@Harmeet89
with 43....sum is 946....1000-946=54
54=27 X 2
what is the digital sum of 4444^4444 ????
@getupsid said:what is the digital sum of 4444^4444 ????
4444^4444 mod 9
=(-2)^4444 mod 9
=8^1481*2 mod 9
=-2 mod 9
=7
so 7
@krum said:4444^4444 mod 9=(-2)^4444 mod 9=8^1481*2 mod 9=-2 mod 9=7so 7
Makes sense. But why did you do mod 9?
Newbie at using shortcuts
Newbie at using shortcuts
@getupsid said:1) what is the when 3^450 is divided by 108 ? 2) P is a natural number.2P has 28 divisors and 3P has 30 divisors.how many divisors of 6P are there?
1) 81 2) 42??
@cynara said:Makes sense. But why did you do mod 9?Newbie at using shortcuts
digital sum of any number is always the remainder of that number with 9. Just a numeric observation, nothing else.
you can take examples and verify yourself.
you can take examples and verify yourself.
@adityaknsit said:digital sum of any number is always the remainder of that number with 9. Just a numeric observation, nothing else.
Thanks!
@cynara said:Makes sense. But why did you do mod 9?Newbie at using shortcuts
Digital Sum = Sum of Digits reduced upto single digit => Rem. with 9.
@getupsid said:what is the digital sum of 4444^4444 ????
4444^4444 mod 9 = 7 .
Digital sum of any number is its remainder when divided by 9 .
Digital sum of any number is its remainder when divided by 9 .
@cynara said:Makes sense. But why did you do mod 9?Newbie at using shortcuts
Digital sum is that digit that comes on recursively adding the digits till we get a single digit.
It also comes out to be remainder by 9.
In general, in any number system of base n, the digital sum is the remainder obtained on division by (n-1)
It also comes out to be remainder by 9.
In general, in any number system of base n, the digital sum is the remainder obtained on division by (n-1)
@cynara said:Thanks!
Ohh and the fact that Sum of Digits reduced upto single digit => Rem. with 9 is arithmetically verifiable too.
For simplicity lets take a two digit number. Let the number be pq
Now, pq=10p +q
Now we have to find the rem with 9.
In 10p+q we can keep q aside for a while (less than or equal to 9 and anyway we can use the additive property of mod)
remaider with 10p=1*p
thus the actual remainder is p+q
which is indeed the digital sum.....
For simplicity lets take a two digit number. Let the number be pq
Now, pq=10p +q
Now we have to find the rem with 9.
In 10p+q we can keep q aside for a while (less than or equal to 9 and anyway we can use the additive property of mod)
remaider with 10p=1*p
thus the actual remainder is p+q
which is indeed the digital sum.....
@catahead said:In general, in any number system of base n, the digital sum is the remainder obtained on division by (n-1)
I remember reading something like this somewhere. Thanks! :)
@getupsid said:1) what is the when 3^450 is divided by 108 ? 2) P is a natural number.2P has 28 divisors and 3P has 30 divisors.how many divisors of 6P are there?
1) - divide the number with 27 & 4.
with 27, R = 0
with 4 , R = 1
So number can be written as 4a + 1 = 27b
only at b = 3, a = 20, 81 satisfy the equation
Hence R = 81.
with 27, R = 0
with 4 , R = 1
So number can be written as 4a + 1 = 27b
only at b = 3, a = 20, 81 satisfy the equation
Hence R = 81.
@getupsid said:please give steps for (1)for 2) 35 is given as correct ans.
P is a natural number.2P has 28 divisors and 3P has 30 divisors.how many divisors of 6P are there?
P = 2^5*3^3
2P = 2^6*3^3
3P = 2^5*3^4
so 6P = 2^6*3^4
hence 7*5=35 divisors
find remainder when 7+77+777+7777+......100 times is divided by 8??
find remainder when 2222^5555 and 5555^2222 is dividived by 7??
@getupsid said:find remainder when 7+77+777+7777+......100 times is divided by 8?? find remainder when 2222^5555 and 5555^2222 is dividived by 7??
1. 7+5+1*98 / 8 => 6 remainder ?
2. If the question is 2222^5555 + 5555^2222 is dividived by 7
=> 2222^5 + 5555^2 / 7
=> 3^5+4^2 / 7
=> 0 remainder ?
2. If the question is 2222^5555 + 5555^2222 is dividived by 7
=> 2222^5 + 5555^2 / 7
=> 3^5+4^2 / 7
=> 0 remainder ?