Since it's a multiple of 24, it has to be a multiple of the factors of 24.
Factors of 24:
2,3,4,6,8,12
You can use some of this knowledge to help create the number.
Since the # needs to be a multiple off 2, the last digit needs to be an 8
All numbers that are multiples of 3 have the property that all of their digits added together have to be a number that is evenly divisible by 3.
so your number will look like:
_ _ _ _ _ 8
so start trying combinations for the other 5 digits that give you a number that is a multiple of 3: 3,6,9,12,15, ect. If you can't find one, then it's impossible
Answer:
Sabemos que:
L es el largo de la avenida.
En la primer etapa se asfalto la mitad, L/2, entonces lo que queda por asfaltar es:
L - L/2 = L/2.
En la segunda etapa se asfalto la quinta parte, L/5, entonces lo que queda por asfaltar es:
L/2 - L/5 = 5*L/10 - 2*L/10 = (3/10)*L
En la tercer etapa se asfalto la cuarta parte del total, L/4, entonces lo que queda por asfaltar es:
(3/10)*L - L/4 = 12*L/40 - 10L/40 = (2/40)*L
Y sabemos que este ultimo pedazo que queda por asfaltar es de 200m:
(2/40)*L = 200m
L = 200m*(40/2) = 4,000m
There are 720 total different combinations, and 24 of those will have the letters DAY consecutively within all those combinations.
<u>Explanation:</u>
FRIDAY is a six letter word.
To form all combinations of letters, we would have six choices of letters to fill the first spot. Whatever letter you choose for the first spot, you then have five choices of letters to fill the second spot. Four choices of letters to fill the third spot. And so on.
So, the total number of different combinations of the six letters would be 6!
6! = 6 X 5 X 4 X 3 X 2 X 1
= 720
DAY could appear consecutively starting in 4 different spots: the 1st, 2nd, 3rd, and 4th spots of all combinations. DAY will appear consecutively 6 times in each of those 4 starting spots.
6 X 4 = 24.
So, there are 720 total different combinations, and 24 of those will have the letters DAY consecutively within all those combinations.
Answer : 18x3+57x2+38x+7
Hope this helps !
