 digits..we get d ans as 64
-------------------------stripped-------------------
guys got 2 problems...
1. wats the reminder when (1!)^3 + (2!)^3 + .... + (1152!)^3 is divided by 1152?
2. Last 2 digits of: 101x102x103x197x198x199 (any hint of how to crack problems of this kind - 'last 2 digits' ?)
Help me out...
Thanks much!
1. wats the reminder when (1!)^3 + (2!)^3 + .... + (1152!)^3 is divided by 1152?
2. Last 2 digits of: 101x102x103x197x198x199 (any hint of how to crack problems of this kind - 'last 2 digits' ?)
Help me out...
Thanks much!
Use eulers..to express it as
2^22k+20
rem(2^22K)/23=1
rem(2^20)/23=6
hence
(2^(2^36))/23=6 mod 23
2^22k+20
rem(2^22K)/23=1
rem(2^20)/23=6
hence
(2^(2^36))/23=6 mod 23
Answer should be 6.