If the LCM of (10!)(18!) and (12!)(17!) is expressed as....

[merlin2007]

Hi, I have a problem that feels more like number theory, but I'm not really sure, because there doesn't seem to be a number theory section, and it doesn't seem all that advanced.

If the least common multiple of (10!)(18!) and (12!)(17!) is expressed in the form a!b!/c!, where a and b are two-digit numbers and c is a one-digit number, find abc.

Thanks

 

[soroban]

Re: Not sure if this is the correct forum

Hello, merlin2007!

What a strange problem!
I found no direct way to solve it.

If the LCM of \(\displaystyle (10!)(18!)\) and \(\displaystyle (12!)(17!)\) is expressed in the form \(\displaystyle \L\frac{a!b!}{c!}\)

where \(\displaystyle a\) and \(\displaystyle b\) are two-digit numbers and \(\displaystyle c\) is a one-digit number, find \(\displaystyle abc.\)

I did it by "brute force" factoring . . . well, sort of.

. . \(\displaystyle (10!)(18!) \:=\10!)\cdot18\cdot(17!)\)

. . \(\displaystyle (12!)(17!) \:=\:12\cdot11\cdot(10!)(17!)\)

Hence, the LCM contains: \(\displaystyle \,10!\) and \(\displaystyle 17!\), and the LCM of \(\displaystyle 18\) and \(\displaystyle 12\cdot11\)

. . The LCM of \(\displaystyle 18\) and \(\displaystyle 11\cdot12\) is: \(\displaystyle \,2^2\cdot3^2\cdot11\)

Hence, the LCM is: \(\displaystyle \10!)(17!)\cdot2^2\cdot3^2\cdot11\)


To get the answer in the desired form, I did some acrobatics.
. . I noted that: \(\displaystyle \,2^2\cdot3^2\cdot11 \:=\:12\cdot11\cdot3\)

Then the LCM is: \(\displaystyle \:\underbrace{12\cdot11\cdot(10!)}(17!)\cdot3\)
. . .and we have: \(\displaystyle \;\;\;(12!)(17!)\cdot3\)


Now, what to do about that extra "3" on the end?

It took a while, but I saw that: \(\displaystyle \:3 \:=\:\frac{18}{6}\:=\:\frac{18}{3!}\)

So the LCM is: \(\displaystyle \12!)(17!)\frac{18}{3!} \;=\;\frac{(12!)(17!)\cdot18}{3!} \;=\;\frac{(12!)(18!)}{3!}\;\) . . . There!


Therefore: \(\displaystyle \:abc\:=\12)(18)(3)\:=\:\fbox{648}\)


Is there a shorter way? . . . I hope so!