shanks4mba Sayshey do u have a gud source on expansions...?
I have a pdf which I can send you, not very good but should be enough here.
I have a pdf which I can send you, not very good but should be enough here.
number of zeros at the end of (2146!) in base 10 is 534
number of zero for 10! is 2 while that of 26! is 5 .. so we have a difference of 3
so number of zeros at the end = 534/3 = 178.. but one cant rule out 177 here .. its definitely a possibility..so i am not sure whether its 177 or 178
ps : obtained this method by sampling it to smaller power like octal to decimal
the answer is 177.
and the solution given by Career Launcher is:
In base 2, 10 means 2 and in base 3, 10 means 3.
Similarly in base 26, 10 means 26.
In base 10, 10 is obtained by multiplying 2 and 5.
Similarly in base 26, it is obtained by multiplying 2 and 13.
To find the number of zeroes in base 26, we need to see the
number of 2s and 13s in 2146!. Since, number of 2s is much
more than 13s, so we count number of 13s, which comes
out to be
(2146/13) + (2146/13^2)
= 165 + 12 = 177
pls explain this...:o
the answer is 177.
and the solution given by Career Launcher is:
In base 2, 10 means 2 and in base 3, 10 means 3.
Similarly in base 26, 10 means 26.
In base 10, 10 is obtained by multiplying 2 and 5.
Similarly in base 26, it is obtained by multiplying 2 and 13.
To find the number of zeroes in base 26, we need to see the
number of 2's and 13's in 2146!. Since, number of 2's is much
more than 13's, so we count number of 13's, which comes
out to be
(2146/13) + (2146/13^2)
= 165 + 12 = 177
pls explain this...:o
was dividing it by 26 rather than 13:banghead:
the answer is 177.
and the solution given by Career Launcher is:
In base 2, 10 means 2 and in base 3, 10 means 3.
Similarly in base 26, 10 means 26.
In base 10, 10 is obtained by multiplying 2 and 5.
Similarly in base 26, it is obtained by multiplying 2 and 13.
To find the number of zeroes in base 26, we need to see the
number of 2's and 13's in 2146!. Since, number of 2's is much
more than 13's, so we count number of 13's, which comes
out to be
(2146/13) + (2146/13^2)
= 165 + 12 = 177
pls explain this...:o
Post #400...well..dude..u've explained whole thing..what else do u need as an explanation...i think you are stuck with the fact...2*13...and 2*5....see...when we calculate number of 2's or 5's in a given factorial..what we do is search for number of 5's in a factorial..coz number of 2's are abundant...so we calculate number of 5's....similarly here we'll calculate number of 13's in 2146....which comes out to be....2146/13 = 165.....and.165/13 = 12...so number of 13's is 165+12 = 177
Originally Posted by orangejawan View Post
What is the answer for this question!!!!
1: Find the remainder when 8^643 is divided by 132
euler of 132 = 132(1-1/2)(1-1/3)(1-1/11) = 20
so 8^643 = 8^640+3
so 8^3 mod 132 = 116.....wats the answer ?????
I'm afraid, 132 and 8 are not co-prime, hence we need to simplify 132 to a lower number, in our case to 33, so as the same becomes a co-prime to the neumerator, in our case 2.
But your answer is correct.
Originally Posted by orangejawan View Post
What is the answer for this question!!!!
1: Find the remainder when 8^643 is divided by 132
I'm afraid, 132 and 8 are not co-prime, hence we need to simplify 132 to a lower number, in our case to 33, so as the same becomes a co-prime to the neumerator, in our case 2.
But your answer is correct.
--wrong post---
divishth Saysi think....132 has to be co-prime with the power...or the number..please cross-check.....
the number.........
Originally Posted by orangejawan View Post
What is the answer for this question!!!!
1: Find the remainder when 8^643 is divided by 132
I'm afraid, 132 and 8 are not co-prime, hence we need to simplify 132 to a lower number, in our case to 33, so as the same becomes a co-prime to the neumerator, in our case 2.
But your answer is correct.
well mam....how stupid i am...sorry naga....u groaned at me...
well the correct approach is ....8^643 = 2^1929 / 2^2 * 33
now euler(33) = 20...
so 2^7 mod 33 = 29....
now remainder is 4*29 = 116....he prabhu mera kya hoga.....
Hi,
can someone please explain Euler therom with example.
Thanks in advance
Hi,
can someone please explain Euler therom with example.
Thanks in advance
a^e(n) mod n=1
where a nd n re co prime to each other
e(n)=n(1-1/p)(1-1/q)........
where p nd q re prime factors of
e(n)is called d euler number
eg
find d remainder when 3^120 mod 100
now 3 nd 100 re coprime so we know euler's theorem can be applied
100=2^2*5^2
e(n)=100*(1-1/2)*(1-1/5)
=40
so by euler's theorem 3^(40*3) mod 100=1.........
hope itz clear now..........
What is the answer for this question!!!!
1: Find the remainder when 8^643 is divided by 132
132=4*33
8^643mod4 =0
8^643mod33 = 8^3 mod 33(E(33)=20 and 640 =20*32) = 512mod33 =17
4a =33b+17
a ~ (b+17)/4 => b=3
Remainder =33*3+17 =116
What is the answer for this question!!!!
1: Find the remainder when 8^643 is divided by 132
8^643 mod 132 ==>
2^1927 mod 33 ==>
2^1927 mod 3 = 2
2^1927 mod 11 = 4.32^385 mod 11 = 7
3k+2 = 11p+7,==>p=2
2^1927 mod 33 = 29
remainder = 29*4 = 116
a^e(n) mod n=1
where a nd n re co prime to each other
e(n)=n(1-1/p)(1-1/q)........
where p nd q re prime factors of
e(n)is called d euler number
eg
find d remainder when 3^120 mod 100
now 3 nd 100 re coprime so we know euler's theorem can be applied
100=2^2*5^2
e(n)=100*(1-1/2)*(1-1/5)
=40
so by euler's theorem 3^(40*3) mod 100=1.........
hope itz clear now..........
Its clear when they are co-prime.Please explain me in the example of : 2^1001 mod 100. I broke 100 into 25 and 4 and now 25 is in co-prime with 2
then 2^(25*4+1)mod 25 and got result as 2 but after that ????
Please help
2^1990 mod 199 = 29
2^1990 mod 5 = 4.
So,199a + 29 = 5b+4 = 2c.
On solving we get 2^1990 mod 1990 = 1507.
Answer to be 1024
Its clear when they are co-prime.Please explain me in the example of : 2^1001 mod 100. I broke 100 into 25 and 4 and now 25 is in co-prime with 2
then 2^(25*4+1)mod 25 and got result as 2 but after that ????
Please help
2^1001 mod 100==>
2^1001 mod 4 = 0
E(25) = 20
2^1001 mod 25 =
2*(2^20)^50 mod 25 = 2
25k+2 = 4p,==>k = 2
remainder = 52
= 5(10^92+10^91+10^90.....10^0) % 98
= 5(2^46+ 2^45(10+1)+ 2^44(10+1)+ 2^43(10+1)......2^0(10+1)) % 98
= 5(2^46+ 11(2^45 + 2^44 + 2^43....+2^0)) % 98
= 5(2^46 + 11(2^46 - 1)) % 98
= 5(12*2^46 -11)) % 98
= (60*2^46 - 55) % 98, solving this
= (78-55) % 98
= 23
can u plz explain bold red part?
tac007 Sayscan u plz explain bold red part?
10^96 mod 98 = (98+2)^48 mod 98 = 2^48
tac007 Sayscan u plz explain bold red part?
tac, kindly refer the below link for explanation
http://www.pagalguy.com/forum/quantitative-questions-and-answers/43728-number-system-ii-30.html#post1652852
Originally Posted by msabhi View Post
How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?
Puys I'm taking a straight route,
nos between 1000000 ~ 9999999 divisible by 9 --> (9999999 - 999999) / 9 = 1000000 (9s)
note: 999999 used for calculation convenience, hence 1 is not added to the result.
In every 100, we have 11, 9s out of which 1 will have 5 @ 10's digit (ex 54, 53, 52 etc)
Hence, every 11 out of 1000000(9s) has 1 multiple of 9 with 5 @ 10's digit
every 1100 ---------------- has 101 multiples of 9 --------------
every 110000-------------- has 10101 multiples of 9-------------
110000 has 10101(9s)
1000000 has 10101*1000000/110000 = 10101 * 9.09 = 91818
pls check and let me know, if you have any observation..
puys wats the answer for the above question? we didnt close this question yet.. pls confirm the answer..
Puys,I m new to Fermat Theorem..But plz help..
Fermat theorem : for any prime number p , (p-1)times,same digit is repeated , then that number formed is exactly divisible by p ..
So if we chk for 9999(4 times),then it shud be exactly divisible by 5..but it is not the case...
Is there any mistake?