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S98-07

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:52
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The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because $$2 \times 2 \times 5 = 20$$. What is the length of 5,950?

A. 1
B. 2
C. 3
D. 4
E. 5

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16 Sep 2014, 01:52
Official Solution:

The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because $$2 \times 2 \times 5 = 20$$. What is the length of 5,950?

A. 1
B. 2
C. 3
D. 4
E. 5

In this problem, we should notice that the "length" of an integer is defined in some way involving prime factors. Therefore, even if we are not 100% sure what "length" means, we should factor 5,950 down to its prime factors and examine what we get.

First, it's easy to take out a factor of 10: $$5950 = 10 \times 595$$.

10 is very simple to factor: $$10 = 2 \times 5$$.

Since 595 ends in 5, we know that 595 is divisible by 5. Actually performing this division, we wind up with $$595 = 5 \times 119$$.

So $$5950 = 2 \times 5 \times 5 \times 119$$.

Now, 119 "looks" prime, but you must test it with primes up to the square root of 119, which is approximately 11. (If 119 is not prime, then at least one of its prime factors must be smaller than the square root of 119.)

It turns out that 119 is divisible by 7. $$119 = 7 \times 17$$.

Thus, we now have the full prime factorization of 5,950: $$5950 = 2 \times 5 \times 5 \times 7 \times 17$$.

Finally, we return to the definition of "length." We are given the example that the length of 20 is 3, since $$2 \times 2 \times 5 = 20$$. So we can see that "length" is just the number of prime numbers in the prime factorization, counting repeats (such as 2 in the example of 20).

Thus, the length of 5,950 is 5.

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30 Dec 2014, 05:56
First of all, this was a very logical solution.

One question I have is whether this would be possible if we divided 5950 in 5000 + 900 + 50.

If this is also possible, how do we reduce the factors?

I am asking because my initial thought was to find the factors for the above mentioned addition.

Thank you.
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31 Dec 2014, 04:48
pacifist85 wrote:
First of all, this was a very logical solution.

One question I have is whether this would be possible if we divided 5950 in 5000 + 900 + 50.

If this is also possible, how do we reduce the factors?

I am asking because my initial thought was to find the factors for the above mentioned addition.

Thank you.

No, you cannot do this way.

Check similar questions to practice:
for-any-integer-k-1-the-term-length-of-an-integer-108124.html
for-any-positive-integer-n-the-length-of-n-is-defined-as-126740.html
for-any-positive-integer-n-n-1-the-length-of-n-is-the-126368.html
the-length-of-integer-x-refers-to-the-number-of-prime-132624.html
the-length-of-a-positive-integer-greater-than-1-is-the-number-of-pri-188734.html

Hope it helps.
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04 Oct 2018, 06:35
Is there any shortcut to this?
I tried multiple combinations to find factors of 119, and then gave up with just a guess.
Re: S98-07   [#permalink] 04 Oct 2018, 06:35
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