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Fiesta28 [93]
3 years ago
5

(79,35+7,32) × 42,2-9543​

Mathematics
1 answer:
vladimir2022 [97]3 years ago
3 0

Answer:

42(79,42,32)

Step-by-step explanation:

I got the solution by using  the  Trigonometric Identities. If somebody else could do a step by step explanation that would be great.  

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Find all the square roots of x2 ≡ 53 (mod 77)
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Answer:

x=\pm\sqrt{77n+53}

Step-by-step explanation:

We have been given an equivalence equation x^2\equiv 53\text{ (mod } 77). We are asked to find all the square root of the given equivalence equation.

Upon converting our given equivalence equation into an equation, we will get:

x^2-53=77n

Add 53 on both sides:

x^2-53+53=77n+53

x^2=77n+53

Take square root of both sides:

x=\pm\sqrt{77n+53}

Therefore, the square root for our given equation would be x=\pm\sqrt{77n+53}.

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How many multiples of 4 are there in {n; 37< n <1001}?
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Let's begin by listing the first few multiples of 4:  4, 8, 12, 16, 20, 24, 28, 32, 36, 38, 40, 44.  So, between 1 and 37 there are 9 such multiples:  {4, 8, 12, 16, 20, 24, 28, 32, 36}.  Note that 4 divided into 36 is 9.

Let's experiment by modifying the given problem a bit, for the purpose of discovering any pattern that may exist:

<span>How many multiples of 4 are there in {n; 37< n <101}?  We could list and then count them:  {40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100}; there are 16 such multiples in that particular interval.  Try subtracting 40 from 100; we get 60.  Dividing 60 by 4, we get 15, which is 1 less than 16.  So it seems that if we subtract 40 from 1000 and divide the result by 4, and then add 1, we get the number of multiples of 4 between 37 and 1001:

1000
   -40
-------
 960

Dividing this by 4, we get 240.  Adding 1, we get 241.

Finally, subtract 9 from 241:  We get 232.

There are 232 multiples of 4 between 37 and 1001.

Can you think of a more straightforward method of determining this number? </span>

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