@Calvin4ever
1 ?
@Calvin4ever said:Find the least possible value of (a+b+c)(1/a + 1/b + 1/c)
AM=(a+b+c)/3
HM=3/(1/a+1/b+1/c)
So given expression=3*AM*(3/HM)>=9*1=9
@Calvin4ever said:Find the least possible value of a+ 1/(b(a-b)). a>b>0
a=2
b=1
Min value=3
Just a guess....
Plz tag me in the OA
@ChirpiBird said:9?
AM=(a+b+c)/3
HM=3/(1/a+1/b+1/c)
So given expression=3*AM*(3/HM)>=9*1=9
HM=3/(1/a+1/b+1/c)
So given expression=3*AM*(3/HM)>=9*1=9
A.M is always greater than or equal to h.m
A.m/h.m minimum is 1
therefore 9
@Calvin4ever said:Find the least possible value of a+ 1/(b(a-b)). a>b>0
a + 1/b(a-b) will have min value when 1/b(a-b) has the least value.
Apply am>=gm inequality on the two terms b and (a-b)
[b+(a-b)] / 2 >= [b(a-b) ]^(1/2)
=> b(a-b)
Thus max value of b(a-b) is a^2 /4.
Our question then becomes a + 4 / (a^2)
Aplly diff and equate to 0 to get minima at a=2
Thus req value is
a+ 4/ (a^2) = 2+ 4 /(2^2)
= 3
Kindly reply with the correct answer and solution.
@Calvin4ever said:Find the least possible value of (a+b+c)(1/a + 1/b + 1/c)
AM>=GM
a+b+c / 3 >= 3/ (1/a + 1/b + 1/c)
So
(a+b+c)(1/a + 1/b + 1/c) >=9
Is there a way to find the last two digits of 77^84^56?
@nramachandran said:AM>=GMa+b+c / 3 >= 3/ (1/a + 1/b + 1/c)So(a+b+c)(1/a + 1/b + 1/c) >=9
No G.M used here it A.M>=H.M that is used
@surajmenonv taking a=2 and b=1minimum changes bro.hence your solution is incorrect.infacr first assumption is wrong
@nramachandran said:Is there a way to find the last two digits of 77^84^56?
remainder by 100
100 = 25*4
remainder by 4 = 1
remainder by 25
2^84^56 mod 25
E(25) = 20
84^56 mod 20 = 16
2^16 mod 25 = 11
25k + 11 = 4n + 1
=> number = 61
@nramachandran said:Approach please
77^4 last 2 digits =41
now find last digit of 4^55 *21^56 it is 4
last digit is 1
digit in tens place=last digit of 4*4=6
therefore last 2 digits=61
now find last digit of 4^55 *21^56 it is 4
last digit is 1
digit in tens place=last digit of 4*4=6
therefore last 2 digits=61
The sum of five real numbers is 7; the sum of their squares is 10. Find the minimum and maximum possible values of any one of the numbers.
@Calvin4ever said:The sum of five real numbers is 7; the sum of their squares is 10. Find the minimum and maximum possible values of any one of the numbers.
@Calvin4ever said:Find the least possible value of (a+b+c)(1/a + 1/b + 1/c)
It will be 9. Use his property AM>= HM for a, b and c.
@Exodia said:Find all numbers of the form of 56x3y that are divisible by 36.
there are only two such numbers: 56232 and 56736 .
@bs0409 said:AM=(a+b+c)/3HM=3/(1/a+1/b+1/c)So given expression=3*AM*(3/HM)>=9*1=9
Wen v put a=b= -1 and c= 1....we get 1... No condition for a,b,c has been mentioned...