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# Divisibility

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Intern
Joined: 31 Oct 2010
Posts: 31

Kudos [?]: 74 [0], given: 25

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03 Dec 2010, 01:20
This is bothering the hell out of me. I know that if the last two digits of a number added together are divisible by 4 then the number is divisible by 4. In addition if a number is divisible by 3 and 4 then it is divisible by 12. I also know from simple math that 96 is divisible by all 3. However, when you add 9 + 6 you get 15. A number not divisible by 4. Is 96 just an exception to this rule, or are there more loop holes in other places as well? Thank you.

Kudos [?]: 74 [0], given: 25

Math Expert
Joined: 02 Sep 2009
Posts: 41660

Kudos [?]: 124307 [1], given: 12077

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03 Dec 2010, 01:42
1
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Expert's post
mmcooley33 wrote:
This is bothering the hell out of me. I know that if the last two digits of a number added together are divisible by 4 then the number is divisible by 4. In addition if a number is divisible by 3 and 4 then it is divisible by 12. I also know from simple math that 96 is divisible by all 3. However, when you add 9 + 6 you get 15. A number not divisible by 4. Is 96 just an exception to this rule, or are there more loop holes in other places as well? Thank you.

The red part is not correct. The number is divisible by 4 if the last two digits form a number divisible by 4.

Sum of the digits rule work only for 3 and 9:
3 - If the sum of the digits is divisible by 3, the number is also.
9 - If the sum of the digits is divisible by 9, so is the number.

For more check on number theory and divisibility check: math-number-theory-88376.html
_________________

Kudos [?]: 124307 [1], given: 12077

Intern
Joined: 31 Oct 2010
Posts: 31

Kudos [?]: 74 [0], given: 25

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03 Dec 2010, 02:13
cheers, apparantly im the a-hole. haha

Kudos [?]: 74 [0], given: 25

Re: Divisibility   [#permalink] 03 Dec 2010, 02:13
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