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Dennis_Churaev [7]
4 years ago
14

How many positive integers less than 2018 are divisible by at least 3, 11, or 61?

Mathematics
1 answer:
Schach [20]4 years ago
7 0
To find all the positive integers less than 2018 that are divisible by 3, 11, and 61, you will use what you know about factors.

3, 11, and 61 are all answers.  So are 33, 183, 671, and 2013.

If you put these in factors, the product will be divisible by them!

3 x 11 = 33
3 x 61 = 183
11 x 61 = 671
3 x 11 x 61 = 2013

Take each number and square it, cube it, etc...
9, 27, 81, 243, 729
121, 1331

9 x 11 = 99
27 x 11 = 297
81 x 11 = 891

121 x 9 =1089
121 x 3 =  363

61 x 9 = 549
61 x 27 = 1647

Everything in bold is a correct answer.
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Answer:

<u>Option C. The product was negative.</u>

Step-by-step explanation:

Let the non-zero rational number is = \frac{x}{y}

So, its additive inverse = -\frac{x}{y}

The result of multiplication the non-zero rational number by its additive inverse = \frac{x}{y} *-\frac{x}{y} =-\frac{x^2}{y^2}

We will check the options:

A. The product was the multiplicative inverse of the original number.

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B. The product was the additive inverse of the original number.

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C. The product was negative.

<u>True</u>

D. The product was positive.

<u>Wrong </u>because the product was negative.

<u>So, the answer is option C.</u>

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