hello everybody,
please explain this one for me.
Q: find the remainder when 3^5^9^13^.....33 is divided by 5??
(A) 1
(B)2
(C)3
(D)4
(E)0
hello everybody,
please explain this one for me.
Q: find the remainder when 3^5^9^13^.....33 is divided by 5??
(A) 1
(B)2
(C)3
(D)4
(E)0
should be 3 ..
3^(4k + 1) ends in 3 ..
hello everybody,
please explain this one for me.
Q: find the remainder when 3^5^9^13^.....33 is divided by 5??
(A) 1
(B)2
(C)3
(D)4
(E)0
Try to first express this in the form 4k+1
5^9^13^.....33/4
=>1
=>5^9^13^.....33/4 = 4K+1
3^4k+1/5
=>(3^4k/5) x 3/5
=>1x3
=>3
yeah answer to is correct but i cudn;t understand after step
=>(3^4k/5)x3/5
=>1x3
=>3.....
yeah answer to is correct but i cudn;t understand after step
=>(3^4k/5)x3/5
=>1x3
=>3.....
for this remember the cyclicity
3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
3,9,7,1...so any number of form 3^4k will have unit digit 1..and now on dividing by 5 it gives 1 as remainder
so 1x3/5
=3
thanx alot...got it....:)
hi,
Q: N! is completely divisible by a 13^52.what is the sum of digits of smallest such number N?
a.11
b.15
c.16
d.19
regards
-abiral
hi,
Q: N! is completely divisible by a 13^52.what is the sum of digits of smallest such number N?
a.11
b.15
c.16
d.19
regards
-abiral
is the answer 15?
hi,
Q: N! is completely divisible by a 13^52.what is the sum of digits of smallest such number N?
a.11
b.15
c.16
d.19
regards
-abiral
is the answer is 16?
answer is 16...:)
abiral3081 Saysanswer is 16...:)
please explain the approach as well... thanks
Q: N! is completely divisible by a 13^52.what is the sum of digits of smallest such number N?
a.11
b.15
c.16
d.19
Solution: N! = 1 * 2 * 3 * .... 13 *... 26 * ... 39 *...* N
In this series of numbers, we need to consider only the multiples of 13.
Most of these multiples have only 13^1 as the factor and not higher powers of 13 like (13^2 or 13^3 as its factor).
But we also need to note that several numbers such as:
169 = 13 * 13; 338 = 2 * 13 * 13; 507 = 3 * 13 * 13 = 39 * 13.
Thus, all these numbers contain 2 powers of 13 or in other words, divisible by 13^2.
So in 507!, we have 39 + 3 = 42 powers of 13 (Since the 3 numbers above contain 2 powers of 13 instead of just 1).
But N! should be divisible by 13^52, hence we need to extend the value of N to include 10 more multiples of 13, thus N needs to be 10 multiples of 13 more than 507 = 507 + 130 = 637.
Then, we can be sure that N! = 637! contains 52 powers of 13 or in other words, 637! is exactly divisible by 13^52.
Sum of digits of N = 6 + 3 + 7 = 16, which is the final answer.
Cheers,
Raj Narap.
Answer :c
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
Answer:a
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
Answer:a
here u can express 16!^1777 as 4k
=>38^4k/17
=>1
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
Answer:a
As, 38 and 17 are co-prime, and the Euler number of 17 is 16, any power of 38 which is a multiple of 16 will give remainder equal to 1.
So, option (a)
The equation 4x-Ay=B has a number of integral soln.If HCF of (A and 4) is 1 and the number of solution (x,y)for 0
Some Specifications of a password
a) 7 digit odd no.
b) 1st 2 digits = 16 or 18 ;
c) 6 occurs twice ; 8 occurs once .
Max trials to crack such password =
a) 7680
b) 1250
c) 3840
d) 5760
Success ....
its 5760 😃
The equation 4x-Ay=B has a number of integral soln.If HCF of (A and 4) is 1 and the number of solution (x,y)for 0Puys the answer is 1. A=11 ...Please post ur approach ...
We can see value of x will increase by 11 and y by 4
so , last term = 11 + (45 - 1 ) 11 = 495
11 would be the answer here !!
now u'll say y did i choose 11 . well star from 5 , 7 , 9, 11 as these are smalles no's coprime to 4 . 11 will satisfy your no. of solutions.
Suja
