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how to solve fast: divisibility

This topic has 4 expert replies and 2 member replies
buoyant Master | Next Rank: 500 Posts Default Avatar
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how to solve fast: divisibility

Post Thu Dec 18, 2014 8:08 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?

    N is divisible by 3
    N is divisible by 7

    OA: B

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    Post Thu Dec 18, 2014 8:39 pm
    Hi buoyant,

    This question is based on a couple of Number Properties and the basic rules/patterns of division:

    We're told than K is a POSITIVE INTEGER LESS than 10. This means than N could be 1, 2, 3, 4....9 (only 9 possibilities).

    Next, we're told that N = 4,321 + K. This means than N could be 4322, 4323, 4324....4330 (only 9 possibilities).

    We're asked for the value of K.

    Fact 1: N is divisible by 3.

    There's a math rule called "the rule of 3", which means that if the DIGITS of a number sum to a total that is divisible by 3, then the original number is divisible by 3.

    eg. 18 --> the digits 1+ 8 = 9. 9 is divisible by 3 so 18 is divisible by 3.
    eg. 19 --> the digits 1 + 9 = 10. 10 is NOT divisible by 3 so 19 is NOT divisible by 3.

    In this question, the first possible value of N that is divisible by 3 is:
    4323, so K = 2

    But there are other values in this range that are also divisible by 3:
    4326, so K = 5
    4329, so K = 8

    Fact 1 is INSUFFICIENT.

    Fact 2: N is divisible by 7

    Here we need to do a little "hand math":

    4321/7 = 617 remainder 2

    By increasing 4321 by 5 we would have a 0 remainder...

    4326/7 = 618 remainder 0

    So 4326 is the first value that fits this information. The next value would be 7 greater (4333), but that number is not possible here (it's outside the range defined by the prompt). This means that 4326 is the ONLY answer and K = 5.

    Fact 2 is SUFFICIENT.

    Final Answer: B

    GMAT assassins aren't born, they're made,
    Rich

    _________________
    Contact Rich at Rich.C@empowergmat.com

    Thanked by: buoyant
    Post Fri Dec 19, 2014 7:35 am
    buoyant wrote:
    If K is a positive integer less than 10 and N = 4,321 + K, what is the value of K?

    N is divisible by 3
    N is divisible by 7

    OA: B
    Here's another approach:

    Target question: What is the value of K?

    Given: K is a positive integer less than 10 and N = 4,321 + K
    If K is a positive integer less than 10, then K = 1, 2, 3, 4, ... or 9
    So, 4321 + K (aka N) can have 9 possible values.
    In other word, N = 4322, 4323, 4324, ..., 4330 (9 consecutive integers)

    Statement 1: N is divisible by 3
    IMPORTANT RULE: Among a set of consecutive integers, every 3rd integer will be divisible by 3.
    Likewise, every 4th integer will be divisible by 4.
    Every 5th integer will be divisible by 5.
    Etc.
    So, in a set of consecutive integers, about 1/3 of them will be divisible by 3.

    Since N can be any of the 9 consecutive integers from 4322 to 4330, and since 1/3 of these 9 integers are divisible by 3, we can conclude that N could have 3 possible values.
    Aside: We need not determine what these 3 possible values are, but if we were to find them, we'd see that N could equal 4323, 4326 or 4329
    Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

    Statement 2: N is divisible by 7
    Using our rule from earlier, about 1/7 of the 9 integers are divisible by 7, so in this case, it's possible that there's 1 value for N or possibly 2 values.
    So, we need to check.
    We'll look for a multiple of 7 among the possible values of N (4322, 4323, 4324, ..., 4330)
    We know that 4200 is divisible by 7
    So, 4270 is divisible by 7 (added 70 to 4200)
    Which means 4340 is divisible by 7 (added 70 to 4270)
    Oh, we've gone to far.
    4333 is divisible by 7 (subtracted 7 from 4340)
    4326 is divisible by 7 (subtracted 7 from 4333) BINGO

    Since every 7th integer is divisible by 7, we can see that N could equal ... 4212, 4219, 4326, 4333, 4340,...
    Since only one of these values is in the range of possible values of N (4322, 4323, 4324, ..., 4330), we can be certain that N = 4326
    Since we can answer the target question with certainty, statement 2 is SUFFICIENT

    Answer = B

    Cheers,
    Brent

    _________________
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    Last edited by Brent@GMATPrepNow on Wed Apr 12, 2017 4:06 am; edited 1 time in total

    Thanked by: buoyant
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    Zoser Junior | Next Rank: 30 Posts
    Joined
    16 Jan 2017
    Posted:
    22 messages
    Post Wed Apr 12, 2017 3:20 am
    Quote:
    Statement 1: N is divisible by 3
    IMPORTANT RULE: Among a set of integers, every 3rd integer will be divisible by 3.
    Likewise, every 4th integer will be divisible by 4.
    Every 5th integer will be divisible by 5.
    Does this rule work all the times?

    I tried number 9223 and it is not divisible by 3!

    Thanks

    Post Wed Apr 12, 2017 4:08 am
    Zoser wrote:
    Quote:
    Statement 1: N is divisible by 3
    IMPORTANT RULE: Among a set of integers, every 3rd integer will be divisible by 3.
    Likewise, every 4th integer will be divisible by 4.
    Every 5th integer will be divisible by 5.
    Does this rule work all the times?

    I tried number 9223 and it is not divisible by 3!

    Thanks
    Sorry, that should have read "Among a set of consecutive integers, every 3rd integer will be divisible by 3"

    This statement is true.
    For example: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13...

    _________________
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    Zoser Junior | Next Rank: 30 Posts
    Joined
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    Posted:
    22 messages
    Post Wed Apr 12, 2017 4:17 am
    Quote:
    Sorry, that should have read "Among a set of consecutive integers, every 3rd integer will be divisible by 3"

    This statement is true.
    For example: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13...
    What about:
    2,3,4,5,6,7,8,9

    4 and 7 are not divisible by "3" eventhough 4 is the 3rd consecutive integer!

    Also, in 4,5,6,7,8,9,10,11,12,13 7 is the 4th integer but not divisible by 4!

    Post Wed Apr 12, 2017 4:30 am
    Zoser wrote:
    Quote:
    Sorry, that should have read "Among a set of consecutive integers, every 3rd integer will be divisible by 3"

    This statement is true.
    For example: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13...
    What about:
    2,3,4,5,6,7,8,9

    4 and 7 are not divisible by "3" eventhough 4 is the 3rd consecutive integer!

    Also, in 4,5,6,7,8,9,10,11,12,13 7 is the 4th integer but not divisible by 4!
    I don't mean to say that, if we have some consecutive integers, the 3rd integer will be divisible by 3, the 6th integer will be divisible by 3, the 9th integer will be divisible by 3, etc.

    I means that out of any three consecutive integers, 1 will be divisible by 3.
    Likewise, out of any four consecutive integers, 1 will be divisible by 4.
    etc

    Cheers,
    Brent

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course

    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!

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