pandajee wrote:
VeritasKarishma wrote:
MSDHONI wrote:
When I see more than one '=', my instinct is to make them equal to k to get to a 'common ground'.
\(3x = 4y = 7z = k\)
\(x = \frac{k}{3}; y = \frac{k}{4}; z = \frac{k}{7}\)
Now, all x, y and z are positive integers so k should be divisible by 3, 4 and 7. The minimum value of k will be 3*4*7.
\(x + y + z = \frac{k}{3} + \frac{k}{4} + \frac{k}{7} = \frac{3*4*7}{3} + \frac{3*4*7}{4} + \frac{3*4*7}{7} = 61\)
VeritasKarishmaMa'm, why have you replaced "K" by 3x4x7 (confused by this equality-- 3x=4y=7z=k)
You have three different terms, all equal to each other. We bring it equal to k to bring everything in terms of a single variable k. Then we have x, y and z, all in terms of k and we can easily find their ratio.
x = k/3
y = k/4
z = k/7
Since x, y and z all need to be integer, k/3 should be an integer so k should have 3 as a factor. k/4 should be an integer so k should have 4 as a factor. k/7 should be an integer so k should have 7 as a factor.
So k should have 3, 4 and 7 as factors so minimum value of k = 3*4*7
So min value of x = 28, of y = 21 and of z = 12
Alternatively, we can focus on one pair of equation at a time:
3x = 4y
x/y = 4/3 = 28/21
4y = 7z
y/z = 7/4 = 21/12
x : y: z = 28:21:12
Since x, y and z are integers, least value of their sum = 28 + 21 + 12 = 61