Number System - Questions & Discussions

MBA CAT 2022
784
guys just a few questions i am posting pls help me !!!!

Q.1 Find the remainder when 6666.....2002 times is divided by 2002 ?

(1) 663 (2) 660 (3) 0 (4) 66 (5) None of these



Q.3 Find the last two digits of the number 3^15^17^19^21

(1) 01 (2) 07 (3) 17 (4) 41 (5) 81


my take for
1st 660
3rd 07..
785
hey i dnt need the ans bro i know them...pls tell me the method to solve them
!!!!!


3)last two digit of odd power ends with 3, 7 .. only two option satisfies that.. also tenth digit is even

so 07

2) direct result .. remainder 1

1) 2002 = 2*7*11*13

so we can use fermat theorem to get individual remainder .. then calculate final remiander by chinese remain der theorem

answer is 660
786
hey i dnt need the ans bro i know them...pls tell me the method to solve them
!!!!!

updated my post with explanation.
787
1.
option 2) 660
3333......2002 mod 7 =
3333 mod 7 = 1
333.....2002 mod 11 = 0
3333....2002 mod 13 =
3333 mod 13 = 5
11k = 7p+1 = 13q+5
remainder = 330 *2 = 660


can u please elaborate after the red part
788
sapient1989 Says
can u please elaborate after the red part

take 2 at a time
7p+1 = 13q+5
p = q + (6q+4)/7,==> q = 4
91w + 57 = 11k
k = 8w + 5 + (3w + 2)/11, ==>w = 3
remainder = 273 + 57 = 330

twice of which will be the answer
789


1) 2002 = 2*7*11*13

so we can use fermat theorem to get individual remainder .. then calculate final remiander by chinese remain der theorem

answer is 660

Is finding individual remainder not very time consuming?
790

hi to all
which is d largest four digit no. which when divided by 6 leaves remainder 5 and when divided by 5 leaves remainder 3
:shocked:

791
Abhinav90 Says
Is finding individual remainder not very time consuming?



no not at all...if u know fermat , euler and chinese remainder thm very well , its a cake walk
792
hi everybody
which is d largest no. that always divided d product of 3 consecutive multiples of 2
plzzzzzzzzzzz replyyyyyyyyyyyyyyyyyyy



is it 6?
add n-2,n.n+2= 3n
and n multiple of 2
793
no champ
ans is 48 as they r asking for largest no.
options r
1} 8
2}16
3}24
4}48
5}36
hey :w00t: bt we can find out by option as d no should b the multiple of 6
n 48 is the largest no. in option
i hope i m going on right way:splat:



its like this

take three consecutive numbers
2n, 2n-2, 2n+2
product will be
2(n^3-n)


somehow we need to show that n^3-n is always divisible by 24.
best way is to go by options......................


EDITED
take three consecutive numbers
2n, 2n-2, 2n+2
product will be
8(n^3-n)
794
its like this

take three consecutive numbers
2n, 2n-2, 2n+2
product will be
2(n^3-n)


somehow we need to show that n^3-n is always divisible by 24.
best way is to go by options......................

any three consecutive number are divisible by 3 and 2.so there multiplication is divisible by 6.
suppose they are n-1,n,n+1.
multiplying them by 2.
2(n-1),2n,2(n+1).
so,2(n-1)*2n*2(n+1)
= 8* (n-1)*n*(n+1).
so we can say it is divisible by 8*6=48.
am i right??
795

take the smallest three multiples of 2 viz. 2,4,6.multiplying them we get 48 which is the ans since any no. greater than that will not divide (2*4*6)..

796

Q) which is the biggest prime no.??

797
ranamitabh Says
Q) which is the biggest prime no.??

man something is missing in the Q,,,,
798

yeah...how about ..options?

799
drkunjan Says
man something is missing in the Q,,,,

puys..dere z nothing wrong with ques...actually there is nothing to calculate in this prob....u have to remember it...
ans is
2^13466917-1
800

find the maximum value of a^4 + b^4 + c^4 if 2a+3b+4c=18 ??

801
barclays_boss Says
find the maximum value of a^4 + b^4 + c^4 if 2a+3b+4c=18 ??


is 273 the answer??
802
is 338 the answer??

please post ur approach ......
803
is 273 the answer??

drkunjan Says
please post ur approach ......


again played with a bit logic..
to maximize a^4+b^4 + c^4
where
2a+3b+4c=18
we have to take min value for 4.. took 1.
now b has to be an odd no.. so 3
and a have to be max at 4..
so 4^4 + 2^4 + 1^4
= 256+16+1
= 273..
hope u got it..
:-P