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factors theory of nb

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by kvcpk » Thu Jun 17, 2010 7:43 am
112 = 16 * 7
33 = 11 * 3

a * 43 * 62 * 1311 = (2 ^4 * 7) * k1
a * 43 * 31 * 2 * 1311 = (2^4 *7) * k1
a * 43 * 31 * 2 * 3 *437 = (2^4 *7) * k1

therefore minimum value of a shud be 2^3 * 7 = 56

for the second factor,

a * 43 * 62 * 1311 = (11 * 3) * k2
a * 43 * 31 * 2 * 3 *437 = (11 * 3) * k2

so min = 11


on the whole it shud be 56 * 11 = 616

let me know OA?
Last edited by kvcpk on Thu Jun 17, 2010 7:52 am, edited 1 time in total.
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by amising6 » Thu Jun 17, 2010 7:44 am
If both 112 and 33 are factors of the number a * 43 * 62 * 1311, then what is the smallest possible value of a?

a*43*62*1311
a*43*2*31*3*19*23

now we know 33 is a factor this means a should be multiple of 11

now we also know that 112 is a factor
112=2*2*2*2*7
so we can see that a*43*2*31*3*19*23 contains only one 2 and no 7
so a should also contain 2*2*2 and one 7
so a=2*2*2*7*11=616
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by amising6 » Thu Jun 17, 2010 7:46 am
kvcpk wrote:112 = 4 * 11 * 3
33 = 11 * 3 this is not required.. bcos 112 is factor implies that 33 is factor(how can you say this)

a * 43 * 62 * 1311 = (4 * 11 * 3) * k1
a * 43 * 31 * 2 * 1311 = (4 * 11 * 3) * k1
a * 43 * 31 * 2 * 3 *437 = (4 * 11 * 3) * k1

therefore minimum value of a shud be 2 * 11 = 22


let me know OA?
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by kvcpk » Thu Jun 17, 2010 7:48 am
amising6 wrote:
kvcpk wrote:112 = 4 * 11 * 3
33 = 11 * 3 this is not required.. bcos 112 is factor implies that 33 is factor(how can you say this)

a * 43 * 62 * 1311 = (4 * 11 * 3) * k1
a * 43 * 31 * 2 * 1311 = (4 * 11 * 3) * k1
a * 43 * 31 * 2 * 3 *437 = (4 * 11 * 3) * k1

therefore minimum value of a shud be 2 * 11 = 22


let me know OA?
Sorry!! My mistake.. messed up 112 with 132... thanks for noting it.. will modify now..
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by jube » Thu Jun 17, 2010 8:11 am
112 =2^4 * 7
33=11*3

factors which the number already has:
43
62 = 2*31
1311 = 3*19*23

therefore a has to have 2^3*7*11 in it or a=616
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by francoisph » Thu Jun 17, 2010 2:29 pm
sorry it is 11^2 and 3^3
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by sk818020 » Thu Jun 17, 2010 3:01 pm
I also got 616 as the answer for similar reasoning as posted above. Could you please post what the official explanation was if you saying the OA is 11^2 and 3^3 so that we might all learn from this problem?

Thanks,

Jared
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by francoisph » Thu Jun 17, 2010 3:24 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?

11^2 is a factor of the given number.
The number does not have a power or multiple of 11 as its factor.
Hence, "a" should include 11^2
3^3 is a factor of the given number. 6^2 is a part of the number. 6^2 has 3^2 in it.
Therefore, if 3^3 has to be a factor of the given number a * 43 * 62 * 1311, then we will need at least another 3.
Therefore, "a" should be at least 11^2 * 3 = 363 if the given number has to have 11^2 and 3^3 as its factors.
The smallest value that "a" can consequently take is 363.

Could someone explain clearly please?
I didnt get the official explanation
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by sk818020 » Thu Jun 17, 2010 3:31 pm
francoisph wrote: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?

11^2 is a factor of the given number.
The number does not have a power or multiple of 11 as its factor.
Hence, "a" should include 11^2
3^3 is a factor of the given number. 6^2 is a part of the number. 6^2 has 3^2 in it.
Therefore, if 3^3 has to be a factor of the given number a * 43 * 62 * 1311, then we will need at least another 3.
Therefore, "a" should be at least 11^2 * 3 = 363 if the given number has to have 11^2 and 3^3 as its factors.
The smallest value that "a" can consequently take is 363.

Could someone explain clearly please?
I didnt get the official explanation
The prime factorization of a*43*62*1311 = 2*3*19*23*31*43*a

The prime factorization of 112 = 2^4*7

The prime factorization of 33 = 3*11

Because a*43*62*1311 has 2 and 3, the smallest possible number a could be = (2^3)*7*11 = 616

Hope that helps.

Thanks,

Jared
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by francoisph » Thu Jun 17, 2010 3:41 pm
the official explanation

A. 121
B. 3267
C. 363
D. 33
E. None of the above

The correct choice is (C) and the correct answer is 363.
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by sk818020 » Thu Jun 17, 2010 3:46 pm
francoisph wrote:the official explanation

A. 121
B. 3267
C. 363
D. 33
E. None of the above

The correct choice is (C) and the correct answer is 363.
That is why your throwing everyone off, lol!

You incorrectly put 112 and 33 in your original post when you meant, 11^2 and 3^3.
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by francoisph » Thu Jun 17, 2010 3:54 pm
lol o:) power mate
11^2 and 3^3 are factors of the number a * 43 * 62 * 1311
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by sk818020 » Thu Jun 17, 2010 3:55 pm
francoisph wrote:lol o:) power mate
11^2 and 3^3 are factors of the number a * 43 * 62 * 1311
I also think its supposed to be 6^2. Please confirm.
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by francoisph » Thu Jun 17, 2010 3:58 pm
apologies guys

11 power 2 and 3power 3 are factors of the number a * 4power 3 * 6 power 2 * 13power 11
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