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Lelechka [254]
3 years ago
14

If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n st ?

I. \small s^{t} II. \small \left ( st \right )^{2} III. \small s+tA) NoneB) 1 onlyC) 2 onlyD) 3 onlyE) 1 and 2
Mathematics
1 answer:
yulyashka [42]3 years ago
4 0

Answer:

A) None

Step-by-step explanation:

1) s^t shoudnt neccesarily be a factor of nst, for example, if s = 3, t = 4, and n = 12, then both s and t are factors of n, but 3^4 = 81 is not a factor of nst = 144.

2) (st)^2 shoudnt neccesarily be a factor of nst. Let s be 4, let t be 6, and let n be 12. Then n is a factor of both s and t, but (st)^2 = 24^2 is not a factor of nst = 12*24. In fact, it is a greater number.

3) Again, s+t isnt necessarily a factor of nst, let s be 2 and t be 3. Then both s and t are factor of n = 12. However 5 = s+t is not a factor of nst = 72.

So, neither of the three options is guaranteed to be a factor of nst. In fact, for s = 4, t = 6, and n = 12, none of the three options are valid.

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katrin [286]

Answer:

20,475\ ways

Step-by-step explanation:

we know that

<u><em>Combinations</em></u> are a way to calculate the total outcomes of an event where order of the outcomes does not matter.

To calculate combinations, we will use the formula

C(n,r)=\frac{n!}{r!(n-r)!}

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In this problem

n=28\\r=4

substitute

C(28,4)=\frac{28!}{4!(28-4)!}\\\\C(28,4)=\frac{28!}{4!(24)!}

simplify

C(28,4)=\frac{(28)(27)(26)(25)(24!)}{4!(24)!}

C(28,4)=\frac{(28)(27)(26)(25)}{(4)(3)(2)(1)}

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3 years ago
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pshichka [43]

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2 years ago
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Answer: P=5y+11x


Step-by-step explanation:

To solve this problem you must apply the proccedure shown below:

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8 0
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