LCM

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LCM

by HeyArnold » Fri Aug 12, 2011 3:55 am
Which of the following is the lowest positive integer that is divisible by 2, 3, 4, 5, 6, 7, 8, and 9?

A) 15,120
B) 3,024
C) 2,520
D) 1,890
E) 1,680

[spoiler]OA: C ... I got 5040 as an answer, in this case you will get 2^7, so I treated each "matched set" of 2's as one, and for LCM also multiply against the unmatched, so 2^6 becomes just 2 x 2 x2, and then 1 unmatched 2, ... so 2^4 x 3^2 x 5 x 7 ?[/spoiler]
Source: — Problem Solving |

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by Anurag@Gurome » Fri Aug 12, 2011 4:05 am
HeyArnold wrote:Which of the following is the lowest positive integer that is divisible by 2, 3, 4, 5, 6, 7, 8, and 9?
Calculating the LCM of these numbers is a time consuming and cumbersome process. Instead it'll be pretty much easier to check for the individual divisibility.

Let's start with the divisibility by bigger ones as it is quiet expected that all of the options are divisible by the smaller ones like 2 or 3.

Divisibility by 9 --> The digits of option E add upto 15 --> Rejected
Divisibility by 8 --> The last three digits of option D is not divisible by 8 --> Rejected
Divisibility by 5 --> The last digit of option B is neither 0 or 5 --> Rejected

Only two option remains A and C. Out of which option C is smaller. Let's check whether it is divisible by 7 or not. 2520 = 7*360

Hence, out of the five options 2520 is the lowest positive integer that is divisible by 2, 3, 4, 5, 6, 7, 8, and 9.

The correct answer is C.
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by GmatKiss » Fri Aug 12, 2011 4:14 am
Hi Anurag,

Divisibility by 9 --> The digits of option E add upto 15 --> Rejected

Also could you share us the rules for all other digits as well.

TIA,
GK

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by Anurag@Gurome » Fri Aug 12, 2011 4:19 am
GmatKiss wrote:...could you share us the rules for all other digits as well
Please have a look in the second table of the article here : https://www.mathgoodies.com/lessons/vol3 ... ility.html

For a more advanced discussion go through this Wikipedia article : https://en.wikipedia.org/wiki/Divisibility_rule
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