One more: Find all values of n>= 1 for which n!+(n+1)!+(n+2)! is equal to a perfect square.
Team BV - Vineet
@bodhi_vriksha said:One more: Find all values of n>= 1 for which n!+(n+1)!+(n+2)! is equal to a perfect square.Team BV - Vineet
None?
Edit: One. Read the question as n > 1 instead of n >=1. My bad :banghead:
regards
scrabbler
Edit: One. Read the question as n > 1 instead of n >=1. My bad :banghead:
regards
scrabbler
@scrabbler said:8?The difference is 2^4 * 3^2 * 5^1 * 11^1 which can be split into co-prime pairs in 2^(4-1) = 8 ways...regardsscrabbler
Bro, Why did u take the difference here?
@scrabbler said:None?
regardsscrabbler
Nope. give reasoning for saying so..Try again....
Team BV- Vineet
if k is a natural number such that f(f(f(f(k)))) =1 and k>f(k)>f(f(k))>f(f(f(k)))>1.. what is the least number of digits that k can have??
@bodhi_vriksha said:Evening Teaser: (Easy) LCM and HCF of two numbers is 11! and 7! respectively. How many such pairs of two numbers are possible?Team BV - Kamal Lohia
8??
@nramachandran said:Bro, Why did u take the difference here?
Difference in terms of ratio I meant...multiplicative difference....wrong choice of word sorry :(
See if I want to take two numbers whose HCF is 10 and LCM is 120, then I can say LCM / HCF = 12. Any possible division of 112 x co-primes will work. Now 12 = 12 * 1 or 4 * 3 so we will have two such pairs, 120 & 10 , and 40 & 30.
Similarly if HCF =2 and LCM = 210 (say) then the ratio is 105 which can be written as product of co-primes as 1 * 105 or 3 * 35 or 5 * 21 or 7 * 15 i.e. 4 ways.
Hence took the ratio, and tried to express it as co-primes.
regards
scrabbler
See if I want to take two numbers whose HCF is 10 and LCM is 120, then I can say LCM / HCF = 12. Any possible division of 112 x co-primes will work. Now 12 = 12 * 1 or 4 * 3 so we will have two such pairs, 120 & 10 , and 40 & 30.
Similarly if HCF =2 and LCM = 210 (say) then the ratio is 105 which can be written as product of co-primes as 1 * 105 or 3 * 35 or 5 * 21 or 7 * 15 i.e. 4 ways.
Hence took the ratio, and tried to express it as co-primes.
regards
scrabbler
@scrabbler said:None?regardsscrabbler
try n = 1 dear...
1! + 2! + 3! = 1 + 2 + 6 = 9 = a perfect square :)
Team BV - Kamal Lohia
1! + 2! + 3! = 1 + 2 + 6 = 9 = a perfect square :)
Team BV - Kamal Lohia
@pavimai said:if k is a natural number such that f(f(f(f(k)))) =1 and k>f(k)>f(f(k))>f(f(f(k)))>1.. what is the least number of digits that k can have??
2?
@iLoveTorres said:at n=1 you get 1+2+6 = 9 which is a perfect square
Ok I read greater than 1, it is greater than or equal to 😛 My bad!.
At 1 it works of course...the thing I got is it will be a perfect square only if n! is a perfect square which happens only at 1.
regards
scrabbler
At 1 it works of course...the thing I got is it will be a perfect square only if n! is a perfect square which happens only at 1.
regards
scrabbler
@pavimai said:if k is a natural number such that f(f(f(f(k)))) =1 and k>f(k)>f(f(k))>f(f(f(k)))>1.. what is the least number of digits that k can have??
3
@pavimai said:if k is a natural number such that f(f(f(f(k)))) =1 and k>f(k)>f(f(k))>f(f(f(k)))>1.. what is the least number of digits that k can have??
Just giving a start...
f(f(f(k))) has to be smallest natural number which is greater than 1 and have sum of digits equal to 1, right?
I am sure you can easily guess this number. Yes it is 10.
Now next f(f(k)) has to be smallest natural number which is greater than 10 and have sum of digits equal to 10. Can you guess it correctly this time?
It is 19. I hope you must have managed it easily.
Going same way, f(k) has to be smallest natural number which is greater than 19 and have sum of digits equal to 19. Find this number now.
Apply the same sequence once again to get the value of k. :)
Team BV - Kamal Lohia
f(f(f(k))) has to be smallest natural number which is greater than 1 and have sum of digits equal to 1, right?
I am sure you can easily guess this number. Yes it is 10.
Now next f(f(k)) has to be smallest natural number which is greater than 10 and have sum of digits equal to 10. Can you guess it correctly this time?
It is 19. I hope you must have managed it easily.
Going same way, f(k) has to be smallest natural number which is greater than 19 and have sum of digits equal to 19. Find this number now.
Apply the same sequence once again to get the value of k. :)
Team BV - Kamal Lohia
OA 23.
min value of k is 1 followed by 22 9s..
so minimum is 23 digits
@pavimai said:if k is a natural number such that f(f(f(f(k)))) =1 and k>f(k)>f(f(k))>f(f(f(k)))>1.. what is the least number of digits that k can have??
Can I define f(k) as 0.5 * k? Or even (0.5^0.25) * k
:P
regards
scrabbler
:P
regards
scrabbler
One more: What is the largest n such that 2^n divides [10^30/(10^10 + 3)], where [] represents greatest integer function?
Team BV - Kamal Lohia
@bodhi_vriksha said:Evening Teaser: (Easy) LCM and HCF of two numbers is 11! and 7! respectively. How many such pairs of two numbers are possible?Team BV - Kamal Lohia
let two nmbrs be a and b.
a=7!*k ... k can be any number.
b=7!*11*10*9*8*m.... m can be anynmbr
m!=k
basically.. difference is 11*10*9*8
writing them in the form of prime numbers.. 11*2^4*3^2*5
number of primes =4.
2^(k-1) will give ways where they can be written as product of two coprimes
2^(4-1) = 8
a=7!*k ... k can be any number.
b=7!*11*10*9*8*m.... m can be anynmbr
m!=k
basically.. difference is 11*10*9*8
writing them in the form of prime numbers.. 11*2^4*3^2*5
number of primes =4.
2^(k-1) will give ways where they can be written as product of two coprimes
2^(4-1) = 8