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Susan123456
 Post subject: Prove that there exists a 3-digit natural number
PostPosted: Sun, 27 Sep 2009 00:03:42 UTC 
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Hello,

Hope someone can help me.

Problem: Prove that there exists a 3-digit natural number less than 400 w/distinct digits, such that the sum of the digits is 17 and the product of the digits is 108.

My setup: I have 2 equations:

(1) xyz = 108
(2) x +y +z = 17.

I did some substitution and the last step I got was x^2z +xz^2 -17xz +108. This is where I got stuck. I can't factor by grouping. Could someone give me a hint?

Thank you.

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dmoran
 Post subject: Re: Prove that there exists a 3-digit natural number
PostPosted: Sun, 27 Sep 2009 00:16:56 UTC 
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Susan123456 wrote:
Hello,

Hope someone can help me.

Problem: Prove that there exists a 3-digit natural number less than 400 w/distinct digits, such that the sum of the digits is 17 and the product of the digits is 108.

My setup: I have 2 equations:

(1) xyz = 108
(2) x +y +z = 17.

I did some substitution and the last step I got was x^2z +xz^2 -17xz +108. This is where I got stuck. I can't factor by grouping. Could someone give me a hint?

Thank you.


Have you learned Lagrange Multipliers?

Dave

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Susan123456
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PostPosted: Sun, 27 Sep 2009 00:58:54 UTC 
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Yes, I have but it's been a while. I wonder if I can use Algebra for this proof. Thank you.

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Susan
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dmoran
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PostPosted: Sun, 27 Sep 2009 01:12:21 UTC 
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Susan123456 wrote:
Yes, I have but it's been a while. I wonder if I can use Algebra for this proof. Thank you.


I don't think that'd be the best way to go.

Dave

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outermeasure
 Post subject: Re: Prove that there exists a 3-digit natural number
PostPosted: Sun, 27 Sep 2009 02:54:18 UTC 
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Susan123456 wrote:
Hello,

Hope someone can help me.

Problem: Prove that there exists a 3-digit natural number less than 400 w/distinct digits, such that the sum of the digits is 17 and the product of the digits is 108.

My setup: I have 2 equations:

(1) xyz = 108
(2) x +y +z = 17.

I did some substitution and the last step I got was x^2z +xz^2 -17xz +108. This is where I got stuck. I can't factor by grouping. Could someone give me a hint?

Thank you.


Since 108=2^2\times 3^3, x,y,z can only be 1,2,3,4,6,8 or 9. Because you want your 3-digit nmber to be less than 400, and there are no 2 single-digit numbers that give a product >81, the hundred digit (presumably x?) must be 2 or 3. Now examine each case.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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alstat
 Post subject: Re: Prove that there exists a 3-digit natural number
PostPosted: Sun, 27 Sep 2009 07:18:06 UTC 
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Lagrange Multipliers? That's opening a peanut with a sledgehammer!

How about basic algebra?

Spoiler:
Since 100x+10y+z<400, x<4 so that x=1, 2 or 3

x=1 means yz=108 (impossible for one-digit numbers).

x=3 means yz=36 and y+z=14 (no such combination of integers exists).

x=2 means yz=54 and y+z=15, thus (y,z) = (6,9) or (9,6)

so 269 and 296 are the numbers we seek.
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outermeasure
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PostPosted: Sun, 27 Sep 2009 08:35:34 UTC 
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Actually, it is easier to establish one of the digits is 6.

From the factorisation of 108, you know the three digits must contain at least one even number. Also, since the sum of all three digits is odd, we know the three digits must be 2 even and one odd. Now look at how the three 3s can distribute, and you are forced to conclude one of the digit must be 6.

So you are reduced to solving for two numbers whose sum is 11, and whose product is 18. So solving gives you the two numbers.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Susan123456
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PostPosted: Sun, 27 Sep 2009 17:44:57 UTC 
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Thanks everyone for your help.

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Susan
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