Number System - Questions & Discussions

MBA CAT 2022
1858
abhi108890 Says
sirjee could not understand the part in bold.plz explain


nth term = (n + 4n + 1)n! = {(n + 2)(n + 1) + (n + 1) - 2}*n!
= (n + 2)(n + 1)*n! + (n + 1)*n! - 2*n!
= (n + 2)! + (n + 1)! - 2*n!

Now, summation will look like
S = (3! + 2! - 2) + (4! + 3! - 2*2!) + ..... + {(n + 2)! + (n + 1)! - 2*n!}
= (n + 2)! + 2*(n + 1)! - 4
1859

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

friend can u elaborate it bit more
1861
abhi108890 Says
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.

Say, there were n questions in total .

Then 1st one answered wrongly by 2^(n-1)
2nd one answered wrongly by 2^(n-2)
and thus,
nth one answered wrongly by 2^(1-1)
So it's 2^(n-1)+2^(n-2)+.....+2^0=4095
2^(n-1)+2^(n-2)+....+2^0=2^n - 1 (1+2=4-1,1+2+4=8-1 and so on)
Now 4095 can be written as 2^12-1 and here n=12
So answer is 12.
If a questions say like this , I mean 2^n-j then total sum can only be in the form (2^n)-1 .

sirjee i got you but i am still confused by what was meant by "j or more questions" in the question.
plz clarify that part.
1862

me too, confused...

1863

the number of digits in the cube root of a 29 digit number is
a)10 b)9 c)8 d)cant say

pls help me...wid d xplanation...:lookround:

1864
the number of digits in the cube root of a 29 digit number is
a)10 b)9 c)8 d)cant say

pls help me...wid d xplanation...:lookround:

The smallest 10 digit no starts with 1 and is followed by 9 zeroes. When we take the cube of it, we get a 28 digit no.

Also, the smallest 11 digit no. starts with 1 and is followed by 10 zero's. When we take the cube of it, we get a 31 digit no.

Hence, the no. of digits in the cube root of a 29 digit number should be a) 10.
1865

hey guys ...i am in need of totalgadha number system ebook.
can anybody please give me the book. i will pm him my email-id for the book....plz!!!:lookround:

1866

although i understood the problem but i am not able to simplify it to 2^n -1.not able to understand (1+2=4-1,1+2+4=8-1 and so on)
in
2^(n-1)+2^(n-2)+....+2^0=2^n - 1 (1+2=4-1,1+2+4=8-1 and so on)
please help me friends

1867
ankitripathi Says
friend can u elaborate it bit more

Say, there were n questions in total .

Then 1st one answered wrongly by 2^(n-1)
2nd one answered wrongly by 2^(n-2)
and thus,
nth one answered wrongly by 2^(1-1)
So it's 2^(n-1)+2^(n-2)+.....+2^0=4095
2^(n-1)+2^(n-2)+....+2^0=2^n - 1 (1+2=4-1,1+2+4=8-1 and so on)
Now 4095 can be written as 2^12-1 and here n=12
So answer is 12.
If a questions say like this , I mean 2^n-j then total sum can only be in the form (2^n)-1 .

sirjee i got you but i am still confused by what was meant by "j or more questions" in the question.
plz clarify that part.

deepshii Says
me too, confused...

although i understood the problem but i am not able to simplify it to 2^n -1.not able to understand (1+2=4-1,1+2+4=8-1 and so on)
in
2^(n-1)+2^(n-2)+....+2^0=2^n - 1 (1+2=4-1,1+2+4=8-1 and so on)
please help me friends


Hey friends!
It seems this question is creating a lot of confusion. To end it all, here is my take on this.

2^(n-j) student answered j or more questions wrongly.
For j = 1, 2^(n-1) students answered 1 or more questions wrongly.
For j = 2, 2^(n-2) students answered 2 or more questions wrongly.
It means the number of students who answered exactly 1 question wrong is 2^(n-1)-2^(n-2).
Similarly, the number of students who answered exactly 2 questions wrong = 2^(n-2)-2^(n-3) and so on.
Now total wrong answered question is 4095 which we get due to the contribution of exactly 1 wrong, exactly 2 wrong ... and so on.
It means
1*(2^(n-1)-2^(n-2))+2*(2^(n-2)-2^(n-3))+....+(n-1)*(2^(n-(n-1))-2^(n-n))+n*2^(n-n) = 4095
2^(n-1)+2^(n-2)+....+2^1+1=4095
2^n-1=4095
2^n=4096
n=12.
1868
the number of digits in the cube root of a 29 digit number is
a)10 b)9 c)8 d)cant say

pls help me...wid d xplanation...:lookround:


Number of digits can be easily found out with the help of logarithm.
The characteristics of a number is one less than the number of digits in that number. It is based on the scientific notation of the numbers.
12345 = 1.2345*10^4, so the characteristics is 4 here. When you take the logarithm on base 10 of 12345, it will be 4...., the digits after the decimal decided by the logarithmic table and called mantissa.
Now, x = (a 29 digit number)^(1/3).
Taking logarithm (on base 10) of both sides,
log x = (1/3)*log(a 29 digit number)
log x = (1/3)*(28.mantissa)
log x = 9.something
So x is a 10 digit number.
1869

duplicate post, sorry

1870
Number of digits can be easily found out with the help of logarithm.
The characteristics of a number is one less than the number of digits in that number. It is based on the scientific notation of the numbers.
12345 = 1.2345*10^4, so the characteristics is 4 here. When you take the logarithm on base 10 of 12345, it will be 4...., the digits after the decimal decided by the logarithmic table and called mantissa.
Now, x = (a 29 digit number)^(1/3).
Taking logarithm (on base 10) of both sides,
log x = (1/3)*log(a 29 digit number)
log x = (1/3)*(28.mantissa)
log x = 9.something
So x is a 10 digit number.


well d answer is right for dis one..bt i m nt sure dis method works accurately evrytym..!

Q)no. of digits in (2PQR)^4..where 2PQR is a 4 digit no.
->acc to ur logic,
logx=4log(2PQR)
or,logx=4*3.mantissa
or,logx=12.mantissa
or,x=13 digit no..
whereas d answer is 'a 14 digit no.'

alternatively,by another method posted in dis thread..
for smallest no. considerin d constraint given=(2000)^4
it's a 14 digit result..=1600......
and for d largest no. takin d constraint given.=(2999)^4
~(3000)^4
=8100....
=14 digit no.
so any no. of d form (2PQR)^4 is a 14 digit no.
1871
the number of digits in the cube root of a 29 digit number is
a)10 b)9 c)8 d)cant say

pls help me...wid d xplanation...:lookround:

well d answer is right for dis one..bt i m nt sure dis method works accurately evrytym..!

Q)no. of digits in (2PQR)^4..where 2PQR is a 4 digit no.
->acc to ur logic,
logx=4log(2PQR)
or,logx=4*3.mantissa
or,logx=12.mantissa
or,x=13 digit no..
whereas d answer is 'a 14 digit no.'

alternatively,by another method posted in dis thread..
for smallest no. considerin d constraint given=(2000)^4
it's a 14 digit result..=1600......
and for d largest no. takin d constraint given.=(2999)^4
~(3000)^4
=8100....
=14 digit no.
so any no. of d form (2PQR)^4 is a 14 digit no.


The method is never wrong as it is not a guesswork or something like that but a proven fact of mathematics although you can not apply it to solve all kinds of problems relating to finding the number of digits.
In the previous question of cube root, we have to divide the logarithm of the number by 3, and the division results in a number between 9 and 10.
But in your this example, you have to multiply by 4, and the multiplication of the characteristics results in 12, but do not forget to multiply the mantissa by 4 which will give you a number between 1 and 2. So the lagarithm of the number is between 13 and 14, ie, it is 13.something.
So the number of digits will be 14.
1872
The method is never wrong as it is not a guesswork or something like that but a proven fact of mathematics although you can not apply it to solve all kinds of problems relating to finding the number of digits.
In the previous question of cube root, we have to divide the logarithm of the number by 3, and the division results in a number between 9 and 10.
But in your this example, you have to multiply by 4, and the multiplication of the characteristics results in 12, but do not forget to multiply the mantissa by 4 which will give you a number between 1 and 2. So the lagarithm of the number is between 13 and 14, ie, it is 13.something.
So the number of digits will be 14.


ya..dats wat..! d method is definitely rite..bt i ws askin abt d approximation dat ' a number is one less than the number of digits in that number......'

it's hard to guess d frst mantissa digit..in dis case log(2PQR) should hv given '3.3...' or a higher no. as d result (3.3*4=13.2) den belongs to d req answer..

bt how can 1 judge is dis givin '3.3...' or '3.09...'(3.09*4=12.36,as d case ud hv been if d question '1PQR' wer given "log(1234)=3.09")
??

d method's gr8 bt is der a way to minimise d risk involved..??
1873
ya..dats wat..! d method is definitely rite..bt i ws askin abt d approximation dat ' a number is one less than the number of digits in that number......'

it's hard to guess d frst mantissa digit..in dis case log(2PQR) should hv given '3.3...' or a higher no. as d result (3.3*4=13.2) den belongs to d req answer..

bt how can 1 judge is dis givin '3.3...' or '3.09...'(3.09*4=12.36,as d case ud hv been if d question '1PQR' wer given "log(1234)=3.09")
??

d method's gr8 bt is der a way to minimise d risk involved..??


I have already mentioned that may be this can not be applied in all the questions and may be the question is framed itself to be solved by the tricks and calculations taking examples rather than using any mathematical theory.
In this particular case, it is helpful if you remember that log 2 is .3010 and so the multiplication by 4 will definitely result in a number greater than 1.
1874

yup... !
thanx nyways..:thumbsup:

1875

hey guys ...i am in need of totalgadha number system ebook.
can anybody please give me the book. i will pm him my email-id for the book....plz!!!

1876
Hey friends!
It seems this question is creating a lot of confusion. To end it all, here is my take on this.

2^(n-j) student answered j or more questions wrongly.
For j = 1, 2^(n-1) students answered 1 or more questions wrongly.
For j = 2, 2^(n-2) students answered 2 or more questions wrongly.
It means the number of students who answered exactly 1 question wrong is 2^(n-1)-2^(n-2).
Similarly, the number of students who answered exactly 2 questions wrong = 2^(n-2)-2^(n-3) and so on.
Now total wrong answered question is 4095 which we get due to the contribution of exactly 1 wrong, exactly 2 wrong ... and so on.
It means
1*(2^(n-1)-2^(n-2))+2*(2^(n-2)-2^(n-3))+....+(n-1)*(2^(n-(n-1))-2^(n-n))+n*2^(n-n) = 4095
2^(n-1)+2^(n-2)+....+2^1+1=4095
2^n-1=4095
2^n=4096
n=12.


Perfect bhai!!! :cheerio:.No confusion now.
1877

Okay. Am a newbie on this awesome site! I've got a doubt in the following example, which is seemingly easy but I haven't yet understood the concept properly! :lookround:

If A=1^1*2^2*3^3*.........*100^100,then how many zeroes will be there at the end of A?

Thanks in advance puys!! :cheerio:

1878

Hi friends...I'm new here...May i knw where can i get material for cat preparation.?

Thanks
Jatin