but if u take a=4,b=3 and c=1 then 2a+3b+4c is not equal to 18..
abhgud Saysbut if u take a=4,b=3 and c=1 then 2a+3b+4c is not equal to 18..
,made a very silly mistake ..
now edited my post and the correct answer is 273.....
again played with a bit logic..
to maximize a^4+b^4 + c^4
where
2a+3b+4c=18
we have to take min value for 4.. took 1.
now b has to be an odd no.. so 3
and a have to be max at 4..
so 4^4 + 2^4 + 1^4
= 256+16+1
= 273..
hope u got it..
:-P
how did u get this ....( in bold )
for maximising a^4+b^4+c^4 we can use the AM> GM concept to get the answer.....it's given that 2a+3b+4c=18 so write 2a= a/2+a/2+a/2+a/2....similarly write 3b= 3b/4+3b/4+3b/4 and write c=c+c+c+c.....then use the AMGM concept twice......try out this way guys
barclays_boss Saysfind the maximum value of a^4 + b^4 + c^4 if 2a+3b+4c=18 ??
the way to solve these kind of problems
in Max,we have to see the small coeff of variable in a condition ,try to put the MAX value for dat..see if dat satisfies the condition
in Min,,we have to see the large coeff of variable in a condition ,try to put the MAX value for dat..see if dat satisfies the condition
in the ques we have 2a+3b+4c=18
for max,as stated above,take the var whose coeff. is small ,here its a ..try to put the large value of
a=4,b=2,c=1 ,condition satisfies..
now put a^4+b^4+c^4=
4^4+2^4+1^4= 273..
try dis now If 5x + 2y + z = 81, where x, y and z are distinct positive integers, then find the difference between the
maximum and the minimum possible value of x + y + z. :)
the way to solve these kind of problems
in Max,we have to see the small coeff of variable in a condition ,try to put the MAX value for dat..see if dat satisfies the condition
in Min,,we have to see the large coeff of variable in a condition ,try to put the MAX value for dat..see if dat satisfies the condition
in the ques we have 2a+3b+4c=18
for max,as stated above,take the var whose coeff. is small ,here its a ..try to put the large value of
a=4,b=2,c=1 ,condition satisfies..
now put a^4+b^4+c^4=
4^4+2^4+1^4= 273..
try dis now If 5x + 2y + z = 81, where x, y and z are distinct positive integers, then find the difference between the
maximum and the minimum possible value of x + y + z. :)
nice concept friend!
check my solution please.
when we want maximum,Z is greatest.
took x=1,y=2,z=72.
x+y+z= 75.
for minimum value,
we maximize x.
taking x=14,y=4,z=2.
x+y+z=20.
so different between
max(x+y+z)-min(x+y+z)=55
P.S. thnks mani for correction.
barclays_boss Saysfor maximising a^4+b^4+c^4 we can use the AM> GM concept to get the answer.....it's given that 2a+3b+4c=18 so write 2a= a/2+a/2+a/2+a/2....similarly write 3b= 3b/4+3b/4+3b/4 and write c=c+c+c+c.....then use the AMGM concept twice......try out this way guys
pls tell how to use?
if it has been a^4*b^4*c^4 i understand but the above question seems to be totally different!
barclays_boss Saysfind the maximum value of a^4 + b^4 + c^4 if 2a+3b+4c=18 ??
Dudes, if we take b=c=0, then a will be 9 and the max(a^4+...) = 9^4.
Its not stated in the qn reg range of a,b,c. Infact if u take -ive values then the max value will be indeterminate. (4th power in any case will be +ive)
Hw to understand??
::::::if the square of 121 with base n is 14641 with base n , how many values can n assume from first 50 natural numbers.
options are
1. 40
2. 41
3. 42
4. 43
5. 44
please answer as soon as one can
:
if the square of 121 with base n is 14641 with base n , how many values can n assume from first 50 natural numbers.
options are
1. 40
2. 41
3. 42
4. 43
5. 44
please answer as soon as one can
Is the answer 44??
avinav2712 SaysIs the answer 44??
yes u r true :-P , please explain me the question explicitly and implicitly , i would be thankful to u
nice concept friend!
check my solution please.
when we want maximum,Z is greatest.
took x=1,y=2,z=72.
x+y+z= 75.
for minimum value,
we maximize x.
taking x=14,y=3,z=2.
x+y+z=19.
so different between
max(x+y+z)-min(x+y+z)=56
just a small correction dude .. min value will be 20 --> 14,5,1 or 15,4,2 for x,y,z respectively and hence the diff = 55
i hope it helps !!!
if the square of 121 with base n is 14641 with base n , how many values can n assume from first 50 natural numbers.
options are
1. 40
2. 41
3. 42
4. 43
5. 44
please answer as soon as one can
explain this question why we can not take values from 1 to 6:o
hi to all
which is d largest four digit no. which when divided by 6 leaves remainder 5 and when divided by 5 leaves remainder 3
:shocked:
Answer = 9983
gauravpruthi Saysexplain this question why we can not take values from 1 to 6:o
It's pretty obvious why we can't take values till 6, since the square of 121 contains the digit 6 😃 You would probably know that for any base 'n', the highest digit that can be used is 'n-1'. Thus, any base after 6 can be used.
barclays_boss Saysfind the maximum value of a^4 + b^4 + c^4 if 2a+3b+4c=18 ??
I don't know where I am missing...
I think the answer is infinity
If c=-4, b=0,a=9
Like this we will get various values and the value of a^4+b^4+c^4 will be increasing till infinity..
Where I am wrong?:shocked:
Well,I think it was tacit that a,b,c are whole number....
this mite jus seem a very trivial sawaal.. but still..:oops:
can some1 pl tell me how does one find sum of nos between two nos??
for ex) how can we find the sum of nos between 51 to 80??
regards
afshaa
this mite jus seem a very trivial sawaal.. but still..:oops:
can some1 pl tell me how does one find sum of nos between two nos??
for ex) how can we find the sum of nos between 51 to 80??
regards
afshaa
its like if u have no. of terms and the first and last term you can find out using the formula...
s = n/2(first term + last term)
here first term is 52(not including 51) and last term is 79(again not including 80) no. of terms will be 79-52+1 = 28
hence sum = 28/2 * (52+79) = 1834
barclays_boss Saysfind the maximum value of a^4 + b^4 + c^4 if 2a+3b+4c=18 ??
let b and c be 0 then we get a = 9
hence max value = 9^4 = 6561
is a and b and c are not negative..
if thy can take negative value thn it will be infinity