@Faruq said:Find Remainder when 10^ 2 + 11^ 2 + 12^2 + . . . . 28^ 2 is divided by 19
Remainder = (-9)^2 + (-8)^2 + .....(-1)^2 + 1^2 + 2^2 +...9^2
= 2*( 1^2 + 2^2 + 3^2 ...9^2 ) = 2*(10)*(19)/6
= 0
Find remainder when 3^0 + 3^1 + 3^2 + . . . . .201 terms is divided by 13
a. 0 b. 12 c. 3 d. None of these
a. 0 b. 12 c. 3 d. None of these
@Faruq said:Find remainder when 3^0 + 3^1 + 3^2 + . . . . .201 terms is divided by 13 a. 0 b. 12 c. 3 d. None of these
a.0
1 + 3 + 9 = 13
Given series becomes
(1 + 3 + 9) + 27(1 + 3 + 9) + ... upto 201 terms
So the series is divisible by 13
1 + 3 + 9 = 13
Given series becomes
(1 + 3 + 9) + 27(1 + 3 + 9) + ... upto 201 terms
So the series is divisible by 13
@Faruq said:Find remainder when 3^0 + 3^1 + 3^2 + . . . . .201 terms is divided by 13 a. 0 b. 12 c. 3 d. None of these
3^0+3^1+3^2 = 13
Hence Remainder = 0
@Faruq said:Find remainder when 3^0 + 3^1 + 3^2 + . . . . .201 terms is divided by 13 a. 0 b. 12 c. 3 d. None of these
0
Find remainder when 6^(65^56) mod 43
@Koushik98 said:yeah thats ok.....but see... the odd factors of 1000 are 5, 25,125,1Taking n=5,we are getting avg=200,hence 198+ 199+......+202Taking n=25, we are getting avg=40,hence 28+29+30+.....+52But for avg =62.5,we are getting n=16 on what logic are we taking n=16 or vice versa avg=62.5? is it by trial and error??? Hope my query is clear now.......
Ok let me explain you from scratch. Here we go...
As i already explained earlier that there could be two cases: odd number of terms and even number of terms.
I dont think you have any issues understanding with odd number of factors. Still there is an important point to make here. The odd factors of 1000 are 1,5,25, and 125. We cannot take n as 125 because then the consecutive numbers will not be natural numbers and will involve some negative integers as well. Also, we are supposed to consider odd factors greater than 1. So only valid values of n are 5 and 25.
When there are even number of terms, since these numbers are consecutives and since the middle term must be the average of two middle terms, which are consecutive terms, hence the middle term is a rational number of the form X.5
The least divisor, i.e. n, of 1000, such that we get the quotient in the form X.5 is 16. The we could also have other values of n as 80 and 400 such that the quotient are 12.5 and 2.5 respectively. But the problem with the latter values of n =80 and 400 is that we will not be able to consecutive natural numbers as the nubers will extend towards negative side on the number line. So, the only valid value of n is 16.
So there is not trial and error, but pure logic. Hope you get a transparent picture now :)
@Faruq said:Find remainder when 6^(65^56) mod 43
6 ?
6^(65^56) mod 43
E(43) = 42
65^56 mod 42 = 25
6^(42k + 25) mod 43 = 6^25 mod 43 = 6 ..
@Faruq said:Find Remainder when 888222888222. . . . . up to 9235 digits divided by 53Find Remainder when 10^ 2 + 11^ 2 + 12^2 + . . . . 28^ 2 is divided by 19
For the second one the remainder is 0.
@Faruq said:Find remainder when 6^(65^56) mod 43
remainder is 6
cyclicity lagaya...
6/43 r=6
6^2/43 r =36
6^3/43 r =1
the cycle repeats..so cyclicity of 3
65^56/3
E(3) = 2
so R=1
6^65^56 is of the form 3k+1
hence remainder = 6
@Faruq said:Find remainder when 6^(65^56) mod 43
6?
65^56 mod 42 = 23^56 mod 42
23^56 mod 6 = (-1)^56 = 1 mod 6
23^56 mod 7 = 23^54 * 23^2 = 1 * 2^2 = 4 mod 7
6a + 1 = 7b + 4
6a = 7b + 3
a = 4
b = 3
6a + 1 = 25
65^56 mod 42 = 25
6^25 mod 43 = (36^12 * 6) mod 43
= (-7)^12 * 6 mod 43
= 49^6 * 6 mod 43
= 6^6 * 6 mod 43
= 36^3 * 6 mod 43
= (-7)^3 * 6 mod 43
= 49 * -7 * 6 mod 43
= 6 * 6 * -7 mod 43
= 36 * -7 mod 43
= -7 * -7 mod 43
= 49 mod 43
= 6 mod 43
65^56 mod 42 = 23^56 mod 42
23^56 mod 6 = (-1)^56 = 1 mod 6
23^56 mod 7 = 23^54 * 23^2 = 1 * 2^2 = 4 mod 7
6a + 1 = 7b + 4
6a = 7b + 3
a = 4
b = 3
6a + 1 = 25
65^56 mod 42 = 25
6^25 mod 43 = (36^12 * 6) mod 43
= (-7)^12 * 6 mod 43
= 49^6 * 6 mod 43
= 6^6 * 6 mod 43
= 36^3 * 6 mod 43
= (-7)^3 * 6 mod 43
= 49 * -7 * 6 mod 43
= 6 * 6 * -7 mod 43
= 36 * -7 mod 43
= -7 * -7 mod 43
= 49 mod 43
= 6 mod 43
@Faruq said:Find Remainder when 888222888222. . . . . up to 9235 digits divided by 53Find Remainder when 10^ 2 + 11^ 2 + 12^2 + . . . . 28^ 2 is divided by 19
Guys, i solved this question about 4 years back and the problem read the "remainder when divided by 5^3". I can see that this problem has been misquoted by someone (the original power sign being omitted)and over the time has been wrongly transmitted on several forums and question banks.
While the original question was on the lines of actual CAT, testing the divisibility criteria of 125 for which you need to check just the last three digits of the given series. So all you needed to deduce was the last three digits of the expression.
This altered version of the problem is a futile attempt in that it does not help you much in preparing for CAT or any other known test prevalent globally. One should not use other tools such as Euler's theorem just for the heck of it when the problem itself is bad.
Always verify the source and authenticity of the problem before practicing other wise you will end up wasting your valuable time on nonsensical problems.
STAY AWAY FROM THIS PROBLEM.
@Faruq said:Find remainder when 3^0 + 3^1 + 3^2 + . . . . .201 terms is divided by 13 a. 0 b. 12 c. 3 d. None of these
3 aayega ..
cycle check kiya...
3^0/13 r=1
3^1/13 r=3
3^2/13 r =4
3^3/13 r =1..the repeats so on...
hence 1+3+4 + 1+3+4 +1+3+4 +......till 201/3 terms
8*67/13
16/13 , so R=3
OA kya hai?
@Faruq said:Find Remainder when 10^ 2 + 11^ 2 + 12^2 + . . . . 28^ 2 is divided by 19
0?
10^2 + 11^2 + ............28^2 mod 19 = (-9)^2 + (-8)^2 + ...............+ (9)^2 mod 19
= 2*(5 + 7 + 11 + 17 + 6 + 16 + 9 + 4 + 1) mod 19 = 152 mod 19 = 0
@bodhi_vriksha Excellent memory brother...
4 years back and you still remember it..
Yahan toh 4 din pahle ka yaad nai rehta..
Yahan toh 4 din pahle ka yaad nai rehta..
@bodhi_vriksha was pondering over this question for the last 10 mins...good that you pointed it out..
P.S. guys please see that you post correct questions..it wastes a lot of time trying questions that are not even correct....
@Faruq said:Find remainder when 3^0 + 3^1 + 3^2 + . . . . .201 terms is divided by 13 a. 0 b. 12 c. 3 d. None of these
3^0 + 3^1 + 3^2 mod 13 = (1 + 3 + 9)(1 + 3 + 9).....67 pairs mod 13 = 0 ?
@saurav205 said:@bodhi_vriksha was pondering over this question for the last 10 mins...good that you pointed it out..P.S. guys please see that you post correct questions..it wastes a lot of time trying questions that are not even correct....
Happy to help :)