Please tell me how to solve:
(3003^9000) / 9000
Please tell me how to solve:
(3003^9000) / 9000
Is the remainder 0?
aditya131025 Saysis the answer option d??
yes it is option d
find sum of the remainders obtained when a number n is divided by 9 and 7 successively.If n is the smallest number that leaves respective remainders of 4,6,9 when divided succesively by13,11,15
Rahul895 Saysfind sum of the remainders obtained when a number n is divided by 9 and 7 successively.If n is the smallest number that leaves respective remainders of 4,6,9 when divided succesively by13,11,15
is the answer 12??
The units digit of 2358^784 is the same as as units digit of 8^784.
cyclicity of 8 is 4. 784%4=0. so units digit of 8^784= units digit of 8^0=1
plz tell me if this is correct.
cyclicity of 8 is 4. 784%4=0. so units digit of 8^784= units digit of 8^0=1
plz tell me if this is correct.
no the answer is 6.
aditya131025 Saysis the answer 12??
sorry answer is 6.
Rahul895 Saysno the answer is 6.
can you jus explain me wr i went wrong??
check the attachment that is the solution... can u plz explain how they got 1369.
Rahul895 Sayscheck the attachment that is the solution... can u plz explain how they got 1369.
if u look at the data given divisors are 13,11,15 which upon successive division leaves remainders 4,6,9. the procedure we adopt is, write the numbers in the following format
13 11 15 (divisors)
4 6 9 (remainders)
we start at the bottom right corner 9 and go from 2nd row to the 1st diagonally to the left multiplying. we get 9*11=99, then we come down to the 2 nd row adding, we get 99+6=105. again multiplying, diagonally left upwards , we get 105*13. coming down to 2nd row adding, we get 1365+4=1369. hope this clears
Rahul895 Sayscheck the attachment that is the solution... can u plz explain how they got 1369.
i made a mistake in the last step. instead of adding 4, i multiplied with 4
if u look at the data given divisors are 13,11,15 which upon successive division leaves remainders 4,6,9. the procedure we adopt is, write the numbers in the following format
13 11 15 (divisors)
4 6 9 (remainders)
we start at the bottom right corner 9 and go from 2nd row to the 1st diagonally to the left multiplying. we get 9*11=99, then we come down to the 2 nd row adding, we get 99+6=105. again multiplying, diagonally left upwards , we get 105*13. coming down to 2nd row adding, we get 1365+4=1369. hope this clears
awesome... thanks for the explanation. 😃
Find the units digit of 34563^20359+2358^784.
Rahul895 SaysFind the units digit of 34563^20359+2358^784.
is the answer 13??
Rahul895 SaysFind the units digit of 34563^20359+2358^784.
The question reduces to 3^59 + 8^84
as 84 is factor of 4, hence 8^84 ends with 6.
3^59 = 3^(4K+2) = 9
hence 9+6 = 15.
Updated later when I was in my senses:
.8^84 ends with 6
3^59 = 3^(4k+3) = 7
Hence the number ends with 13.
Thanks for correcting me.
The question reduces to 3^59 + 8^84
as 84 is factor of 4, hence 8^84 ends with 6.
3^59 = 3^(4K+2) = 9
hence 9+6 = 15.
can u explain how 3^59=3^(4k+2). thought it t be of the form 4k+3'n hence end with 7'n hence the sum 13
yes the sum shd be 13.
aditya131025 Sayscan u explain how 3^59=3^(4k+2). thought it t be of the form 4k+3'n hence end with 7'n hence the sum 13
theairbender Saysyes the sum shd be 13.
OMG :shocked:.
I guess my super best buddy (read silly mistake) is back again:shocked: or I was very sleepy when I posted it.
Here is a problem I'm facing.
( 3^5^7 ) / 4 remainder?
I'm trying to figure it out by getting cycle or 3/4 and then whether 5^7 is of the form 3n / 3n+1 / 3n+2 etc
I'm getting 1 as answer.
But in a note it says the answer is 3 and I am not being able to understand where I'm making the mistake.
Please help!
Use Euler's Theorem:
It says Rem = 1 --- eqn (1)
=> We have to find the rem of 5^7 by 2 which is clearly 1.
Hence, we have to find Rem
Now, Rem = 1 using eqn (1)
=> Rem = 3.