## Sunday, October 4, 2009

### Problem 56 - GMAT Divisibility

1. OA E
resultant outcome of this eqn is
23times 9 followed by 440 i.e.
9999999....(23times)...440
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