The number 9990 can be written as the product of 5 an equal positive integers are greater than 1 what is the largest possible integer that could be included
To find the largest possible integer that could be included in the product of 9990 as equal positive integers greater than 1, we need to factorize the number.
The prime factorization of 9990 is:
9990 = 2 * 3 * 3 * 5 * 7 * 19
Now, in order to find the largest integer that could be included as a factor, we need to distribute these prime factors as evenly as possible among the equal positive integers.
The idea is to distribute the prime factors from largest to smallest until we run out of factors or cannot distribute them any further.
Let's start with the largest prime factor, 19. We can include it as a factor along with the other prime factors until we reach the point where the remaining prime factors are smaller than 19.
9990 ÷ 19 = 526.32...
Since we can't have a fraction of a factor, we stop here with 19.
Now, let's continue with the next largest prime factor, 7.
526 ÷ 7 = 75.14...
Again, we stop with 7.
Moving on to the next largest prime factor, 5.
75 ÷ 5 = 15.
We stop with 5.
Next is 3.
15 ÷ 3 = 5.
Lastly, we have 2.
5 ÷ 2 = 2.
At this point, we have distributed all the prime factors.
Now, let's calculate the product of these factors:
19 * 7 * 5 * 3 * 2 = 13380.
Therefore, the largest possible integer that could be included in the product of 9990 as equal positive integers greater than 1 is 13380.
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9990 = 9*10*111
That may help to answer the question, since I'm not sure what
5 an equal positive integers
means