Number System - Questions & Discussions

MBA CAT 2022
1837
Since, 100 = 4*25

1) N = 101*102*103*197*198*199 = 1*2*3*(-3)*(-2)*(-1) (mod 25) = 14 (mod 25)
N = 101*102*103*197*198*199 = 0 (mod 4)

=> N = 25k + 14 = 4n
=> n = 6k + 3 + (k + 2)/4, n will be integer for k = 2
=> N = 100m + 64
So, last two digits will be 64

2) N = 65*29*37*63*71*87 = 1*1*1*3*3*3 (mod 4) = 3 (mod 4)
N = 65*29*37*63*71*87 = 15*4*12*13*21*12 (mod 25) = 60*156*252 (mod 25) = 10*6*2 (mod 25) = 20 (mod 25)
=> N = 4n + 1 = 25k + 20
=> N = 100m + 95
=> Last two digits will be 95

hey 4n + 1 = 25k + 20 is fine..but how did u conclude to 100m + 95
1838
hey thank you so much..the concept is really getting into my head now..this means that fermets theorem is a special case of the above stated theorem which says /p (where p is a prime number) has a remainder equal to 1.
so eulers concept can be applied to every other problem of remainders involving big values????


Yes it can be applied, but some cases, it is advisable to use options.

For example the question like:

21^140 mod 100.

Either use options or use the last two digits rule. I guess the second option is faster.
1839
dntcheatme Says
hey 4n + 1 = 25k + 20 is fine..but how did u conclude to 100m + 95


4n + 1 = 25K + 20 => 4n = 25K + 19

Dividing RHS by 4 and taking the remainder, 4n1 = K + 3 , here K = 1 would satisfy the equation.

Putting the value of K in the original equation, 4n + 1 = 25K + 20, K =1, n = 11.

Now this equation will satisfy for the an AP of K with a common difference of coefficient of the other equation, that is, AP with an Common Difference of 4. And same is true for n with a common difference of 25.

hence 1st set of value is K =1, n = 11
2nd solution is K = 5, n = 36 , you can try to substitute and check:lookround:.
Similarly other values for K = 9, n = 61.

The combined equation can be written as LCM (of 4 and 25) + (1st solution of the equation)

hence 100A + 45, I guess it should be 45 instead of 95.

@chillfactor : Correct me if I am wrong anywhere.

@dntcheatme: The above is "chinese Remainder Theorem".

And also the coefficient has to be Coprimes and thats why Chillfactor has split it into 25 and 4 and NOT 50 and 2. :thumbsup:
1840
dntcheatme Says
hey 4n + 1 = 25k + 20 is fine..but how did u conclude to 100m + 95


Sorry for the typo, actually its N = 4n + 3 = 25k + 20

=> 4n + 3 = 25k + 20
=> n = (6k + 4) + (k + 1)/4, least value of k for which n is integer is 3

=> N is of form 100m + (25*3 + 20) or 100m + 95

@KinjiPG ....yeah you are right, sorry for the typo.
1841

hey dis thread is going blank..please post some good questions on number system for further discussion...

1842
#1 Let S be the set of all positive integers that are smaller than or equal to 21600 and are co prime to it. Let k be the number of elements in S. Let s be the sum of all these elements. Then find the value of (s + k).
1843
chillfactor Says
#1 Let S be the set of all positive integers that are smaller than or equal to 21600 and are co prime to it. Let k be the number of elements in S. Let s be the sum of all these elements. Then find the value of (s + k).

All the numbers dat do not have a factor as 2,3,5 will be co-prime to 21600
now,
numbers divisible by 2 = 10800
numbers divisible by 3 = 7200
numbers divisible by 5 = 4320
numbers divisible by 6 = 3600
numbers divisible by 10 = 2160
numbers divisible by 15 = 1440
numbers divisible by 30 = 720

now the total number of integers that are divisble by 2,3,5 =
{(10800+7200+4320)-(3600+2160+1440)+720}
= 15840
therefore the total number of positive integers less than 21600 dat are co prime to 21600 are (21600-15840) i.e 5760
therefore the value of k is 5760

Cannot understand how to find the value of s..
and is the above value of k correct???
1844
All the numbers dat do not have a factor as 2,3,5 will be co-prime to 21600
now,
numbers divisible by 2 = 10800
numbers divisible by 3 = 7200
numbers divisible by 5 = 4320
numbers divisible by 6 = 3600
numbers divisible by 10 = 2160
numbers divisible by 15 = 1440
numbers divisible by 30 = 720

now the total number of integers that are divisble by 2,3,5 =
{(10800+7200+4320)-(3600+2160+1440)+720}
= 15840
therefore the total number of positive integers less than 21600 dat are co prime to 21600 are (21600-15840) i.e 5760
therefore the value of k is 5760

Cannot understand how to find the value of s..
and is the above value of k correct???

do i have to proceed in the same way as above for calculating the value of "s" and if yes then this involves a calculation of very big numbers...
is there a smarter way of doing dis problem???
1845
All the numbers dat do not have a factor as 2,3,5 will be co-prime to 21600
now,
numbers divisible by 2 = 10800
numbers divisible by 3 = 7200
numbers divisible by 5 = 4320
numbers divisible by 6 = 3600
numbers divisible by 10 = 2160
numbers divisible by 15 = 1440
numbers divisible by 30 = 720

now the total number of integers that are divisble by 2,3,5 =
{(10800+7200+4320)-(3600+2160+1440)+720}
= 15840
therefore the total number of positive integers less than 21600 dat are co prime to 21600 are (21600-15840) i.e 5760
therefore the value of k is 5760

Cannot understand how to find the value of s..
and is the above value of k correct???


brother there is an easy method to calculate the value of k , first do the prime factorization of the number which in this case is 21600= 2^5x3^3x5^2.
then use the following Euler's formula 21600(1-1/2)(1-1/3)(1-1/5) = 5760 i'm also struck in finding the value of s..!!!
1846
chillfactor Says
#1 Let S be the set of all positive integers that are smaller than or equal to 21600 and are co prime to it. Let k be the number of elements in S. Let s be the sum of all these elements. Then find the value of (s + k).

All the numbers dat do not have a factor as 2,3,5 will be co-prime to 21600
now,
numbers divisible by 2 = 10800
numbers divisible by 3 = 7200
numbers divisible by 5 = 4320
numbers divisible by 6 = 3600
numbers divisible by 10 = 2160
numbers divisible by 15 = 1440
numbers divisible by 30 = 720

now the total number of integers that are divisble by 2,3,5 =
{(10800+7200+4320)-(3600+2160+1440)+720}
= 15840
therefore the total number of positive integers less than 21600 dat are co prime to 21600 are (21600-15840) i.e 5760
therefore the value of k is 5760

Cannot understand how to find the value of s..
and is the above value of k correct???

brother there is an easy method to calculate the value of k , first do the prime factorization of the number which in this case is 21600= 2^5x3^3x5^2.
then use the following Euler's formula 21600(1-1/2)(1-1/3)(1-1/5) = 5760 i'm also struck in finding the value of s..!!!


friends, s = sum of all positive integers less than 21600 and co-prime to it
= (21600/2)*5760
Now u can proceed further.
It is a theorem that sum of all co-primes less than N = (N/2)*number of co-primes less than N:cheerio:
1847
brother there is an easy method to calculate the value of k , first do the prime factorization of the number which in this case is 21600= 2^5x3^3x5^2.
then use the following Euler's formula 21600(1-1/2)(1-1/3)(1-1/5) = 5760 i'm also struck in finding the value of s..!!!

do i have to proceed in the same way as above for calculating the value of "s" and if yes then this involves a calculation of very big numbers...
is there a smarter way of doing dis problem???

friends, s = sum of all positive integers less than 21600 and co-prime to it
= (21600/2)*5760
Now u can proceed further.
It is a theorem that sum of all co-primes less than N = (N/2)*number of co-primes less than N:cheerio:


Yeah right, it is a theorem that sum of all co-primes less than N = (N/2)*number of co-primes less than N

If anyone interested in the proof, then its as follows:-

For any number N = (a^x)*(b^y)*(c^z)*...
We need to find the number of numbers less than or equal to N such that they are coprime to N, which means they are coprime to a, b, c, ....
Hence among every abc... numbers there will be abc ab bc ac + a + b + c 1 such numbers
So, total such numbers is given by:-
(N) = N(abc ab bc ac + a + b + c 1)/abc = N(1 1/a)(1 1/b)(1 1/c)..
This is Eulers number, so Eulers number of any number represents the number of numbers less than or equal to N and coprime to N.

21600 = 2^5*3^3^5^2
So, k = (21600) = 21600(1/2)(2/3)(4/5) = 5760

Now suppose p is one such number than (N p) will also be a such number.
Hence, s = p + q + r + ...... + (N r) + (N q) + (N p)
s = (N p) + (N q) + (N r) + ..... + r + q + p
Adding them we will get,
2s = N + N + N + .... + N (k times, as there k such numbers)
So, s = kN/2 = N* (N)/2
So, s(21600) = 5760*21600/2
Hence, s + k = 62213760
1848
Yeah right, it is a theorem that sum of all co-primes less than N = (N/2)*number of co-primes less than N

If anyone interested in the proof, then its as follows:-

For any number N = (a^x)*(b^y)*(c^z)*...
We need to find the number of numbers less than or equal to N such that they are coprime to N, which means they are coprime to a, b, c, ....
Hence among every abc... numbers there will be abc - ab - bc - ac + a + b + c - 1 such numbers
So, total such numbers is given by:-
(N) = N(abc - ab - bc - ac + a + b + c - 1)/abc = N(1 - 1/a)(1 - 1/b)(1 - 1/c)..
This is Euler's number, so Euler's number of any number represents the number of numbers less than or equal to N and coprime to N.

21600 = 2^5*3^3^5^2
So, k = (21600) = 21600(1/2)(2/3)(4/5) = 5760

Now suppose p is one such number than (N - p) will also be a such number.
Hence, s = p + q + r + ...... + (N - r) + (N - q) + (N - p)
s = (N - p) + (N - q) + (N - r) + ..... + r + q + p
Adding them we will get,
2s = N + N + N + .... + N (k times, as there k such numbers)
So, s = kN/2 = N* (N)/2
So, s(21600) = 5760*21600/2
Hence, s + k = 62213760

thanks brother..i did know the eulers concept..but didnt know what does it actually represent..but fr the start i am happy dat atleast i got a part of it rite..used the rite concept although a pretty long one:)
1849

some more good questions.....

1850
Find sum of the following series for n terms:-

6*1! + 13*2! + 22*3! + 33*4! + 46*5! + .... (n terms)
1851
Find sum of the following series for n terms:-

6*1! + 13*2! + 22*3! + 33*4! + 46*5! + .... (n terms)

i reached to a point where i have to calculate the sum of n!..am i gng rite??and what is the sum of (1! + 2! +.............n!)
1852

In a certain examination paper, there are n questions. For j = 1, 2 n, there are 2^n-j students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then find the value of n.

1853
Find sum of the following series for n terms:-

6*1! + 13*2! + 22*3! + 33*4! + 46*5! + .... (n terms)

@chillfactor

the general term is (n^2+4n+1)*n!
now summation of n*n! can be calculated as (n+1)-1.
But no idea how to get the summation of other two terms:lookround::lookround:
1854
@chillfactor

the general term is (n^2+4n+1)*n!
now summation of n*n! can be calculated as (n+1)-1.
But no idea how to get the summation of other two terms:lookround::lookround:


(n + 4n + 1) = (n + 2)(n + 1) + (n + 1) - 2
=> (n + 4n + 1)*n! = (n + 2)! + (n + 1)! - 2(n!)

=> Summation will be (n + 2)! + 2{(n + 1)!} - 4
1856
(n + 4n + 1) = (n + 2)(n + 1) + (n + 1) - 2
=> (n + 4n + 1)*n! = (n + 2)! + (n + 1)! - 2(n!)

=> Summation will be (n + 2)! + 2{(n + 1)!} - 4
sirjee could not understand the part in bold.plz explain
1857

1+2+4+........+2^11=4095
12 questions in total.


sirjee plz explain a little