Finding Factors

Continuing Bernardo's solution for an explanation of why $129600$ is the minimum value:

View this as an inequality problem:

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Minimize $a+1+b+1+c+1$ where $a$ , $b$ , and $c$ are not necessarily integers* such that $(a+1)(b+1)(c+1)=105$ .

*It doesn't matter right now, it will simply prove a point.

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Apply the AM-GM Inequality to get

$\dfrac{a+1+b+1+c+1}{3} \geq \sqrt[3]{(a+1)(b+1)(c+1)}$

Substituting, this is

$\dfrac{a+1+b+1+c+1}{3} \geq \sqrt[3]{105}$

The minimum is obviously attained when $a=b=c=\sqrt[3]{105}-1$ .

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The result of course is very trivial, but it's all we need for the problem. What we've shown is that as long as $a+1$ , $b+1$ , and $c+1$ are as close as possible to equal, we'll have minimized their sum.

Why do we want to minimize their sum? Because the less their sum is, the less we'll be multiplying our number to make it greater. Convince yourself that this is true, it's also rather trivial.

Looking at the prime factorization of $105$ , which coincidentally is

$3 \times 5 \times 7,$

we notice how $3$ , $5$ , and $7$ are the most condensed possible $a+1$ , $b+1$ , and $c+1$ (convince yourself this is true). Our exponents, then, will be $3-1=2$ , $5-1=4$ , and $7-1=6$ .

But how do we distribute them? Should we raise the $5$ to the $6$ th power?

No. We want to have as little $5$ 's in our prime factorization as we possible can. So we assign $2$ to our $5$ . Similarly, we raise the $3$ to the $4$ , and lastly the $6$ to the $2$ . Our number is thus

$2^6 \times 3^4 \times 5^2=\boxed{129,600}.$

Great problem @Katie Gardner ! :D

d(n)=103+2=105 d(n)=(a+1)(b+1)(c+1)=105=(3)(5)(7)

notice that: 2^{a} <3^{a} < 5^{a} so we choose : a=6 , b=4 , c=2 from all possible choices..

and finally : min(n)=(2^{6} )(3^{4} )(5^{2} )=129600

p.s: i am muslim..islam is best in all of the universe.

I can pass this case.But i think this case made me Confused.Because the the number has 103 proper factors Not including 1 or the number itself(i think Its mean that actually the number factors is 104 )

First attemp and second attemp i get wrong answered.So the third Attemp,I assume the factors is 105 sothat i can answered this case.Despite i can do this case.I hope Katie Gardner Repaired the word of this

2^a x 3^b x 5^c have the factors

(a+1)x(b+1)x(c+1)=105

(a+1)x(b+1)x(c+1)=5 x 3 x 7=105

so all of you can guess the value of a=6 b=4 c=2 I hope this is useful :)

The number of proper factors is the number of divisors - 2.

The number of divisors would be (a + 1)(b + 1)(c + 1).

We want to get a, b and c such that (a + 1)(b + 1)(c + 1) - 2 = 103.

(a + 1)(b + 1)(c + 1) - 2 = 103

(a + 1)(b + 1)(c + 1) = 105

(a + 1)(b + 1)(c + 1) = (3)(5)(7)

To minimize this number, the biggest exponent should be a. Then b and, subsequently, c.

a = 6; b = 4; c = 2;

2 ^ 6 x 3 ^ 4 x 5 ^ 2 = 129600

Note that we could also have, for instance,

a = 14; b = 6; c = 0;

But this is 11943936.

I didn't wrote a proof that 129600 is THE MINIMUM. Just showed a way to get to it.

With Both Unity And The Number We Have No. Of Divisors....

(a+1)(b+1)(c+1) = 105 = 7x5x3..........

And Thus The Answer.........

{{Note: we give 2 : 6....3 : 4.....and 5:6 to get lowest answer....

there will be (a+1)x(b+1)x(c+1)-2 proper fraction.... (a+1)x(b+1)x(c+1)- 2 = 103.... (a+1)x(b+1)x(c+1)=105 for no. to be minimum, a>b>c... also 105 is odd, so a,b,c, should be even, put c=2, b=2 .... no integer value of a is possible put c=2, b=4 ==> a=6 also c=4 ,b=2 will also give same value of a =6 but that will not give minimum value ..... so put a=6 ,b=4 and c=2 ans .....129600