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12 Mar 2007, 04:59
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If x can be 1,2,3 and 4 and y can be 2, 4, 8 and 16. Then the product of XY can have how many different values
a. 8
b. 9
c. 10
d 11
e. 12
Kindly explain
a. 8
b. 9
c. 10
d 11
e. 12
Kindly explain
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Thank you for understanding, and happy exploring!
(C) for me 
To me, I start to count each possibility... For each value of X, I check how many of values of Y create an unseen result for X*Y in the previous calculus.
o If x=1, y could be
> 2 ... x*y=2
> 4 ... x*y=4
> 8 ... x*y=8
> 16 ... x*y=16
o If x=2, y could be
> 16 ... x*y=32
o If x=3, y could be
> 2 ... x*y=6
> 4 ... x*y=12
> 8 ... x*y=24
> 16 ... x*y=48
o If x=1, y could be
> 16 ... x*y=64
To me, I start to count each possibility
... For each value of X, I check how many of values of Y create an unseen result for X*Y in the previous calculus.
o If x=1, y could be
> 2 ... x*y=2
> 4 ... x*y=4
> 8 ... x*y=8
> 16 ... x*y=16
o If x=2, y could be
> 16 ... x*y=32
o If x=3, y could be
> 2 ... x*y=6
> 4 ... x*y=12
> 8 ... x*y=24
> 16 ... x*y=48
o If x=1, y could be
> 16 ... x*y=64
12 Mar 2007, 06:15
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I get D.
The number of total possible values of X*Y = 4*4 = 16
The number of repetitions :
2,2|1,4 ---->2
1,16|4,4|2,8 ---->3
2,4|1,8|4,2 ----> 3
16 - (2+3+3) +3 (1 for each possibility of 4,16,8) = 16-8+3 = 8+3 = 11.
Hmm, I see my answer is different from Fig's.
The number of total possible values of X*Y = 4*4 = 16
The number of repetitions :
2,2|1,4 ---->2
1,16|4,4|2,8 ---->3
2,4|1,8|4,2 ----> 3
16 - (2+3+3) +3 (1 for each possibility of 4,16,8) = 16-8+3 = 8+3 = 11.
Hmm, I see my answer is different from Fig's.
