What number am I? Explain your process of how you got your answer.?

Question

Number One
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My number is even
It is a factor of 198 and a multiple of nine
It is less than 100
What number am I

Number Two
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My nuber is the product of three different prime numbers.
It is an odd number less than 125.
The sum of te digits is a multiple of three
One of my factors is the third prime number
What is my number?

Answer 1

The answer to number one is 18
The answer to number two is 105

Answer 2

For Number 1:
Answer is 18.
Since the number is even, I think of the number 2 which is a factor of 198. Since it must be a multiple of 9 and 2 isn't a factor of 9, I multiply 2 by 9 to get 18.

For Number 2:
xyz = p (number you're looking for)
p <125
also, 125=5x5x5
then x,y,z < or = 5 where x,y,z are prime #s.
which leaves you with x,y,z= 2,3 or 5.
since your # must be odd, forget 2.
So, I pick and choose numbers randomly..
i) x=3, y=3, z=5
xyz=45
4+5=9 which is a multiple of 3, so this works
ii) x=5, y=5, z=3
xyz=75
7+5=12 which is a multiple of 3, so this also works.
iii) x=3, y=3, z=3
xyz=27
2+7=9, works too.

so the answer could be 45 or 75, or 27

Answer 3

factors of 198 are:
1 x 198, 2 x 99, 3 x 66, 6 x 33, 9 x 22, 11 x 18
of those factors, only 9, 18, 99,198 are multiples of 9
only 9, 18, and 99 are under 100
only 18 is even

prime numbers 3,5,7 multiply to 105
105 is an odd number less than 125
the sum of the digits (1 + 0 + 5 = 6) is a multiple of 3

Answer 4

Number 1

Answer = 18
for mutiple of 9...I just did 9 x 2 = 18 then since it has to be a factor of 198 (and it has to be even), I divided 18 into 198 to see if it worked and it did


don't know the 2nd one...sorry

Answer 5

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Answer 6

Thats easy! Your a total zero.....lol