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# M05-04

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Math Expert
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16 Sep 2014, 00:24
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95% (hard)

Question Stats:

40% (02:26) correct 60% (02:28) wrong based on 224 sessions

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What is the 101st digit after the decimal point in the decimal representation of $$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}$$?

A. $$0$$
B. $$1$$
C. $$5$$
D. $$7$$
E. $$8$$

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16 Sep 2014, 00:24
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10
Official Solution:

What is the 101st digit after the decimal point in the decimal representation of $$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}$$?

A. $$0$$
B. $$1$$
C. $$5$$
D. $$7$$
E. $$8$$

$$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...$$.

102nd digit will be 8, thus 101st digit will be 0.

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03 Oct 2014, 16:31
Bunuel,

how do you know that 1/27 and 1/37 are 37/999 and 27/999 respectively.
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03 Oct 2014, 16:44
3
Vaico wrote:
Bunuel,

how do you know that 1/27 and 1/37 are 37/999 and 27/999 respectively.

999 = 27*37:

$$\frac{1}{37} =\frac{1*27}{37*27}=\frac{27}{999}$$.

$$\frac{1}{27} =\frac{1*37}{27*37}=\frac{37}{999}$$.
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07 Oct 2014, 17:57
1
Vaico wrote:
Bunuel,

how do you know that 1/27 and 1/37 are 37/999 and 27/999 respectively.

You can actually take a LCM of all the denominators to determine the factors by which you need to multiply the numerator and have a uniform denominator...
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04 Nov 2014, 08:19
Is this some sort of rule? 508/999 =.508508508

Because I noticed that =408/999 =.408408408
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04 Nov 2014, 09:21
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3
awal_786@hotmail.com wrote:
Is this some sort of rule? 508/999 =.508508508

Because I noticed that =408/999 =.408408408

Yes. Check Converting Decimals to Fractions chapter here: math-number-theory-88376.html

Hope it helps.
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03 Jun 2015, 22:01
Hi Bunuel

I actually did the longer method..... 1/3 = 0.3333.....1/9 = 0.11111......1/37 = 0.027027.......1/27 = 0.037037...

Going by this logic my digits at 101st place will be 3+1+2+3 = 9.....If I take a carryover from 102nd place its 3+1+7+7 = carry over 1........then answer is 9+1 = 10...0 at 101st place.....I made a mistake by not considering 102nd place.....Is it correct?
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04 Jun 2015, 03:14
Hi Bunuel

I actually did the longer method..... 1/3 = 0.3333.....1/9 = 0.11111......1/37 = 0.027027.......1/27 = 0.037037...

Going by this logic my digits at 101st place will be 3+1+2+3 = 9.....If I take a carryover from 102nd place its 3+1+7+7 = carry over 1........then answer is 9+1 = 10...0 at 101st place.....I made a mistake by not considering 102nd place.....Is it correct?

______________
Yes, that's correct.
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18 Aug 2015, 09:14
Hi,

Can someone explain how you got 102nd digit will be 8, thus 101st digit will be 0 from 0.508508....

Thanks,
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18 Aug 2015, 09:45
1
Hi,

Can someone explain how you got 102nd digit will be 8, thus 101st digit will be 0 from 0.508508....

Thanks,

The digits after decimal point repeat in blocks of three {508}{508}... So, third, sixths, ninths, ... digits are 8. So is 102nd digit (102 is a multiple of 3). Previous digit must be therefore 0.
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18 Aug 2015, 12:29
ok..can someone please explain how 333/999 and 111/999 are in the working?
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18 Aug 2015, 21:29
thanks Bunuel
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18 Oct 2015, 07:37
1
Three things:
1. This question is the same as M28-36
2. In the explanation, should read 333/999 + 111/999 + 37/999 + 27/999 = 508/999 instead of 333/999 + 111/999 + 27/999 + 37/999 = 508/999 because it implies that 27/999 corresponds to 1/27 and 37/999 corresponds to 1/37 and it is the other way around. Not written incorrectly, as order does not matter, just confusing if you are trying to figure it out.
3. Would be helpful to include in explanation the following: We are dealing with a repeating decimal in this question. It's helpful to know that there's a way to write these kinds of decimals as a fraction. For example, the repeating decimal 0.444444444(4) may be written as 4/9. So, 5/9, 7/9 and 8/9 will all be repeating decimals. You might check it in your calculator. In order to make two decimal points repeat, you have to divide the two digit number by 99. For example, 23/99=0.232323232323(23). In order to make 3 decimal points repeat, you have to divide the three digit number by 999. For example, 508/999=0.508508508508(508)
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09 Aug 2019, 08:52
1
Bunuel wrote:
What is the 101st digit after the decimal point in the decimal representation of $$\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}$$?

A. $$0$$
B. $$1$$
C. $$5$$
D. $$7$$
E. $$8$$

An alternative approach: This one is for people (like me) who have some number sense but are not as studied or quick on the draw as a math expert. The first three fractions would be easy enough to combine, but that forth denominator throws things off. What about the individual decimals instead?

$$\frac{1}{3}$$ = 0.333333...
$$\frac{1}{9}$$ = 0.111111... (this is one of the more well-known properties of dividing by 9)
$$\frac{1}{27}$$ = 0.037037... (via long division and pattern recognition: once you hit the second 3, you can see that the next digit would be 7)
$$\frac{1}{37}$$ = 0.027027... (again, long division and pattern recognition)

At this point, you can chunk or cluster some of the repeating decimals by finding their sum. I split the four decimals into the first two and second two:

0.333333 + 0.111111 = 0.444444 and
0.037037 + 0.027027 = 0.064064

Adding these chunks to each other yields 0.508508...

Now, we are getting really close. That is, every third digit will be an 8, so we just need to figure out on which digit the 101st spot will land. Since 33 repetitions of 3 gets us to 99, we can pick up the pattern there:

99th digit: 8
100th digit: 5
101st digit: 0

Of course, the 102nd digit would loop back to an 8, the 103rd to a 5, and so on, but the underlying principle is that whether we are looking for the 101st or the 1001st digit, we can use pattern recognition to arrive the correct response fairly efficiently and crack a tough question. Note: I am not saying that this method is better than the official solution proposed above, just that it can help some test-takers mitigate a panic response or a compulsion to guess at random.

Good luck with your studies.

- Andrew
Re: M05-04   [#permalink] 09 Aug 2019, 08:52
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