Number System - Questions & Discussions

abhinav90 · November 4, 2009, 1:31am

X=[579 * 580 * 581 * 582 * ......... * 635] + [ 52 * 57 ]
Find the remainder when X is divided by 57! .

a) 52
b) 0
c) 57! -1
d) None of these.

52 is the answer...
The Approach was:
[579*580*581*........*635] + [52*57] % 57! =
[579*580*581*.....*635]%57! + 52%56!

Now [579*580*581*.....*635] can be written as 635! / 578!
Hence first term reduces to 635! / 578! * 57! = 635 C 57 which will give 0 remainder...
hence we are left with 52%56! and hence 52 is the answer...

thagudu · November 4, 2009, 2:08am

52 is the answer...
The Approach was:
+ % 57! =
%57! + 52%56!

Now can be written as 635! / 578!
Hence first term reduces to 635! / 578! * 57! = 635 C 57 which will give 0 remainder...
hence we are left with 52%56! and hence 52 is the answer...

@ abhinav (good approach reducing it to combinations man).Btw there is a rule which states that the product of any n consecutive natural numbers is divisible by n! is there not?Cos then the first part becomes divisible by 57! leaving 52 as the remainder for the sum. Am a little doubtful bout this could someone clarify or confirm?

nksbits · November 4, 2009, 4:05am

yes it can be proved...

let x be any random natural number
(x+1)*(x+2)........(x+n)=(x+n)!/x!
divide this by n!
we get x+nCn
hence remainder =0

kashyap.rastogi · November 4, 2009, 5:40am

52 is the answer...
The Approach was:
+ % 57! =
%57! + 52%56!

Now can be written as 635! / 578!
Hence first term reduces to 635! / 578! * 57! = 635 C 57 which will give 0 remainder...
hence we are left with 52%56! and hence 52 is the answer...

Didnt get the bold part. Can someone please elaborate a bit..

nksbits · November 4, 2009, 5:48am

can be written as 635! / 578!
so /57! can be written as 635! / 578!*57!
635! / 578! * 57!=635C57
hence remainder =0

kashyap.rastogi · November 4, 2009, 6:01am

can be written as 635! / 578!
so /57! can be written as 635! / 578!*57!
635! / 578! * 57!=635C57
hence remainder =0

Dat i can understand, how combinations are calculated. But, how remainder is ZERO?? Is it that xCy will always have a remainder zero??

nksbits · November 4, 2009, 6:05am

xCy is number of ways of choosing y objects from x.
this is always a natural number

alley9515 · November 4, 2009, 1:40pm

Hmmm...
method is kind of crude and not very elegant..
but i doubt that is the correct approach.
Your answer doesn't match anyways.
The answer given is 195

This is the exact question:

Given that 2^2010 is a 606-digit number whose first digit is 1, how many elements of the set S = {2^0, 2^1, 2^2, , 2^2009} have a first digit of 4?

Solution: The smallest power of 2 with a given number of digits has a first digit of 1, and there are some elements of S with n digits and for each positive integer, n 605, so there are 605 elements of S whose first digit is 1.
Next, if the first digit of 2^k is 1, then the first digit of 2^(k+1) is either 2 or 3, and the first digit of 2^(k+2) is either 4, 5, 6, or 7. Therefore there are 605 elements of S whose first digit is 2 or 3, 605 elements whose first digit is 4, 5, 6, or 7, and 2010 - 3(605) = 195 whose first digit is 8 or 9. Finally, note that the first digit of 2^k is 8 or 9 if and only if the first digit of 2^(k - 1) is 4, so there are 195 elements of S whose first digit is 4.

abhinav90 · November 5, 2009, 1:53am

This is the exact question:

Given that 2^2010 is a 606-digit number whose first digit is 1, how many elements of the set S = {2^0, 2^1, 2^2, , 2^2009} have a first digit of 4?

Solution: The smallest power of 2 with a given number of digits has a first digit of 1, and there are some elements of S with n digits and for each positive integer, n 605, so there are 605 elements of S whose first digit is 1.
Next, if the first digit of 2^k is 1, then the first digit of 2^(k+1) is either 2 or 3, and the first digit of 2^(k+2) is either 4, 5, 6, or 7. Therefore there are 605 elements of S whose first digit is 2 or 3, 605 elements whose first digit is 4, 5, 6, or 7, and 2010 - 3(605) = 195 whose first digit is 8 or 9.

I m not getting this 605 thing...Will this always be true?Please elucidate more,if possible with an example..

Finally, note that the first digit of 2^k is 8 or 9 if and only if the first digit of 2^(k - 1) is 4, so there are 195 elements of S whose first digit is 4.

ok..got this...

alley9515 · November 5, 2009, 7:24am

I m not getting this 605 thing...Will this always be true?Please elucidate more,if possible with an example..

ok..got this...

Yes this will always happen. Lets take a shorter number as 2^x whose first digit will be 1..lets check for x=7, 2^7=128...all the numbers which are of the form, 2^n, where n

amitshanky · November 5, 2009, 4:24pm

1)LCM of first 100 natural numbers is K, then LCM of first 105 natural numbers will be

k,101k,10403k,100k

shanks4mba · November 5, 2009, 4:28pm

1)LCM of first 100 natural numbers is K, then LCM of first 105 natural numbers will be

k,101k,10403k,100k

LCM(1, 100) = all the 25 primes..
so from 101 to 105 we have 101,103 as prime
hence it shud be 101*103k = 10403k

ankurb · November 5, 2009, 6:14pm

1)LCM of first 100 natural numbers is K, then LCM of first 105 natural numbers will be

k,101k,10403k,100k

LCM(first 100 numbers) = product of highest powers of all prime numbers = k
LCM(first 105 numbers) = 101 *103 *k = 10403k

madnikhil · November 5, 2009, 6:51pm

1)LCM of first 100 natural numbers is K, then LCM of first 105 natural numbers will be

k,101k,10403k,100k

EAsy one
101 and 103 are prime no.
so,
10403k will be answer..
:sneaky:

mba_25807360 · November 5, 2009, 8:00pm

AB is a two digit number in base n such that (AB)to base n = 5(BA)to base n.
Which of the following is a possible vale of n?

(1) 12
(2) 14
(3) 16
(4) 21

manojmsr · November 5, 2009, 9:28pm

since for 100numbers the lcm is k for 105 it ll be k*101*103 as they are the oly prime nos in btw 100 to 105....

So i guess the answer is 10403k i guess..

divishth · November 5, 2009, 9:36pm

AB is a two digit number in base n such that (AB)to base n = 5(BA)to base n.
Which of the following is a possible vale of

n?

(1) 12
(2) 14
(3) 16

(4) 21

Answer is 21

abhinav90 · November 6, 2009, 1:33am

divishth Says
Answer is 21

Plz post approach...I m uneasy with bases...

shanks4mba · November 6, 2009, 4:38am

AB is a two digit number in base n such that (AB)to base n = 5(BA)to base n.
Which of the following is a possible vale of n?
(1) 12
(2) 14
(3) 16
(4) 21

given (AB)n = 5(BA)n
An + B = 5Bn + 5A
A(n-5) = (5n - 1)B
A/B = 5n-1/n-5
now for n = 12 we get A/B = 59/7...which not possible as A shub be from 0 to 11
now for n = 14 we get A/B = 69/9 = 23/3 not possible
now for n = 16 we get A/B = 79/11..not possible
now for n = 21 we get A/B = 104/16 = 13/2...which can be possible in the base system of 21..hence ans is 21

abhinav90 · November 6, 2009, 11:39am

given (AB)n = 5(BA)n
An + B = 5Bn + 5A
A(n-5) = (5n - 1)B
A/B = 5n-1/n-5
now for n = 12 we get A/B = 59/7...which not possible as A shub be from 0 to 11
now for n = 14 we get A/B = 69/9 = 23/3 not possible
now for n = 16 we get A/B = 79/11..not possible
now for n = 21 we get A/B = 104/16 = 13/2...which can be possible in the base system of 21..hence ans is 21

how to check?
Why did you accepted 13/2=6.5 and not 23/3 = 7.666 ?
plz edify !