@pavimai said:s(n) is defined as number of integers (1,2,...n)that are relatively prime to n .if n is product of two prime numbers,whose sum is 40 and s(n)=280 then n are??
Is it 319?
@Calvin4ever said:Find the least possible value of (a+b+c)(1/a + 1/b + 1/c)
If i put a=b=c as 1/100, we have a+b+c as 3/100 n 1/a + 1/b + 1/c = 300..Hence, It will be 1/10,000...Coz there are no conditions mentioned abt the nature of a, b n c..
N similarly it can be seen that the expression cannot have a minima.. 9 kaise ho sakta hai minima..i can't understand . 
@prateek987 said:@Joey_Sharma joey..tum ek behad subtle galti kar rahe ho...question ki shuru ki do lines dhyan se padho..
yeah!! i got it now. i considered 2112 as the actual data for this year. thanks!
@IIMAIM said:Since least value is asked making c as -1 makes-3*-3=9
Point taken, but try a = -100, b = -100, c = +1 and see what the value comes?
regards
scrabbler
regards
scrabbler
@pyashraj said:If i put a=b=c as 1/100, we have a+b+c as 3/100 n 1/a + 1/b + 1/c = 300..Hence, It will be 1/10,000...Coz there are no conditions mentioned abt the nature of a, b n c..N similarly it can be seen that the expression cannot have a minima.. 9 kaise ho sakta hai minima..i can't understand .
Question setter forgot to specify "positive integers" so we are all having fun at the expense of the question :D
regards
scrabbler
regards
scrabbler
@scrabbler
Yeah..Mera paas Valentine toh nahi hai..Isliye Question ke saath flirt kar raha hun.. 
Hwever, Even if the positive Integers condition is implied..Answer cannot be 9..1 hoga shayad then.. at a=b=c = 1...It will be 1 then..
@scrabbler said:Question setter forgot to specify "positive integers" so we are all having fun at the expense of the question regardsscrabbler
hmm -infinity
@pavimai said:s(n) is defined as number of integers (1,2,...n)that are relatively prime to n .if n is product of two prime numbers,whose sum is 40 and s(n)=280 then n are??
n=11*29
319
319
@Calvin4ever said:Find the least possible value of (a+b+c)(1/a + 1/b + 1/c)
AM>HM
(a+b+c)/3>3/(1/a+1/b+1/c)
(a+b+c)(1/a+1/b+1/c)>9
minmum value is 9
(a+b+c)/3>3/(1/a+1/b+1/c)
(a+b+c)(1/a+1/b+1/c)>9
minmum value is 9
what should be the values of a and b such that 30a0b03 is divisible by 13?
A 50 digit number has all 7 s .find the remainder when the number is divided by 74
@gnehagarg said:AM>HM(a+b+c)/3>3/(1/a+1/b+1/c)(a+b+c)(1/a+1/b+1/c)>9minmum value is 9
in the question it is never stated a,b,c are positive if they are positive it is 9
-infinity otherwise
-infinity otherwise
a= 8 ,b=1ďťż
or 2 ,3 or
5,2
@pavimai said:A 50 digit number has all 7 s .find the remainder when the number is divided by 74
7/74=7 , 77/74=3, 777/74=37, 7777/37=7
repeating after 3 cycles so 50/3=2
so 3
repeating after 3 cycles so 50/3=2
so 3
@IIMAIM said:77/74=3,777/74=37,7777/37=7,77777/7=3repeating after 3 cycles so 50/3=2so 37
options
3
13
23
33
@pavimai said:A 50 digit number has all 7 s .find the remainder when the number is divided by 74
37 hoga kya?
Maximum number of square possible that can be constructed using 31 pencils of equal length on the table? PLEASE explain the method