find -12121212...................300digits% 999
viveknitw Saysfind -12121212...................300digits% 999
121 + 212 = 333
333* 50 = 16650
16650
16 + 650 = 666
answer is 666
viveknitw Saysfind -12121212...................300digits% 999
0 remainder
viveknitw Saysfind -12121212...................300digits% 999
Is it 666???
121 + 212 = 333
333* 50 = 16650
16650
16 + 650 = 666
answer is 666
great job
well done!
can u explain a bit more!
great job
well done!
can u explain a bit more!
divisibility rule by 999 :
take 3 digits from right at a time , add continuously till u get a sum less than 999..
thats the remainder
The product of three consecutive odd numbers is 531117. What is the sum of the three numbers?
The product of three consecutive odd numbers is 531117. What is the sum of the three numbers?
79, 81. 83
thus sum= 81*3= 243
(7777....300 times )^300 % 19.
What is the largest prime whose cube divides 1!2!...1001!?
a.997
b.991
c.977
d.973
hey puys can u plz tell me the approach hw 2 solve these type of questions??
a.997
b.991
c.977
d.973
hey puys can u plz tell me the approach hw 2 solve these type of questions??
viveknitw Says(7777....300 times )^300 % 19.
is the ans 1??
What is the largest prime whose cube divides 1!2!...1001!?
a.997
b.991
c.977
d.973
hey puys can u plz tell me the approach hw 2 solve these type of questions??
ans option 1 : 997 ??
viveknitw Says(7777....300 times )^300 % 19.
7^300 mod 19 * (100^150 + 100^149 + ...+1) mod 19
1*(5^150 + 5^149 + .. + 1) mod 19
(5^151 - 1) mod 76 => 72 mod 19 = 15
adiG Saysans option 1 : 997 ??
sorry but i dont knw the oa..
can u plz tell the approach??
What is the largest prime whose cube divides 1!2!...1001!?
a.997
b.991
c.977
d.973
hey puys can u plz tell me the approach hw 2 solve these type of questions??
1001!*1000!*999! => 997 appears thrice in this product
see. the given no. is 1!.2!.3!......997!.998!.999!.1000!.1001!
since the no.s 1001 comes only once and 1000 comes only twice (once in 1000! and once in 1001!) they cant be the ans
999 and no. below tht come atleast thrice
but 998 and 997 are not prime.
so the ans would come to finding the biggest prime no. in the given options.
so 997 being the biggest prime, that I think, is the ans....
I am not sure though
7^300 mod 19 * (100^150 + 100^149 + ...+1) mod 19
1*(5^150 + 5^149 + .. + 1) mod 19
(5^151 - 1) mod 76 => 72 mod 19 = 15
i dont understand my solution and i cant find any fault with my approach
..here goes
777777....300 times
when 7777...i times is divided by 19 the remainders are 7,1,17,6,10,12,13,4,,9,....
there are set of 18 remainders (including 0) . so 7777...300 times has the same remainder as
777....12 times.
that comes out to 11.
now 11^300 divided by 19 remainder be r
11^3 = 19k+1
so (19k+1)^100 divided by 19 will have a remainder of 1.
...
where is the fault.
i dont understand my solution and i cant find any fault with my approach
..here goes
777777....300 times
when 7777...i times is divided by 19 the remainders are 7,1,17,6,10,12,13,4,,9,....
there are set of 18 remainders (including 0) . so 7777...300 times has the same remainder as
777....12 times.
that comes out to 11.
now 11^300 divided by 19 remainder be r
11^3 = 19k+1
so (19k+1)^100 divided by 19 will have a remainder of 1.
...
where is the fault.
isnt the process a lengthy one!
is there anyway to find 11 directly ?(euler nnumber etc...........)
isnt the process a lengthy one!
is there anyway to find 11 directly ?(euler nnumber etc...........)
what is the correct answer...
N is a number such that the ratio of sum of its digit to product of its digits is 3: 40. If N is divisible by 37 and N is the smallest such number, how many factors does N have?
a. 8
b. 16
c. 20
d. 30
Can smbdy pls try this???