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[求助]请教NN两个DS难题

请教NN两个DS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

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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


TOP

 

Great, thank you very much. Only one thing,

QUOTE:
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,

              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.

TOP

感谢!!!

果然是数学NN!!!

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