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 your discussions there :)

Cheers,
Shashank

naga25french Says

the question must have a not as it is divisible by all three other than 11.. i forgot to post that :)

I too found it divisible by 2

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 not 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

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

pk_gt1 Says

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)

the question must have a not as it is divisible by all three other than 11.. i forgot to post that

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 ?

this question is wrong....its should be not divisible..then answer will be 11.

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 ?

v!vek@rora Says

yes, divisible by 3 & not by 11...

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)

v!vek@rora Says

yes, divisible by 3 & not by 11...

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

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 ?

yes, divisible by 3 & not by 11...

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 ?

BIJAY PRASAD Says

Can u elaborate please....

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

Can u elaborate please....

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

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...

pk_gt1 Says

plz explain these areas

(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

41! mod 43 = 1
37! mod 43 = p
41*40*39*38p mod 43 = 1
120p-1 = 43k
p = 19
remainder = 19

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

it comes from Wilson's theorem
(p-1)! +1 mod p = 0, where p is prime > 5

By wilson theorem ....
(p-1)! + 1 mod p = 0 when p is prime>5..

so
42! mod 43 = 42
=> 41! mod 43 = 1

how do you get

41! mod 43 = 1

could you explain pls?

it comes from Wilson's theorem
(p-1)! +1 mod p = 0, where p is prime > 5

41! mod 43 = 1
37! mod 43 = p
41*40*39*38p mod 43 = 1
120p+1 = 43k
p = 24
remainder = 24

how do you get

41! mod 43 = 1

could you explain pls?

hiddenname Says

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...

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 last digit....since last digit will be zero anyway

(9*7*3*1*7)%10 *30 = 3 * 30 = 90

41! mod 43 = 1
37! mod 43 = p
41*40*39*38p mod 43 = 1
120p+1 = 43k
p = 24
remainder = 24

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

not every question can be done that way
65x29x37x63x71x87x62 ==> 85*31*94 ==>90*31==>90

how did u do that???:shocked: i mean getting 85,31,94.. cud u lil' explain the process?

TIA

hiddenname Says

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...

not every question can be done that way
65x29x37x63x71x87x62 ==> 85*31*94 ==>90*31==>90

Pls help me out wth this question.
what is the remainder when 37!/43?

41! mod 43 = 1
37! mod 43 = p
41*40*39*38p mod 43 = 1
120p-1 = 43k
p = 19
remainder = 19

Ankurb Says

last 2 digits = 1*2*3*-3*-2*-1 = -36 = 64

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...

Pls help me out wth this question.
what is the remainder when 37!/43?


is the answer 0 ?

can u explain how?

Pls help me out wth this question.
what is the remainder when 37!/43?


is the answer 0 ?

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

thanks a lot

clear now

fuzzyl0gik Says

what about if it is in the form of 196*186*198?

it wyd b -4 * -14 * -2
8 * -14
or 8 * 86 = 688 = rem 8

Pls help me out wth this question.
what is the remainder when 37!/43?

fuzzyl0gik Says

what about if it is in the form of 196*186*198?

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

its nothing but remainder obtained when each term is divided by 100...

hope this helps

what about if it is in the form of 196*186*198?

Ankur,

Could you please explain this method with an example.

its nothing but remainder obtained when each term is divided by 100...

hope this helps

Ankurb Says

last 2 digits = 1*2*3*-3*-2*-1 = -36 = 64

Ankur,

Could you please explain this method with an example.

guys got 2 problems...

1. wats the reminder when (1!)^3 + (2!)^3 + .... + (1152!)^3 is divided by 1152?

Thanks much!

1152 = 2^7*3*2
(4!)^3 and above will contain that.
(1!)^3 + (2!)^3 + 3!)^3 =
1+8+216 = 225

guys got 2 problems...
2. Last 2 digits of: 101x102x103x197x198x199 (any hint of how to crack problems of this kind - 'last 2 digits' ?)

Help me out...

Thanks much!

last 2 digits = 1*2*3*-3*-2*-1 = -36 = 64

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!

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!

(2^(2^36))/23
what is remainder???

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^(2^36))/23
what is remainder???

Answer should be 6.