If m = 2^x*5^y*7^z and both 350 and 280 are factors of m, what is the

Question

If  m = 2^x*5^y*7^z  and both 350 and 280 are factors of m, what is the minimum value of xyz, where x, y and z are positive integers ?

(A) 700
(B) 70
(C) 24
(D) 6
(E) 5

Kinshook · Answer

Asked: If  m = 2^x*5^y*7^z  and both 350 and 280 are factors of m, what is the minimum value of xyz, where x, y and z are positive integers ?

350 = 2×5^2×7
280 = 2^3×5×7

Minimum value of x = 3
Minimum value of y = 2
Minimum value of z = 1
Minimum value of xyz = 6

IMO D

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TarunKumar1234 · Answer

m=2^x∗ 5^y∗ 7^z and 350= 2 * 5^2 * 7^1 and 280= 2^3 * 5^1 * 7^1 For min value of x, y and z will 3, 2, 1. So, xyz= 3*2*1 = 6 So, I think D.

Baps · Answer

To answer this question, I looked for the LCM between 350 and 280 because both numbers are factor of m :

350 : 7*5*5*2
280 : 7*2*2*2*5

Therefore, LCM = 7*5*5*2*2*2

Coming back to our question :

m = 2^x * 5^y * 9^z. Therefore,

The lowest value of m must be : 2^3*5^2*7^1, which means that the lowest value of xyz = 6.

Answer : D

abs8 · Answer

For divisibility questions , prime factorization should come as a first thought .

Prime factorize , 350 =  2*5^2*7 
250 =  2^3*5*7

Since , x,y and z are exponents , minimum value will be the minimum value of bases , 2 which is 3 , of 5 which is 2 and of 7 which is 1.

xyz= 3*2*1 = 6

Option D

Soz · Answer

Shouldn't the min value of x be 1 and y be 1 ...

If m = 2^x5^y7^z and both 350 and 280 are factors of m, what is the