If x, y, and z are positive integers and 3x = 4y = 7z, then the least

given 3x=4y=7z

x+y+z in terms of x

= x+(3x/4)+(3x/7) = 61x/28

now checking with each of the answers and see which value gives a minimum integer value.

A x = 28*33/61 , not an integer

B,C can be ruled out similarly.

D is minimum value as x = 61*28/61 = 28

Answer is D.

3x = 4y = 7z

LCM = 3 * 4 * 7 = 84
84 / 3 = 28
84 / 4 = 21
84 / 7 = 12

28 + 21 + 12 = 61

The LCM of 3, 4, and 7 = 84. (Easy to figure out - since 3, 4, and 7 are prime, the LCM is 3*4*7)

3(x) = 84 -> x = 28
4(y) = 84 -> y = 21
7(z) = 84 -> z = 12

28+21+12 = 61.

Hope this helps.

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We know x,y and z are postive integers so based on given information, we can say
x= 4/3y and z= 4/7y

then x+y+z= 4/3y+y+4/7y-------> (4/3y+4/7y)+ y ------> 40y/21 +y -----> 61 y/21

61y/21= Sum of positive integers x,y and z. This sum also is a postive integer. Let us say it as Integer "a"

therefoee 61y/21= a now this implies y= a*21/61 ----For y to be an integer, we see that "a" has to be multiple of 61 and the smallest multiple of 61 is 1

Hence ans should D i.e 61

LCM of 3, 4 & 7 is 84

Least value of x+y+z

\(= \frac{84}{3} + \frac{84}{4} + \frac{84}{7}\)

= 28 + 21 + 12 = 61

Answer = D

1) We should minimize z while keeping 7z divisible by a multiple of 3 and 4 (which means z must be even, since 7 is odd and we need divisibility by an even number)

2) 7z must be divisible by 12 because 3x and 4y will have 12 as their LCM. First multiple of 7 divisible by 12 is 84, which is trap AC E) --> we need x+y+z NOT the LCM of all 3

3) 84/3 = 28, 84/4 = 21, 84/7 = 12
28 + 21 + 12 = 61, D)

VeritasKarishma
Ma'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

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