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GMATT73
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Which repeating decimal has the longest sequence of 03 Nov 2005, 06:40
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Which repeating decimal has the longest sequence of different digits?

a. 2/11
b. 1/3
c. 41/99
d. 2/3
e. 23/37

I can solve the answer by manual brute force in about 1.5 minutes. Any quick intuitive tricks to shave another 30 seconds or so off the long division?



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nero44
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Which repeating decimal has the longest sequence of 03 Nov 2005, 07:04
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Intuitively I can narrow it down to A and E

B and D are quite obvious.

c is out because a number divided by 99 never leaves different decimals longer than 2 digits!
you can write every 2 digit depeating decimals in the form of x/99. I think I wrote a how-to in some other post.
then you are left with a and e.
I would probably eliminate a because 99 is simply a multiple of 11, however I got no mathematical proof ot that.

maybe someone else has a better approach.
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GMATT73
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Which repeating decimal has the longest sequence of 03 Nov 2005, 07:28
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nero44
Intuitively I can narrow it down to A and E

B and D are quite obvious.

c is out because a number divided by 99 never leaves different decimals longer than 2 digits!
you can write every 2 digit depeating decimals in the form of x/99. I think I wrote a how-to in some other post.
then you are left with a and e.
I would probably eliminate a because 99 is simply a multiple of 11, however I got no mathematical proof ot that.

maybe someone else has a better approach.
Show more


Thanks for that time saving tip Nero :P I had no idea that all numbers divided by 99 never leaves a decimal longer than 2 digits. This tip will trim another 30 seconds off the problem. An example of how a small step makes a big difference. 8-)
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HIMALAYA
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Which repeating decimal has the longest sequence of 03 Nov 2005, 07:47
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GMATT73
Which repeating decimal has the longest sequence of different digits?

a. 2/11
b. 1/3
c. 41/99
d. 2/3
e. 23/37
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yes, this is good one. there should be one quick way...
b and d are quickly eliminated... A too because 2 divided by 11 doesnot bring more different digits...

remain C and E. now by calculation E gives the longest digits after decimal....
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Which repeating decimal has the longest sequence of 03 Nov 2005, 09:12
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GMATT73
Which repeating decimal has the longest sequence of different digits?

a. 2/11
b. 1/3
c. 41/99
d. 2/3
e. 23/37

I can solve the answer by manual brute force in about 1.5 minutes. Any quick intuitive tricks to shave another 30 seconds or so off the long division?
Show more


Matt, your simple question stems a lot of find-outs!!!
1)for 1<=x<=9 and x is not divisible by 99, we have x/99 = 0.0x0x0x0x......
2) 10<=x<=99 and x is not divisible by 99. x/99= 0.xxxxxxxxx.....
3) for number has form of a0b ( a,b are digits) a0b/ 99= a.(a+b)(a+b)(a+b).....
if a+b<= 0 : put 0 infront of (a+b), for example : 108/99= 1.090909....
if a+b<=18 : no need to put 0 in front of (a+b) , for example : 309/99= 3.121212........

hik,there're still many bizarre find-outs ....but don't know how to prove :roll:
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Which repeating decimal has the longest sequence of 03 Nov 2005, 11:43
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I got E.

B and D is out. I did a quick division on A and result is .1818... so this is out too... then i did C quickly and foud it is .4141 laxieqv's tip will be time saver here (at least 30 sec) so I am left with E. got to be the answer choice... no need to calculate...

took me 1 minute and 42 sec....
hope it's not too long



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