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factors theory of nb

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by sk818020 » Thu Jun 17, 2010 4:01 pm
The problem correctly written is;

If both 11^2 and 3^3 are factors of the number a * 4^3 * 6^2 * 13^11, then what is the smallest possible value of a?

The prime factorization of a * 4^3 * 6^2 * 13^11 is = a*2^8*3^2*13^11.

Because a*2^8*3^2*13^11 has no 11 as a factor and 2 threes a factors, then the smallest a could equal is

11^2*3=363.

Thanks,

Jared
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by francoisph » Thu Jun 17, 2010 4:03 pm
jared

thks dude
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by sk818020 » Thu Jun 17, 2010 4:04 pm
francoisph wrote:jared

thks dude
Anything to help! Even though there were a few errors :-p
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by amising6 » Thu Jun 17, 2010 7:28 pm
If both 11^2 and 3^3 are factors of the number a * 43 * 62 * 1311, then what is the smallest possible value of a?

a * 43 * 62 * 1311 i guess this is a * 4^3 * 6^2 * 13^11
now as given 11^2 is a factor but a * 4^3 * 6^2 * 13^11 this does not contain any 11 so a should contain 2 11's
now 3^3 is also factor in a * 4^3 * 6^2 * 13^11 if you take 6^2 it can be written as (3*2)^2 i.e 3^2*2^2

so we can see a * 4^3 * 6^2 * 13^11 contain 2 3's
thus our a should contain one more 3 so 3^3 can become factor of a * 4^3 * 6^2 * 13^11
so a =11*11*3=363
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by kvcpk » Thu Jun 17, 2010 9:51 pm
francoisph wrote:apologies guys

11 power 2 and 3power 3 are factors of the number a * 4power 3 * 6 power 2 * 13power 11
thats ok!! Powers are so confusing.. Anyway we got to solve a new problem too ;)
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