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fermat theorem : for any prime number p , (p-1)times,same digit is repeated , then that number formed is exactly divisible by p ..
so 88(upto 46 digits) is perfectly divisible by 47..
so the same applies to 88(upto 92 digits)
in the question we have 88(upto 89 digits)
so x888 mod 47 = 0
since last digit is 8 , check for 4*47 = 188
last two digit is taken care
this means x8 leaves remainder 1 when divided by 47..
48 mod 47 = 1
so x = 4 ...
remainder is 4..
hope u got that ..revert me if u dont get it ..
cheers
explain this naga----did'nt get tht highlighted part
answers:
1)41
2)28
1)41
2)28
The inventory was reduced by 54 so the difference between the interchanged no and original no is 54.So 41 and 14 are eliminated.
The remaining nos are 82 and 28.Since the inventory was reduced by 54 the original no is smaller no among the 2 so its 28i.e 82-28=54
Since the no of items is 28 the price per item is 1148/28=41
Question says that the the inventory got reduced by 54 after the digits were interchanged which means it was showing more quantity as sold.
if the quantity originally sold = 10a+b then, b-a = 6.
Also, if the price is y then, (10a+b)*y = 1148.
From hit and trial we get 10a+b = 28 => y = 41.
So, actual price per piece = 41.
And the actual quantity sold = 28.
well puys i think as the value of inventory after entering the data in Comp got reduced so the original value must be higher. hence it shud be 82 and not 28. Also actual price = 1148/82 = 14
explain this naga----did'nt get tht highlighted part
x888 divided by 47 gives 0 as reminder
lets put x = 4
47)4888
___47
---------
____188
____188
-----------
_____0
i am doing the reverse of it .. we have last digit as 8
47 * k = **8
only k = 4 satisfies 7*4 = 28
47 * 4 = 88.. so last digit is taken care
so now it leaves us with differnce of {89th and 90th digit thats x8 } and number in bold part {refer to division } should be 1...
only x= 4 satisfies as 48 - 47 = 1
hence x = 4 and remainder is 4
seriously i cant explain more lucid than this
hope u got it.. if no
, pm me 
What is the remainder when 22^33 + 10^35 is divided by 45
a) 2 b)11 c)8 d) none of these
What is the remainder when 22^33 + 10^35 is divided by 45
a) 2 b)11 c)8 d) none of these
45 = 5*3*3
22^33 mod 5 = 2
22^33 mod 9 = 1
5a+2 = 9b+1
put a=7, we get b=4..which gives remainder 37...
10^35 mod 5 = 0
10^35 mod 9 = 1
5a = 9b+1
a=11,b=6
reminder = 55
now 55+37 mod 45 = 92 mod 45 = 2
What is the remainder when 22^33 + 10^35 is divided by 45
a) 2 b)11 c)8 d) none of these
22^33 + 10^35 mod 5 =
2^33 mod 5 = 2
22^33 + 10^35 mod 9 =
4^33 + 1 mod 9 =
64^11 + 1 mod 9 = 2
remainder = 2
Hmm.. I am also getting the same.
its from nishit sinha and the answer given is 8.
Hmm.. I am also getting the same.
its from nishit sinha and the answer given is 8.
answer is indeed 2..
:
Thanx frnds for ur help ,,, I understood the problem ... Will post the new Scrap ASAP ....
What is the remainder when 22^33 + 10^35 is divided by 45
a) 2 b)11 c)8 d) none of these
when 10^n
is divided by 45 remainder is always 10
and by remainder theorem we can see that 22^33 gives a remainder of 37 upon didvided by 45
so the overal remainder will be
(37+10)%45=2
45 = 5*3*3
22^33 mod 5 = 2
22^33 mod 9 = 1
5a+2 = 9b+1
put a=7, we get b=4..which gives remainder 37...
10^35 mod 5 = 0
10^35 mod 9 = 1
5a = 9b+1
a=11,b=6
reminder = 55
now 55+37 mod 45 = 92 mod 45 = 2
little correction remainder for 10^35 is 10 and not 55
although i dont know the process you have adopted if possible please explain..
little correction remainder for 10^35 is 10 and not 55
although i dont know the process you have adopted if possible please explain..
dude..my approach was Chinese Remainder theorem..answer is 2 na...so why so discrepancy.....
and what is 55 mod 45 by the way.....its 10 only...isn't it ???
45 = 5*3*3
22^33 mod 5 = 2
22^33 mod 9 = 1
5a+2 = 9b+1
put a=7, we get b=4..which gives remainder 37...
10^35 mod 5 = 0
10^35 mod 9 = 1
5a = 9b+1
a=11,b=6
reminder = 55
now 55+37 mod 45 = 92 mod 45 = 2
remainder cant be more than 45. so actual remainder: 55-45=10
dude..my approach was Chinese Remainder theorem..answer is 2 na...so why so discrepancy.....
and what is 55 mod 45 by the way.....its 10 only...isn't it ???
hey i did not wanted to point out your flaws rather as i did not knew the method so i wanted to make suer that you have arrived at it properly...and anyways the answer was correct ...please explain the chinese remainder method...(india ka method kaha gaya sob
)
What is the remainder when 22^33 + 10^35 is divided by 45
a) 2 b)11 c)8 d) none of these
the remainder is 2 (37+10/45)as it is done above i'm not posting the way
One from my side:
Q)In an integer d1d2...dk from left satisfy didi+1 for i even. How many integers from 1000 to 9999 have four distinct digits?
Q)In an integer d1d2...dk from left satisfy di
fermat theorem : for any prime number p , (p-1)times,same digit is repeated , then that number formed is exactly divisible by p ..
so 88(upto 46 digits) is perfectly divisible by 47..
so the same applies to 88(upto 92 digits)
in the question we have 88(upto 89 digits)
so x888 mod 47 = 0
since last digit is 8 , check for 4*47 = 188
last two digit is taken care
this means x8 leaves remainder 1 when divided by 47..
48 mod 47 = 1
so x = 4 ...
remainder is 4..
hope u got that ..revert me if u dont get it ..
cheers
Naga, jus applying the theorem to check for lower primes 2, 3, 5, 7 and 11
for 2, 8 is perfectly divisible by 2
for 3, 88 is not perfectly divisible by 3..
for 5, 8888 is not perfectly divisible by 5
for 7, 888888 is perfectly divisible by 7
for 11, 8888888888 is perfectly divisible by 11
are we missing something in the theorem??
Regards
SK
Naga, jus applying the theorem to check for lower primes 2, 3, 5, 7 and 11
for 2, 8 is perfectly divisible by 2
for 3, 88 is not perfectly divisible by 3..
for 5, 8888 is not perfectly divisible by 5
for 7, 888888 is perfectly divisible by 7
for 11, 8888888888 is perfectly divisible by 11
are we missing something in the theorem??
Regards
SK
This theorem is applicable only for prime numbers greater than 5.