Sum of the factors

**skyskiers** · October 1st 2009, 03:14 PM

Hi,

I need some help in finding the sum of the factors of 18000 which are multiple of 8 but are not multiple of 25

i proceeded as below.

18000 = 2^4 * 3^2 * 5^3

Now
factors that are multiple of 8 = 2* 3^2 * 5^3
factors that are multiple of 8and 25 = 2 * 3^2 * 5

sum = sum[factors that are multiple of 8] - [factors that are multiple of 8and 25]

where m i going wrong and how to proceed?

Thanks.

**Soroban** · October 1st 2009, 07:38 PM

Hello, skyskiers!

If they had asked for "the number of factors which are multiples of 8
but not a multiple of 25", there are some theorems we could use.
. . . There are 12.

But they asked for the sum of these factors.
I see no choice but to find all these factors and add them.

It is not a very intellectual problem, is it?

Find the sum of the factors of 18,000 which are multiple of 8 but are not multiple of 25.

There are 60 factors.
I found them by simple division.

. . $\begin{array}{cc}<br /> 1 & 18000 \\ 2 & 9000 \\ 3 & 6000 \\ 4 & 4500 \\ 5 & 3600 \\ 6 & 3000 \\ {\color{red}8} & 2250 \\ 9 & 2000 \\ 10 & 1800 \\ 12 & 1500 \end{array}$ . . . $\begin{array}{cc}15 & 1200 \\ {\color{red}16} & 1125 \\ 18 & 1000 \\ 20 & 900 \\ {\color{red}24} & 750 \\ 25 & {\color{red}720} \\ 30 & 600 \\ 36 & 500 \\ {\color{red}40} & 450 \\ 45 & 400 \end{array}$ . . . $\begin{array}{cc} {\color{red}48} & 375 \\ 50 & {\color{red}360} \\ 60 & 300 \\ {\color{red}72} & 250 \\ 75 & {\color{red}240} \\ {\color{red}80} & 225 \\ 90 & 200 \\ 100 & 180 \\ {\color{red}120} & 150 \\ 125 & {\color{red}144} \end{array}$

As promised, there are 12 such factors.

Their sum is: . $8+16+24+40+48+72+80+120+144+240+360+720 \:=\:\boxed{1872}$

**skyskiers** · October 2nd 2009, 11:40 AM

Hello Soroban,

Right , doesnt seem intellectual...probably that was the only reason i posted it here ..thought if there existed any better way...

Thanks once again..

**aman_cc** · October 2nd 2009, 12:05 PM

Originally Posted by skyskiers

Hello Soroban,

Right , doesnt seem intellectual...probably that was the only reason i posted it here ..thought if there existed any better way...

Thanks once again..

There indeed is
Any factor, with required criteria is of the form

$2^x.3^y.5^z$

where $x \in \{3,4\}$ , $y \in \{0,1,2\}$ and $z \in \{0,1\}$

What do can you say about
$(2^3+2^4)(3^0+3^1+3^2)(5^0+5^1)$

Do you see the argument?
Thanks

**skyskiers** · October 2nd 2009, 02:05 PM

Originally Posted by aman_cc

There indeed is
Any factor, with required criteria is of the form

$2^x.3^y.5^z$

where $x \in \{3,4\}$ , $y \in \{0,1,2\}$ and $z \in \{0,1\}$

What do can you say about
$(2^3+2^4)(3^0+3^1+3^2)(5^0+5^1)$

Do you see the argument?
Thanks

Yes Aman, got the argument thick and fine.... great logic... got me out of the trouble ....

Thanks..

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