+600 Hi, me again, with a similar question.
1.(a) Write 140 as a product of its prime factors
140= 2^2*5*7
(b) Hence find the smallest integer 'n' such that 140n is a perfect square.
I don't get part (b). Plz help.
2. One red bulb, one blue bulb, and one yellow bulb start flashing at the same time.
The red bulb flashes after every 12 seconds, the blue after every 18 seconds and the yellow after every 20 seconds.
Find how many minutes elaspe before the three bulbs flash together again.
Plz help with this too.
For #2 you could also 'Brute Force' it to find the answer
Multiples of 12
12 24 36 48 60 72 84 96 108 120 132 144 156 168 180 192
Multiples of 20
20 40 60 80 100 120 140 160 180 200
Multiples of 18
18 36 54 72 90 108 126 144 162 180
....to see that they are all the same at 180 sec
Or : Prime factor 12 = 2 x 2 x3 unique factors 2x2x3
18 = 2 x 3 x 3 another 3
20 = 2 x 2 x 5 and 5
Take all of the 'unique' factors 2x2x3x3x5 = 180
It's always good to think of all the various ways to solve a problem!
in the first one ,
the prime factors of 140 are 2^2*5*7 if you want to get a perfect square, you have to multiply this number by 5*7 because , 2^2 is a perfect square and we have to equal 5 and 7 to a perfect square ( 5^5 and 7^7 to get 2^2*5^5*7^7 form which is a perfect square) n is 5*7=35
In the second one ,
you have to find the lowest common factor of 12 , 18 and 20 to find when will they flash together again
12 18 20 - 2 ( divide by 2)
6 9 10 - 2
3 9 5 - 3 2^2 * 3^2 * 5 = 180 sec.
1 3 5 - 3
1 1 5 - 5
1 1 1
For #2 you could also 'Brute Force' it to find the answer
Multiples of 12
12 24 36 48 60 72 84 96 108 120 132 144 156 168 180 192
Multiples of 20
20 40 60 80 100 120 140 160 180 200
Multiples of 18
18 36 54 72 90 108 126 144 162 180
....to see that they are all the same at 180 sec
Or : Prime factor 12 = 2 x 2 x3 unique factors 2x2x3
18 = 2 x 3 x 3 another 3
20 = 2 x 2 x 5 and 5
Take all of the 'unique' factors 2x2x3x3x5 = 180
It's always good to think of all the various ways to solve a problem!
