Sunday, October 4, 2009

Problem 56 - GMAT Divisibility


  1. OA E
    resultant outcome of this eqn is
    23times 9 followed by 440 i.e.
    sum of this number is 8 not divisible by 3..Hence OA E
    11,8,5,4 divisible

  2. Yep, good job Bhushan, answer is E. And also, if you are in a hurry you can quickly eliminate B, C, and D, because the number is gonna end in a zero, so it will be divisible by 5 and will have the two 2s and three 2s to make it divisible by 8.

  3. i agreed with the explanation above
    can u tell me why answer is not 11??

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  6. The logic is cryptic:
    1000-560=440 div
    10000-560=9440 not div
    10^5-560=99440 div
    10^6-560=999440 not div
    10^(2k+1)-560 is div by 11
    10^(2*12+1) - 560=9...(22*times)440 is div by 11
    Basic relation is #of 9's=#10's-3
    because 10000-560 gives the 1st 9
    What you need to remember is you need an even number of 9's in this case for div by 11.

    Man - I had to correct my post twice !

  7. divisibility by 11= difference between odd and even placed numbers .
    Here 10^25-560= 9999.(22times)440.
    All 9's will cancel with each other as even number of terms. Then we have 440. So, 4+0-4 =0
    Hence divisible by 11