The lest number

For students of class 9-10 (age 14-16)
Rezaul Haque
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Joined:Sat Jan 01, 2011 3:45 pm
The lest number

Unread post by Rezaul Haque » Sat Jan 01, 2011 3:56 pm

Well my problem is pretty much simple! But the problem is that I can't deduce the logical steps required. So please help. Now let's return to the problem: What is the least number which when divided by 18, 24, & 36 yield remainders of 12,18, & 30 respectively!!!

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Labib
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Location:Dhaka, Bangladesh.

Re: The lest number

Unread post by Labib » Sat Jan 01, 2011 11:39 pm

The solution is
$66$
.

At first you need to find $lcm(18,24,36)$

then try all $36p+30$ for $p=1,2,3,....$ in $lcm(18,24,36)$ range for a possible solution.

It's my method. other people may have better solutions! :D
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Avik Roy
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Re: The lest number

Unread post by Avik Roy » Sun Jan 02, 2011 2:58 am

There certainly is a better way. Observe that the desired remainders are just $6$ less than the divisors.
$12 = 18 - 6$
$18 = 24- 6$
$30 = 36 -6$
The desired number is just $-6 modulo 18, 24, 36$. This means this numbers can be represented in the form $18a - 6$ or, $24b - 6$ or, $36c -30$, where $a,b,c$ are positive integers.
Such a number is $LCM(18,24,36) - 6 = 66$
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

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Moon
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Re: The lest number

Unread post by Moon » Sun Jan 02, 2011 3:25 am

I was wondering why the moderators did not move this post to correct forum...Anyway, Rezaul: please post your problems in correct forum. (Read the forum guide for more info).

Welcome, anyway. :)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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