Official Quant thread for CAT 2013

MBA CAT 2024
19666
@pavimai said:
there is an easy funda to find itsum of all numbers formed from given digits=(n-1)!*{sum of digits}*{1111...n times}so sum of all numbers formed from digits {1,3,5,7,9} will be= 4!{1+3+5+7+9}*{11111} = 24*25*11111=6666600
yea i know
but i dont follow any formulas
and i took the sum as 16 by mistake
19667
@pavimai said:
remainder when 1^3+2^3+3^3...+1000^3 is divided by 13 is???
(1000*1001/2)^2 mod 13 = 0 ?
as 1001 is divisible by 13
19668
@pavimai said:
remainder when 1^3+2^3+3^3...+1000^3 is divided by 13 is???
0....
Add the numbers...the result is divisible by13...not considering the cube...just the numbers..
In office hence short explanation...
19669
@saurabhlumarrai said:
Apply rule of divisibility of 13,you would get the answer:-(3*1)+(b*-4)+(a*3)+(3*1)=multiples of 13=>3-4b+3a+3=0 or 3-4b+3a+3=13 or 3-4b+3a+3=26 or 3-4b+3a+3=391st Case:3a-4b=-6 =>a=2,b=3 and a=6,b=62nd Case:3a-4b=7 =>a=5,b=2 and a=9,b=53rd Case:3a-4b=20 =>a=8,b=14th case:3a-4b=33 =>no solution
I think the concept for divisibility by 13 is sum.of triplets at odd places - sum of triplets at even places...suppose this comes out to be a number x..then for the number to be divisible by 13 ,x shoild be a miltiple of 13....
And in case x is not a multiple of 13..the remainder you get by dividing x by 13 is the actual remainder....
19670
@pavimai said:
remainder when 1^3+2^3+3^3...+1000^3 is divided by 13 is???
0 is the answer
ye hai QUANT wala thread

19671
@pavimai said:
remainder when 1^3+2^3+3^3...+1000^3 is divided by 13 is???
Other way - write as (1^3+ 1000^3) + (2^3+999^3) +...each bracket mein a^3 + b^3 is divisible by a+b which is 1001 = 13 x 11 x 7. Hence remainder is 0.

regards
scrabbler
19672
@saurabhlumarrai Hi..my doubt is how have u equated the equation= 0,13,26,39
enlighten me plz..
19673
@loksroh said:
@surajmenonv taking a=2 and b=1minimum changes bro.hence your solution is incorrect.infacr first assumption is wrong
my findings say
(1) b(a-b)
(2) a + 1/ b(a-b) is then equivalent to a + 4/ a^2

Now acc to ur argument, a= 2, b=1
Plug in these values in (1) and (2).
You will get
(1) 1
(2) answer = 3
which is exactly wat i derived.

Now in case u actually meant the reverse that a=1, b=2 then there is a condition in the question which says a>b>0.

Hope i have made it clear.
If u still think my logic is flawed, ur comments are welcome...
19674
@pavimai said:
remainder when 1^3+2^3+3^3...+1000^3 is divided by 13 is???
observe that 1^3 + 1000^3 = (1001) x something [a^3 + b^3 formula]
similarly 2^3 + 999^3 = (1001) x something
.
.
.
thus question reduces to finding remainder wen [(1001) x something big] is divided by 13 = 0
since 1001 divisible by 13.

19675
A natural no. N is 100 times the no. of its factors.Find the sum of digits of N ,given N has only 2 prime factors.
19676
@maroof10 said:
@saurabhlumarrai Hi..my doubt is how have u equated the equation= 0,13,26,39
enlighten me plz..
For the equation to satisfy correctly-the result should be multiple of 13,which can be 13*0=0,13*1=13, and so on..
19677
@maroof10 said:
A natural no. N is 100 times the no. of its factors.Find the sum of digits of N ,given N has only 2 prime factors.
N= p^a . q^b
Acc to question, N = p^a . q^b = 100 (a+1) (b+1)
=> N=2^2 . 5^2 . (a+1) (b+1)
=> now a=2 and b=5 (or equivalently a=5 and b=2 since N has only 2 prime factors).
Thus N=1800
Thus answer = 1+8+0+0 = 9
19678
@maroof10 said:
A natural no. N is 100 times the no. of its factors.Find the sum of digits of N ,given N has only 2 prime factors.
2?

Trial and error, 2000 seems to satisfy this.

regards
scrabbler
19679
@surajmenonv said:
N= p^a . q^bAcc to question, N = p^a . q^b = 100 (a+1) (b+1)=> N=2^2 . 5^2 . (a+1) (b+1)=> now a=2 and b=5 (or equivalently a=5 and b=2 since N has only 2 prime factors).Thus N=1800Thus answer = 1+8+0+0 = 9
Er, 1800 is divisible by 3 also?

regards
scrabbler
19680
@maroof10 said:
A natural no. N is 100 times the no. of its factors.Find the sum of digits of N ,given N has only 2 prime factors.
1?
19681
@scrabbler said:
Er, 1800 is divisible by 3 also?regardsscrabbler
stumped...u c the flaw??? o_O
19682
@surajmenonv if a=2 and b=5, then N= 2^2 + 5^5....which is not 1800
19683



@ChirpiBird said:
1?
lol.. sry it's 2.
100=(2^2)*(5^2)
2 prime numbers are 2,5

N=2^(p+2)5^(q+2)

number of factors will be (p+3)(q+3)
put p,q different values and check..
works for 2000. p=2,q=1

@scrabbler : sir itte sahi hits kaise lagate ho... !!!

19684
@surajmenonv said:
N= p^a . q^bAcc to question, N = p^a . q^b = 100 (a+1) (b+1)=> N=2^2 . 5^2 . (a+1) (b+1)=> now a=2 and b=5 (or equivalently a=5 and b=2 since N has only 2 prime factors).Thus N=1800Thus answer = 1+8+0+0 = 9
i think i got the mistake.
N = 2^2 . 5^2 (a+1) (b+1)
=> a=1, b=4 (since N has only two prime factors (here 2,5))
Thus N=(2^3) . (5^3)=1000
Thus answer is 1+0+0+0 = 1
@maroof10 pls clarify with ryt answer
19685
@surajmenonv said:
i think i got the mistake.N = 2^2 . 5^2 (a+1) (b+1)=> a=1, b=4 (since N has only two prime factors (here 2,5))Thus N=(2^3) . (5^3)=1000Thus answer is 1+0+0+0 = 1@maroof10 pls clarify with ryt answer
i think u r calculating it wrong..

N=1000
N=2^3 * 5^3
number of factors will be 16.
16*100 is not equal to 1000.

i did the same thing before. 😃 only 2000 satisfies...

2000 = 2^4 * 5^3
numbr of factors = 20
and 20*100 gives u the number N = 2000
hence sum of digits =2.