@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.
2000 = 2 + 0 + 0 + 0 = 2 ?
N = 100*20 = 2^4*5^3
no. of factors = 5*4 = 20
@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
sorry guys for all the wrong posts....finally figured it out (i think !!!)
N=(100)(a+1)(b+1)
=>N=2^2 . 5^2 (a+1)(b+1)
Here (a+1) and (b+1) can be any powers of 2 and 5.
Thus N= 100 * 2^something * 5^something
=>N =1000*something
check for 1000,2000,3000...
u will get answer as 2000 (ie 2+0+0+0 =2)
phew !!
@scrabbler said:Why not put c also as -1? In fact let's make them all -100....kyun nahin?regardsscrabbler
coz min value has been asked !
@Incognita123 said:coz min value has been asked !
See my later post on the same, can go much lower if we take -100, -100, +1. Can go to -inf technically!
regards
scrabbler
regards
scrabbler
@ChirpiBird said:lol.. sry it's 2.100=(2^2)*(5^2)2 prime numbers are 2,5N=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... !!!
Practice makes a man perfect. Women are already perfect.*
And by the time I was doing trial and error I had figured out that (a) the prime are 2 and 5 and (b) their powers must be such that (a+1) and (b+1) must also be powers of 2 and/or 5...uske baad 2nd or 3rd hit and trial gave answer!
* I'm only saying that because it's V-day :P
regards
scrabbler
And by the time I was doing trial and error I had figured out that (a) the prime are 2 and 5 and (b) their powers must be such that (a+1) and (b+1) must also be powers of 2 and/or 5...uske baad 2nd or 3rd hit and trial gave answer!
* I'm only saying that because it's V-day :P
regards
scrabbler
Here's an easy one:
The no. of triplets (a,b,c) where a,b,c are positive integers,which satisfy the simultaneous equations
(1) ab+bc=44
(2) ac+bc=23 is /are
(1) 1 (2) 2 (3) 4 (4) infinitely many
Here's an easy one:The no. of triplets (a,b,c) where a,b,c are positive integers,which satisfy the simultaneous equations
(1) ab+bc=44
(2) ac+bc=23 is /are
(1) 1 (2) 2 (3) 4 (4) infinitely many
(1) ab+bc=44
(2) ac+bc=23 is /are
(1) 1 (2) 2 (3) 4 (4) infinitely many
@surajmenonv said:Here's an easy one:The no. of triplets (a,b,c) where a,b,c are positive integers,which satisfy the simultaneous equations (1) ab+bc=44(2) ac+bc=23 is /are(1) 1 (2) 2 (3) 4 (4) infinitely many
is it 2???
@surajmenonv said:method pls....dat goes without saying in mathematics
ab+bc=44
b(a+c)=44
so b should be a multiple of 44
so possible values of b is 2,4,11,22,44
2nd condition
ac+bc=23
c(a+b)=23
so c should be a multiple of 23
so possible value of c is 1 and 23 but 23 cannot be in that case so only possible value is 1.
and a+b=23.
apply value of c=1, b=2,4,11,22,44 in eq 1 and satisfy condition also a+b=23....
you will get 2 such triplets..
@surajmenonv said:Here's an easy one:The no. of triplets (a,b,c) where a,b,c are positive integers,which satisfy the simultaneous equations (1) ab+bc=44(2) ac+bc=23 is /are(1) 1 (2) 2 (3) 4 (4) infinitely many
b(a+c) = 44
c(b+a) = 23
either c = 23 & b+a =1( not possible) or c=1, b+a = 23
=> b(a+1) = 44 & b+a = 23
=> b(23-b+1)=44
=> b^2-24b+44 = 0
=> b=22, 2
hence, two simultaneous solutions.
@surajmenonv said:Here's an easy one:The no. of triplets (a,b,c) where a,b,c are positive integers,which satisfy the simultaneous equations (1) ab+bc=44(2) ac+bc=23 is /are(1) 1 (2) 2 (3) 4 (4) infinitely many
c(a+b)=23 =>c=1 and a+b=23
Now b(a+c) = 44 =>b(a+1) = 44
So a=22,b=1 or a=1,b=22
hence (2)
Now b(a+c) = 44 =>b(a+1) = 44
So a=22,b=1 or a=1,b=22
hence (2)
yup 2 triplets is the answer.....
AAo Valentine ke pawan avsar par...thoda factorial pe haath saaf kare :P
Q1:highest power of 30 in 50! ????
Q2:Find the no of zeroes present at the end of 100!
Q3:Find the number od divisors of 15!.
Q3:Find the number od divisors of 15!.
Q3:find the number of divisors of 15!
Q4:rightmost non zero digit in 15!
@pakkapagal said:AAo Valentine ke pawan avsar par...thoda factorial pe haath saaf kare Q1:highest power of 30 in 50! ????
is it 9????