if 'a' is the remainder when (30!)*(12!) is divided by 43 and 'b' is the remainder when (47!)*(11!) is divided by 59. Find the value of a+b.
(1) 1
(2) 2
(3) 4
(4) 0
(5) 43
if 'a' is the remainder when (30!)*(12!) is divided by 43 and 'b' is the remainder when (47!)*(11!) is divided by 59. Find the value of a+b.
(1) 1
(2) 2
(3) 4
(4) 0
(5) 43
Answer is 5) 43
(30!)*(12!) mod 43 = 42
(47!)*(11!) mod 59 = 1
a+b = 43
find-28! mod 31
my approach-
by wilson theorm
30!/31=30
so 29!*30/31 = 30
30/31 is 30 so 29!/31 will be 1.
so 28!*29/31= 1
29/31 =29 : so 28!/31 =.......
i m stuck
pls help
find-28! mod 31
my approach-
by wilson theorm
30!/31=30
so 29!*30/31 = 30
30/31 is 30 so 29!/31 will be 1.
so 28!*29/31= 1
29/31 =29 : so 28!/31 =.......
i m stuck
pls help
28! mod 31 = p
29! mod 31 = 1
29p mod 31 = 1
29p = 31k + 1
p = k + (2k + 1)/29 ==> k = 14
p = 15
find-28! mod 31
my approach-
by wilson theorm
30!/31=30
so 29!*30/31 = 30
30/31 is 30 so 29!/31 will be 1.
so 28!*29/31= 1
29/31 =29 : so 28!/31 =.......
i m stuck
pls help
Look let 28! mod 31 = x
then we know that
30! mod 31 = 30
(30 * 29 * 28!) mod 31 = 30
(30 * 29) mod 31 * (28! mod 31) = 30
2 * x = 30
x=15........
find the last three digits, we need to find 7 ^ 64 % 1000.
Please find the remainder when (23^6 + 27^6) is divided by 50.
viveknitw Saysfind the last three digits, we need to find 7 ^ 64 % 1000.
My Ans: 401.
Waiting for some quick method. mine is a bit lengthy
My Ans: 401.
Waiting for some quick method. mine is a bit lengthy
answer is correct 401..
i had also used a bit lengthy method by finding the complete series of 3 digits..
plz post any short cut if possible..
:-P
cyborg5021a SaysPlease find the remainder when (23^6 + 27^6) is divided by 50.
23^6/50= -27^6/50
thus exp= 2*27^6/50=27^6/25=2^6/25=14
thus remainder is 14*2=28
m i correct?
answer is correct 401..
i had also used a bit lengthy method by finding the complete series of 3 digits..
plz post any short cut if possible..
:-P
pls share the method!
ur answer is correct!
cyborg5021a SaysPlease find the remainder when (23^6 + 27^6) is divided by 50.
my answer is coming out 2 be 14...
is it correct..
:-P 😲
23^6/50= -27^6/50
thus exp= 2*27^6/50=27^6/25=2^6/25=14
thus remainder is 14*2=28
m i correct?
yup correct 😃
my answer is coming out 2 be 14...
is it correct..
:-P :o
Ans is 28. Please see the post above
pls share the method!
ur answer is correct!
just found the series which is following..
7^4 = 401
7^5 = 807
7^6 = 649
7^7 = 543
7^8 = 901
7^9 = 307
7^10 = 043
7^11 = 401
the series is repeating after every 8 steps leaving the first 3..
so question given is 7^640
if we leave 1st 3 it remains 7^537..
nw dividing 537 by 8 leaaves remainder 1..
so answer is 401..
hope u got it..
:o
Whats the remainder when 6^57 is divided by 61?
cyborg5021a SaysWhats the remainder when 6^57 is divided by 61?
6^ 57/61= 6^-3/61=6^-3*306^3/61 =51^3/61=-10^3/61=-24
thus = 61-24 = 37
m i correct?
i m getting 24. please post ur approach
cyborg5021a Saysi m getting 24. please post ur approach
is 37 the correct answer?
i have edited ma post.......................
viveknitw Saysfind the last three digits, we need to find 7 ^ 64 % 1000.
answer is correct 401..
i had also used a bit lengthy method by finding the complete series of 3 digits..
plz post any short cut if possible..
:-P
pls share the method!
ur answer is correct!
just found the series which is following..
7^4 = 401
7^5 = 807
7^6 = 649
7^7 = 543
7^8 = 901
7^9 = 307
7^10 = 043
7^11 = 401
the series is repeating after every 8 steps leaving the first 3..
so question given is 7^640
if we leave 1st 3 it remains 7^537..
nw dividing 537 by 8 leaaves remainder 1..
so answer is 401..
hope u got it..
:o
7^64 mod 1000 = 7^64 mod 8*125
7^64 mod 8 = 1
7^64 mod 125 = 26
8a+1 = 125b + 26
Put b=3 or a = 50...we will get
Remainder = 401..
Hope This Helps..