let f m n 45 m 36 n where m n ve or -ve int what is minimum value for

ans:9
if we put m=1,n=-1
then 45-36=9
the first no should be given positive value,since the minimum positive value is being asked
10 years agoHelpfull: Yes(46) No(2)

f=45*m+36*n
f=9(5*m+4*n)
f/9=5*m+4*n

for integer values of n and m f should be in multiples of 9.
ie, minimum possible value of f is 9.
8 years agoHelpfull: Yes(17) No(1)
if both m and n are large negative nos..then we will get least value(more negative value)...??anyone please post correct solution
10 years agoHelpfull: Yes(2) No(5)
go by option.
10 years agoHelpfull: Yes(2) No(3)
They have asked the"minimum positive value". So large negative values wont come under that condition. So i suppose 9 is the answer.
10 years agoHelpfull: Yes(1) No(1)
I think the ans could be 0 as well..... taking the LCM of 45,36 ie 180 and put m=4,n=5 that will results f(4,5)=180-180=0......
9 years agoHelpfull: Yes(1) No(6)
guys 0 is not a positive number.So 9 is the correct answer.
9 years agoHelpfull: Yes(1) No(1)
by options we can solve
8 years agoHelpfull: Yes(1) No(0)
d=gcd(45,36)
d=gcd(32∗5,22∗32)
d=32
d=9
8 years agoHelpfull: Yes(1) No(0)
the question is saying what is the minimum value of f(m,n), not m and n!
minimum value of f(m,n)=0

as we can substitute m=4 and n=-5
45*(4)+36*(-5)
180-180=0

What so many wrong ques?? tcs
9 years agoHelpfull: Yes(0) No(6)
for larger negative values of m,n doesnt make f(m,n) negative ,,,, i think they should say min positive integer value
8 years agoHelpfull: Yes(0) No(0)
ans=9
f(m,n)=45*m+36*n
4 years agoHelpfull: Yes(0) No(1)

let f(m,n)=45m+36n where m & n(+ve or -ve int)what is minimum value for f(m,n) for all m,n(this may be achived for various values of m,n)