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sukhopar [10]
2 years ago
12

Find integers a, b, and c such that a) a | (b - c) but a - b and a - c b) a | (2b + 3c) but a - b and a - c

Mathematics
1 answer:
seraphim [82]2 years ago
5 0

Answer:

Part a) (a,b,c)=(10,19,9)

Part b) (a,b,c)=(9,3,1)

Please let be know if I have interpreted your question correctly. Thank you.

Step-by-step explanation:

Part a)

This is the way I interpret part a:

Find integers a,b,c such that a|(b-c) but a does not divide b and a does not divide c.

There are many triples satisfying part a.

One such triple is (a,b,c)=(3,10,7).

3 divides 10-7

But 3 does not divide 10 and 3 does not divide 7.

Another example is (a,b,c)=(2,7,5).

2 divides 7-5

But 2 doesn't divide 7 and 2 doesn't divide 5.

I will give another example.

(a,b,c)=(10,19,9)

10 divides 19-9

But 10 doesn't divide 19 and 10 doesn't divide 9

Part b)

This is the say I interpret part b:

a | (2b + 3c) but a does not divide b and a does not divide c

So lets let b=2 and c=5, then 2b+3c=19.

We just need to choose an a such that it divides 19 but not 2 and 5. That should be easy. 19 itself will do that.

So one (a,b,c) could be (19,2,5).

Let's try for another example.

Let b=3 and c=1.

Then 2b+3c=9.

So we just need to find an a that divides 9 but not 3 and not 1.

9 works.

So another example is (9, 3,1).

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At the first draw, we only selected one of the 10 prizes and so 1 draw gives a success

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After the first prizes have being drawn, we now interested in the first time success that we draw a different prize from the first 2

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After the first three prizes have been drawn, we are now interested in the success of a first time draw for the 4th different prize

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