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Dvinal [7]
3 years ago
8

I'm thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is d

ivisible by n for each n from 1 to 10. What is my number?
Mathematics
1 answer:
klemol [59]3 years ago
5 0
The problem here takes a brilliant mind to answer this. This problem can easily be answered using programming because we can not then and there push all the possibilities using paper and pen.

The answer is <span>3816547290. 
</span>
Trying all the possibilities starting form 1000000080 (we are sure that the last number should be 0). Then traversing that number until <span>9999999990. Each traverse, check the number if its divisible to n, so on and so forth.
</span>
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What is 2 and 5/14 divided by (2 and 5/8 times 1 and 3/7)?
Lina20 [59]
(2 5/14) / (2 5/8 * 1 3/7)....turn them all into improper fractions
(33/14) / (21/8 * 10/7)
(33/14) / (105/28)...when dividing fractions, flip what u r dividing by, then multiply
33/14 * 28/105 = 132/210 reduces to 22/35 <==
3 0
4 years ago
Ab - 0.5b when a = 1 and b = 5
Korolek [52]

Answer:

2.5

Step-by-step explanation:

a = 1

b = 5

ab

= 1 x 5

= 5

0.5b

= 0.5 x 5

= 2.5

ab - 0.5b

= 5 - 2.5

= 2.5

4 0
3 years ago
Read 2 more answers
A vehicle arriving at an intersection can turn right, turn left, or continue straight ahead. The experiment consists of observin
Temka [501]

Answer:

a) Sample space (S) = {TR, TL, Cs}

b) Pr(Vehicle Turns) = 2/3.

Step-by-step explanation:

a) By sample space (S) we mean the list all possible outcome of an event. From the illustration in the question, only three (3) outcome is possible:

i) Vehicle Turn Right (TR)

ii) Vehicle Turn Left (TL)

iii) Continue Straight ahead (Cs)

And since the experiment consists of observing the movement of a single vehicle through the intersection, then, the sample space (S) is:

==> {TR, TL, Cs}

b) Since we are assuming that all sample points are equally likely, it imply that they have equal probability of occurring. And by probability:

Pr(TR) = 1/3

Pr(TL) = 1/3

Pr(Cs) = 1/3

Meanwhile, the question wants us to find the probability that the vehicle turns. This means, the vehicle turning either right or left. Thus;

Pr(vehicle turns) = Pr(TR) or Pr(TL) = Pr(TR) + Pr(TL)

Pr(vehicle turns) = (1/3) + (1/3) = 2/3

4 0
3 years ago
Help please (-5)-20-(-49)
Tom [10]
It's just -5-20-49 so the answer would be -74
8 0
3 years ago
Please help will brainliest all please
Shalnov [3]
Font know hope you figure it out
7 0
3 years ago
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