For Qa please participate in this thread also to get your fundas clear...
http://www.pagalguy.com/discussions/how-to-crack-section-i-of-the-cat-25082497
ps- Number system week
Please explain the approach of the below questions.
1 126 natural numbers are put side by side in ascending order to create a large number 12345................125126. What will be the remainder when N is divided by 5625 ?
2.Remainder when 123456789101112.......798999 is divisible by 11
3.If A is the product of first 1000 multiples of 8,i.e. A=8*16*24*....8000.How many zeroes will be there at the end of A?
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Please explain the approach of the below questions.
1 126 natural numbers are put side by side in ascending order to create a large number 12345................125126. What will be the remainder when N is divided by 5625 ?
2.Remainder when 123456789101112.......798999 is divisible by 11
3.If A is the product of first 1000 multiples of 8,i.e. A=8*16*24*....8000.How many zeroes will be there at the end of A?
1) My take is 126.
5625 = 625*9.
5126 mod 625 = 126.
126*127/2 mod 9 = 0.
Hence remainder will be of the form - 625k+126 = 9p,ie, 126.
2) My take is 4.
Just subtract sum of alternate digits.
3) My take is 249.
8*1,8*2,...,8*1000 = 8^1000(1*2*...1000) = 8^1000*1000!.
=> 1000/5 = 200/5 = 40/5 = 8/5 = 1. So, 249.
Can someone please explain the solution:
Consider the sets Tn={n,n+1,n+2,n+3,n+4}, where n=1,2,3,4...96. How many of these sets contain 6 or any integral multiple thereof (i.e any one of the numbers 6, 12, 18...)?:cheers:
80 sets
What is the OA?
Can someone please explain the solution:
Consider the sets Tn={n,n+1,n+2,n+3,n+4}, where n=1,2,3,4...96. How many of these sets contain 6 or any integral multiple thereof (i.e any one of the numbers 6, 12, 18...)?:cheers:
We can use numbers like - 2,3,4,5,6,8,9,10,11,12,14,15,...,96.
Hence, We have 96 - 96/6 = 80 such sets.
80 
Yes, 80 is the OA. Thanks Enceladus for the explanation.
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1) My take is 126.
5625 = 625*9.
5126 mod 625 = 126.
126*127/2 mod 9 = 0.
Hence remainder will be of the form - 625k+126 = 9p,ie, 126.
2) My take is 4.
Just subtract sum of alternate digits.
3) My take is 249.
8*1,8*2,...,8*1000 = 8^1000(1*2*...1000) = 8^1000*1000!.
=> 1000/5 = 200/5 = 40/5 = 8/5 = 1. So, 249.
In the third problem...i got 248 as answer. When 8 multipled with 625, it doesn't provide any extra zero. So i dnt understand why you divide that 8 by 5.
1) My take is 126.
5625 = 625*9.
5126 mod 625 = 126.
126*127/2 mod 9 = 0.
Hence remainder will be of the form - 625k+126 = 9p,ie, 126.
2) My take is 4.
Just subtract sum of alternate digits.
3) My take is 249.
8*1,8*2,...,8*1000 = 8^1000(1*2*...1000) = 8^1000*1000!.
=> 1000/5 = 200/5 = 40/5 = 8/5 = 1. So, 249.
In the third problem...i got 248 as answer. When 8 multipled with 625, it doesn't provide any extra zero. So i dnt understand why you divide that 8 by 5.
well i came through this question while solving today, since i am new to cat preparations a help from odrs might make a big difference to me..
n*(n^2 -1)*(5n+2) is always divisible by? please expalin your answers for n being both odd as well as even.
well i came through this question while solving today, since i am new to cat preparations a help from odrs might make a big difference to me..
n*(n^2 -1)*(5n+2) is always divisible by? please expalin your answers for n being both odd as well as even.
The best way to solve this type of questions is to put values (e.g 1,3 and 2,4) and check with options.
This expression is always divisible by 3.
(n^2-1) is always divisible by 3 for all values which are not 3 or its multiples
as for n = 3, 6, 9... they are anyway multiples of 3.
can nyone plz solve???
what is the remainder when 5^1000 is divided by 61??
a) 13 b) 22 c) 31 d) 40
can nyone plz solve???
what is the remainder when 5^1000 is divided by 61??
a) 13 b) 22 c) 31 d) 40
My Take : Option a) 13
Approach:
5^1000 mod 61
=> 5*5^999 mod 61
=> 5*125^333 mod 61
=> 5*3^333 mod 61
=> 5*3^3*(243)^66 mod 61
=> 5*27*(-1)^66 mod 61
=> 135 mod 61
=> 13
For the number 2450 find .
1. the sum and number of all factors.
2. the sum and number of all even factors.
3. the sum and number of all odd factors.
4. the sum and number of all factors divisible by 5.
5. the sum and number of all factors divisible by 35.
6. the sum and number of all factors divisible by 245.
Please tell the approach .
Q2. In how many ways 15! Can be written as product of two coprime numbers?
I am not sure if this is the right approach to solve Qns like thee but here is how I solved it :
15! can be broken into prime factors as 15! = 2^a * 3^b * 5^c * 7^d * 11^e * 13^f =A*B*C*D*E*F
where a,b,c,d,e,f are some natural nos.
and let A=2^a , B=3^b , C=5^c... and F=13^f.
To write 15! as a prod of two co-prime numbers, we can group the above multilples in
1) (1 multiple) * (5 multiples)
2) (2 multiples) * (4 multiples)
3) (3 multiples) * (3 multiples)
For (1) (1 multiple) * (5 multiples) : There are 6 ways to write the product viz.
*
* , * , * , ...... *
For (2) (2 multiples) * (4 multiples) : There are 5+4+3+2+1 ways to write the product
* , , , , .................... 5 ways
, , , ................. 4 ways
, . .............3 ways
, ... 2 ways
.... 1 way
For (3) (3 multiples) * (3 multiples)
, , , .........4 ways
, , .............3 ways
.............2 ways
................ 1 way
From now on if you will try to group in (2 multiples)(4 multiples) you will again get the same figures as in (2) above.
So total no. ways in which we can write 15! as a product of two co-primes = 6+ (5+4+3+2+1) + (4+3+2+1) = 31 ways
Is it th OA? Pls confirm.
Note : We cannot break powers of prime nos. here ( Why?)
This is the reason we are taking the prime number raised to a specific power as a single entity.
I am sure there must be a shortcut method or a trick to solve such Qns. May be somebody will post a solution done by a shortcut.
I am not sure if this is the right approach to solve Qns like thee but here is how I solved it :
15! can be broken into prime factors as 15! = 2^a * 3^b * 5^c * 7^d * 11^e * 13^f =A*B*C*D*E*F
where a,b,c,d,e,f are some natural nos.
and let A=2^a , B=3^b , C=5^c... and F=13^f.
To write 15! as a prod of two co-prime numbers, we can group the above multilples in
1) (1 multiple) * (5 multiples)
2) (2 multiples) * (4 multiples)
3) (3 multiples) * (3 multiples)
For (1) (1 multiple) * (5 multiples) : There are 6 ways to write the product viz.
*
* , * , * , ...... *
For (2) (2 multiples) * (4 multiples) : There are 5+4+3+2+1 ways to write the product
* , , , , .................... 5 ways
, , , ................. 4 ways
, . .............3 ways
, ... 2 ways
.... 1 way
For (3) (3 multiples) * (3 multiples)
, , , .........4 ways
, , .............3 ways
.............2 ways
................ 1 way
From now on if you will try to group in (2 multiples)(4 multiples) you will again get the same figures as in (2) above.
So total no. ways in which we can write 15! as a product of two co-primes = 6+ (5+4+3+2+1) + (4+3+2+1) = 31 ways
Is it th OA? Pls confirm.
Note : We cannot break powers of prime nos. here ( Why?)
This is the reason we are taking the prime number raised to a specific power as a single entity.
I am sure there must be a shortcut method or a trick to solve such Qns. May be somebody will post a solution done by a shortcut.
For these type of questions first of all find the prime factors of 15! which are 2,3,5,7,11 & 13
=2^5 is the ans
no ways of writing a no as a product of two nos is=2^(n-1)
where n is no of prime factors