Number System - Questions & Discussions

MBA CAT 2022
3316
@PURITAN said:V is product of first 41 natural numbers and A = V + 1. Find the number of prime number among A + 1, A + 2, A + 3, A + 4, €Ś, A + 39, A + 40.
a
1
b
2
c
3
d
4
e
0

none

A=41!+1

a+1 will be divisible by 2 atleast

a+2 will be divisible by 3 atleast

a+40=41!+41hence will be divisble by 41

3317

someone plz help solving the qs:

what is the remainder when 67raised to the power of 53 is divided by 101?
(67^53 divided by 101)

Plz provide detailed solns ...

Thanks 😃

3318
@smittt said: help me solve dis ..
(2^133)/133
whats d remainder ???
is it 2 ??
3319

last two digits of 2^222.... kindly help ppl..


3320

@[547670:koyal1990] answer should be 04

3321

an isn't that imp.... what matters is the method... if u cud elaborate on that, wud be of great help.... TIA


3322

@[547670:koyal1990] well the method i use ......to find the last two digits of 2 is following:-

there two basic tricks or u can say rule for this
1) last two digits 0f (2 ^10) ^ even number = 76
2). last two digits of (2^10) ^ odd number =24
to demonstrate how to use this rule i will take example of your question
2^222 = (2^ 220 ) * (2^2) = (2^10)^22 *(2^2)
as 22 is even last two digits of (2^10)^22 will be 76 ....... furthermore 2^2 =04
therefore last two digits calculation (2^10)^22 *(2^2) equates to 76 * 04 = 304
therefore last two digits is o4.
3323
@swapnil.puy said: @koyal1990 well the method i use ......to find the last two digits of 2 is following:-
there two basic tricks or u can say rule for this
1) last two digits 0f (2 ^10) ^ even number = 76
2). last two digits of (2^10) ^ odd number =24
to demonstrate how to use this rule i will take example of your question
2^222 = (2^ 220 ) * (2^2) = (2^10)^22 *(2^2)
as 22 is even last two digits of (2^10)^22 will be 76 ....... furthermore 2^2 =04
therefore last two digits calculation (2^10)^22 *(2^2) equates to 76 * 04 = 304
therefore last two digits is o4.
well thanks a tonn mate.... really helped me.. do u know any such similar rules for other numbers?? so for other numbers what is the general approach to find last 2 digits???. TIA..
3324
@challenger90 said: acc to second condition in question won't it be 2x=k(f1+f2)???? since additional force is being applied on existing one
yaaaa... i also cnt understand this part.... shudn't it be f1+f2???
3325
@koyal1990 said:
well thanks a tonn mate.... really helped me.. do u know any such similar rules for other numbers?? so for other numbers what is the general approach to find last 2 digits???. TIA..
yeah there is a set of rules for odd numbers like for any number ending with 1 it is as follows:-( i will use some examples to demonstrate)

e.g.(31)^42 for the last two digits at unit place write 1
now to find digit at tens place
take the tenth place digit of the base number (e.g. 3 in 31 ) and multiply it with unit digit of power(e.g 2 in 42) so for tenth place digit of (31)^42 , multiply 3 *2 =6....take the unit digit i.e 6 and write it tenth place
so the last two digits of 2 at tenth digit place and 1 at unit place = 61

e.g (61) ^59 = unit digit ->1 tenth digit -> 6*9 =54 take unit digit of 54 ..4....therefore tenth digit 4 ...last two digits 41

e.g (71)^124252556246 = last digit 1 ..tenth digit 7*6 =42 take 2 therefore last two digits 21

e.g (121515164651)^2476125735489 = last digit 1 ...tenth digit 5*9 =45...take 5
therefore last two digits 51.
similarly there are rules for other odd numbers too. :-)


3326

@ swapnil.puy- thanks for the help!!!! form where will I get to know the rules for other numbers apart from 1 and 2? 5 as a matter of fact will always have 25 as the last two digits, but what about the others? if you cud tell me from where will I get to know the other rules, wud be of great help.. thanks a zziillion anyway!!! tc.. 😃

3327
@koyal1990 said: @ swapnil.puy- thanks for the help!!!! form where will I get to know the rules for other numbers apart from 1 and 2? 5 as a matter of fact will always have 25 as the last two digits, but what about the others? if you cud tell me from where will I get to know the other rules, wud be of great help.. thanks a zziillion anyway!!! tc..
@ koyal1990 ....well about 5, last two digits happen to be 25 or 75, not only 25.........whereas for other odd numbers like 3,7,9......they can be deduced using the rule which i mentioned for last digit 1 ..........and rest of the digits i.e even digits except 2 like 4,6,8 can be deduced by writing them as multiple of 2 and an odd number......like 6 is 2*3 ....so use the rule to find the last two digits of 2^n and then find last two digits of 3^n ...multiply and get for 6^n.....
now if u require that how i m deducing the rule of 1 to use for rule 3,7, 9 ......drop me a PM ......i will try reply asap...:-)
3328

ya acc to me it should be f1+f2..
but dnt knw the answer...can someone plz tell

@koyal1990 said:
yaaaa... i also cnt understand this part.... shudn't it be f1+f2???
3329

how (2^220) factored as (2^10)^22???

@swapnil.puy said: @koyal1990 well the method i use ......to find the last two digits of 2 is following:-
there two basic tricks or u can say rule for this
1) last two digits 0f (2 ^10) ^ even number = 76
2). last two digits of (2^10) ^ odd number =24
to demonstrate how to use this rule i will take example of your question
2^222 = (2^ 220 ) * (2^2) = (2^10)^22 *(2^2)
as 22 is even last two digits of (2^10)^22 will be 76 ....... furthermore 2^2 =04
therefore last two digits calculation (2^10)^22 *(2^2) equates to 76 * 04 = 304
therefore last two digits is o4.
3330
@challenger90 said: how (2^220) factored as (2^10)^22???
@swapnil.puy said: @koyal1990 well the method i use ......to find the last two digits of 2 is following:-
there two basic tricks or u can say rule for this
1) last two digits 0f (2 ^10) ^ even number = 76
2). last two digits of (2^10) ^ odd number =24
to demonstrate how to use this rule i will take example of your question
2^222 = (2^ 220 ) * (2^2) = (2^10)^22 *(2^2)
as 22 is even last two digits of (2^10)^22 will be 76 ....... furthermore 2^2 =04
therefore last two digits calculation (2^10)^22 *(2^2) equates to 76 * 04 = 304
therefore last two digits is o4.
buddy i suggest do have a look at rule exponents- (b^m)^n = b ^(m*n)
above i have put the snapshot from the wikipedia page on exponents to let u check the rule of exponents without any hassles (see the second rule)
therefore by above rule (2^10)^22 = 2^(10*22) = 2^220.......i think it now it clears ur doubt....:-)
3331

@[602051:swapnil.puy]
ya i got it..thanks:)

3332

what will be the highest power of 12 that will divide 5^36-1

3333

que) find max value of n such that 157! is perfectly divisible by 10^n??

also plz explain the concept of finding number of zeroes in an expression like 49!,52!,59!....

and also how to find the highest power of K when divisor is not a prime number..

someone plz explain...TIA:)

3334
@challenger90 said: que) find max value of n such that 157! is perfectly divisible by 10^n??also plz explain the concept of finding number of zeroes in an expression like 49!,52!,59!....and also how to find the highest power of K when divisor is not a prime number..someone plz explain...TIA

10=2*5. Now multiples of 2 in 157! will 157/2=78 + 78/2=39+ 39/2=19+ 19/2=9 +9/2=4 +4/2=2 +2/2=1. Therefore 78+39+19+9+4+2+1=152 occurrences of 2.
S//ly for 5 it will 157/5=31/3=6/5=1 . Therefore number of 5s are 31+6+1=38.(Ans)
As 10 consist of 2*5. Therefore 38 2s and 38 5s is the Answer.
3335

@[593895:challenger90]-

for calculating the highest power of 10,we need to look at the powers of 2 and 5.certainly the no of 2's is going to be much more than the no of 5's in 157!....so we only need to calculate the no of 5's in 157! which is 31+6+1 = 38.


As far as calculating the no. of zeroes in a factorial is concerned,we only need to look at the no of 5's and 2's ,whichever is more.

Hope it helps..