请教NN两个DS的解题思路
1. If M, N, P are three different prime integers and X, Y, Z are three positive integers, is (M)x (N)y (P)z > 100?
(1) x + y + z = 5
(2) M × N × P = 30
答案是A
2. If K is the product of the first 30 positive integers and d is a positive integer, what is the value of d?
(1) (10)d is a factor of K
(2) d > 6
答案是C
1. If M, N, P are three different prime integers and X, Y, Z are three positive integers, is (M)x (N)y (P)z > 100?
From 1) ==> X, Y, Z are 1, 2, 2 =>Smallest (M)x (N)y (P)z = (2)x (3)y (5)z = (2)^2 (3)^ 2 (5)^1 =180 >100;
or 1, 1, 3 ==>Smallest (M)x (N)y (P)z = (2)x (3)y (5)z = (2)^3 (3)^ 1 (5)^1 =120 >100;
From 2) ==> M, N, P are 2, 3, 5
If X=Y=Z=1 => (M)x (N)y (P)z = (2)x (3)y (5)z = (2)^1 (3)^ 1 (5)^1 =30 <100
If X=Y=Z=10 => (M)x (N)y (P)z = (2)x (3)y (5)z >100
2. If K is the product of the first 30 positive integers and d is a positive integer, what is the value of d?
(1) (10)d is a factor of K
(2) d > 6
Since K= 30! , from (1) we know that d could equal to 1, 2, 3, or more. because K's factor includes at least 10, 20, 30.
(2) alone just not eough;
For (1)+(2) ==> 30! = X1*(5x10x15x20x25x30) =X2* (10^7)
note: we can borrow enough 2 from X1 to alter 5 to 10; 25 = 5x5
From (1)+ (2) ==>The smallest d is 7.
==> d=7 answer is C
Great, thank you very much. Only one thing,
note: we can borrow enough 2 from X1 to alter 5 to 10; 25 = 5x5
From (1)+ (2) ==>The smallest d is 7.
==> d=7 answer is C,
note: we can borrow enough 2 from X1 to alter 5 to 10; 25 = 5x5
From (1)+ (2) ==>The smallest d is 7.
==> d=7 answer is C,
K=30!, from (1)==>the range of possible d is from 1~7, then (1)+(2) suggests the answer should be the biggest d, 7.
If I have any misunderstanding, please correct me. Thanks.
感谢!!!
果然是数学NN!!!
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