If each of the following fractions were written as a repeati

Question

If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?

A) 2/11
B) 1/3
C) 41/99
D) 2/3
E) 23/37

coelholds · Accepted Answer

I have taken 15 seconds to resolve this problem! For this problem specifically, there are some very fast tricks that you can do remembering the properties of recurring decimals. You just need a little patience to understand a first time. Then you probably will never forget.

Here is a brief description useful in a some GMAT exercises:

First - We use the number that is repeating and the denominator has as many "9" digits as there are different digits in the block that repeats. e.g.
\(0.555555 = 5/9\)
\(0.13131313 = 13/99\)
\(0.432432432432 = 432/999\)

Second - If the sequence starts to repeat after some zeros, add the same number of zeros in the denominator. e.g.
\(0.005555 = 5/900\)
\(0.013131313 = 13/990\)
\(0.0004324324324... = 432/999000 = 54/124875 = 2/4625\)

Third - Terminating decimals + Repeating Decimals. e.g.
\(2.31555555 = 2.31 + 0.00555555 = 231/100 + 5/900\)
\(0.745454545 = 0.7 + 0.0454545 = 7/10 + 45/990\)

Last one - The reciprocal of a prime number "p", except 2 and 5, has a repeating sequence of p-1 digits, or a factor of p-1 digits. e.g.
\(1/7 = 0.142857 142857 142857...\) - As you can notice, the sequence has 6 digits = p-1 = 7-1 = 6 digits.
And if you multiply this fraction by a number you will only change the beginning of the sequence. e.g.
\(4/7 = 4*1/7 = 0.57 142857 142857 ...\)

****************
Once you have all these little rules in mind it is very fast to work in this problem.
A) 2/11 -> 11 is a factor of 99, so there are 2 repeating digits.
B) 1/3 -> 3 is a factor of 9, so there is 1 repeating digit.
C) 41/99 -> 99, so there are 2 repeating digits.
D) 2/3 -> 3 is a factor of 9, so there is 1 repeating digit..
E) 23/37 -> As I still don't have one item that is winning in repeating digits, this must be CORRECT

\(E\)

Good studies!

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Silver89 · Answer

TIP If the denominator is 9, 99, 999 or another number equal to a power of 10 minus 1, then
the numerator gives you the repeating digits  :D
A) 2/11= 18/99 = 0.181818
B) 1/3 you should know that it is 0.333333333
C) 41/99 according to the tip above it should be 0.4141414141
D) 2/3 you should know that it is 0.66666666
E) 23/37 you solve then you see :D
it didn't take more than a minute
 please consider giving me kudos if you see it was helpful in any way

walker · Answer

E

After reading of the problem I pick E by intuition: I thought that:
1. nominators do not influence on length of sequence. (for right fraction)
2.the largest prime in denominator lead to the largest sequence).

But I did not know a general rule...

\(1/a=0.(b)=b*(10^{-n}+10^{-2n}+10^{-3n.....}+...)=b*\frac{10^{-n}}{1-10^{-n}}=b*\frac{1}{10^{n}-1}=\frac{b}{99...99_n}\)
where n - the length of sequence (the number of digits of b)

Therefore, \(a\) have to be a factor of \(99...9_n\)

1. n=1, 9: B,D out, because of 9=3*k
2. n=2, 99: A,C out, because of 99=11*k=99*k

E remains.
It took about 2 min but I think we have now a good rule

Check:

2/11=0.(18) n=2. ok
1/3=0.(3) n=1. ok
41/99=0.(41) n=2. ok
1/3=0.(6) n=1. ok
23/37=0.(621) n=3. 999=27*37. ok

Check: 2/11=0.(18) n=2. ok 1/3=0.(3) n=1. ok 41 /99=0.( 41 ) n=2. ok 1/3=0.(6) n=1. ok 23/37=0.(621) n=3. 999=27*37. ok

EncounterGMAT · Answer

A. 2/11 = 0.181818..... 2 repeating numbers after decimal
B. 1/3 = 0.3333.....1 repeating numbers after decimal
C. 41/99 = 0.41414.......2 repeating numbers after decimal
D. 2/3 = 0.666666.....1 repeating numbers after decimal
E. 23/37 = 0.621621..........3 repeating numbers after decimal... ANSWER!

P.S: Option C and E involved big numbers. So, for option C, I calculated 40/100 = 0.40...then assumed than since prime is divided by 99, the decimal answer would be closer to 0.40. Also, 99 which is 11*9 will always give non-terminating repeating decimal. 11 gives out 2 repeating decimals so 99 would do the same. The remaining and last suspect option was answer E = prime/prime. Even checked it by calculating.

rgajare14 · Answer

I cant think of anything shorter. But, the way I do it is 
for 2/11. I make it 20/11. So the quotient will be 1.8181..Then I shift the decimal so it is 0.18

10/3 = 3.333 = 0.3
410/99 = 4.11 = 0.41
20/3 = 6.666 = 0.6
230/37 = 6.21621 = 0.621

So 23/37 has the longest sequence of digits

walker · Answer

I think a fast way for such problem is:

1. Test for right fractions.
2. Exclude nominators
3. Exclude 2 and 5 from a denominators.
4. 3,9 - n=1, 3,11,33,99 - n=2
5. POI
6. find 1/denominator for remained variants.

For example,
A.3/22
B.5/198

1. ok
2. 1/22, 1/198
3. 1/11, 1/99
4. n=2, n=2

walker · Answer

E

1. reduce fraction if it is possible. Here we have all proper fractions.
2. because numerator does not influence on period of sequence, set all numerators to 1: 1/11, 1/3, 1/99, 1/3, 1/37
3. transform all fraction to the denominator such as 9, 99, 999 .....: 9/99, 3/9, 1/99, 3/9,  1/37 
4. 1/37 we cannot write out as a fraction with denominator of 9 or 99. (actually we can with 999 but this is not necessary here) So, E is a winner.

JeffTargetTestPrep · Answer

If a fraction (in lowest terms) can be written as a repeating decimal and the number of decimal digits repeating is n, then the denominator must be a factor of 10^n - 1. 
 
For example, 1/3 (in choice B) has a denominator of 3, which is a factor of 9 = 10^1 - 1 = 9; thus, 1/3 has 1 decimal digit repeating. Sure enough, 1/3 = 0.333… Another example, 2/11 (in choice A), has a denominator of 11, which is a factor of 99 = 10^2 - 1 = 99; thus, 2/11 has 2 decimal digits repeating. Sure enough, 2/11 = 0.181818… 
 
Using this fact, we can see that 2/3 (in choice D) and 41/99 (in choice C) have 1 and 2 decimal digits repeating, respectively. Thus, the fraction 23/37 has the longest sequence of repeating decimal digits, since 37 is neither a factor of 9 nor a factor of 99. Regardless of the exact number of repeating decimal digits 23/27 has, it will certainly be more than 2.
 
Answer: E

Utki01 · Answer

Answer is E

Trick: If a fraction consists of two co-prime numbers, then the larger the denominator, the greater is the sequence of different digits before repetition.

Note that a fraction consists of two co-prime numbers when it is cannot be reduced further (Eg: 2/4 can be reduced to 1/2; but 1/2 cannot be reduced further). 
Here, all fractions are in their lowest form. Hence, all these fractions consist of two co-prime numbers each.

Since 37 is the largest denominator, it is the answer.

ScottTargetTestPrep · Answer

The number of unique digits that are repeating in a repeating decimal depends on whether the denominator in its fractional equivalent is a factor of a number with all digits of 9 (i.e, 9, 99, 999, and so on). If it is a factor of 9 (in which 9 is the smallest number such that it is a factor), then it has 1 unique digit repeating. If it is a factor of 99 (in which 99 is the smallest number such that it is a factor), then it has 2 unique digits repeating. If it is a factor of 999 (in which 999 is the smallest number such that it is a factor), then it has 3 unique digits repeating.

For example, 3 is a factor of 9, so both 1/3 and 2/3, when expressed as a repeating decimal, have 1 unique digit repeating (notice that 1/3 = 0.333… and 2/3 = 0.666…). Similarly, since 11 and 99 are factors of 99, so both 2/11 and 41/99, when expressed as a repeating decimal, have 2 unique digits repeating (notice that 2/11 = 0.181818… and 41/99 = 0.414141…). Finally, 37 is neither a factor of 9 nor a factor of 99, 23/37, when expressed as a repeating decimal, must have more than 2 unique digits repeating.

Answer: E

IanStewart · Answer

This is not true at all, as you can easily see by comparing the decimal expansion of 1/7 and of 1/9, say. The size of the denominator has very little to do with the length of the repeating pattern of digits. Scott has correctly explained the theory in the post above mine.

KarishmaB · Answer

If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?

A) 2/11
B) 1/3
C) 41/99
D) 2/3
E) 23/37

We already know a few fraction to decimal equivalents.

1/3 = .33333 
1/11 = .090909 (so 2/11 = .18181818...)
1/9 = .1111111 
1/99 = .010101... (so 41/99 = .41414141...)

So two of the options have 1 repeating digit and two have 2 repeating digits.

Since there can be only 1 answer, answer must be the leftover option (E).

sujoykrdatta · Answer

The number of digits in the decimal depends on the number of 9s in the denominator. For example:

5/9 = 0.555555... => single digit repeating
13/99 = 0.13131313.... => 2 digits repeating
137/999 = 0.137137137... => 3 digits repeating, and so on

A) 2/11 = 18/99 = 0.18181818... => 2 digits repeat
B) 1/3 = 3/9 = 0.3333... => 1 digit repeats
C) 41/99 = 0.414141... => 2 digits repeat
D) 2/3 = 6/9 = 0.6666... => 1 digit repeats
E) 23/37 = (23*3)/(37*3) = 69/111 = (69*9)/(111*9) = 621/999 = 0.621621621... => 3 digits repeat (the longest sequence)

Hence, the correct answer is E

Kinshook · Answer

Asked: If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?

A) 2/11 = 0.1818 : 2 different digits
B) 1/3 = 0.333 : 1 different digit
C) 41/99 = 0.3131 : 2 diffrent digits
D) 2/3 0.6666 : 1 different digit
E) 23/37 = 69/111 = 0.621621 : 3 different digits

IMO E

ArnauG · Answer

To determine which fraction would have the longest sequence of different digits when written as a repeating decimal, we need to convert each fraction into its decimal form.

A) 2/11:
To convert 2/11 into a decimal, we divide 2 by 11:
2 ÷ 11 = 0.181818...
The decimal representation of 2/11 is a repeating decimal with a sequence of 18.

B) 1/3:
To convert 1/3 into a decimal, we divide 1 by 3:
1 ÷ 3 = 0.333333...
The decimal representation of 1/3 is a repeating decimal with a sequence of 3.

C) 41/99:
To convert 41/99 into a decimal, we divide 41 by 99:
41 ÷ 99 = 0.414141...
The decimal representation of 41/99 is a repeating decimal with a sequence of 41.

D) 2/3:
To convert 2/3 into a decimal, we divide 2 by 3:
2 ÷ 3 = 0.666666...
The decimal representation of 2/3 is a repeating decimal with a sequence of 6.

E) 23/37:
To convert 23/37 into a decimal, we divide 23 by 37:
23 ÷ 37 ≈ 0.621621621...
The decimal representation of 23/37 is a repeating decimal with a sequence of 621.

From the given fractions, we can see that the fraction with the longest sequence of different digits when written as a repeating decimal is 23/37.

Therefore, the correct answer is (E) 23/37.

FarhinR · Answer

If denominator is 9, 99, 999 or any number equal to power of 10 minus 1, then numerator tell you the repeating digit. For example, 23÷99= 0.23232323...

(A) 2÷11= 18÷99= 0.181818......
(B) 1÷3= 3÷9= 0.3333.......
(C) 41÷99= 0.414141....
(D) 2÷3= 6÷9= 0.6666...
(E) 23÷37= 0.621621621....

Therefore, the answer is (E)

A_Nishith · Answer

Trick: 
 If in a fraction (a/b) cannot be reduced further then the fraction which has the largest prime in denominator has the longest sequence of different digits when written as a repeating decimal.

E) 23/37:
To convert 23/37 into a decimal, we divide 23 by 37:
23 ÷ 37 ≈ 0.621621621...
The decimal representation of 23/37 is a repeating decimal with a sequence of 621.

From the given fractions, we can see that the fraction with the longest sequence of different digits when written as a repeating decimal is 23/37.

Therefore, the correct answer is (E) 23/37.

egmat · Answer

These are repeating decimal problem - these can definitely be tricky at first, but once you understand the pattern, they become much more manageable. Let me walk you through this one.

The Key Insight

When we're looking for the "longest sequence of different digits" in a repeating decimal, we're actually looking for what mathematicians call the  period length  - that's the number of digits that repeat as a group.

For example:
-  \frac{1}{3} = 0.333...  has a period of 1 (just "3" repeats)
-  \frac{1}{7} = 0.142857142857...  has a period of 6 (the group "142857" repeats)

Here's what you need to see:  The period length depends on the denominator of the fraction when it's in lowest terms. Specifically, we need to find the smallest  n  where  10^n - 1  is divisible by the denominator.

Let's work through our options:

First, let's make sure all fractions are in lowest terms:
-  \frac{2}{11}  - already simplified, denominator = 11
-  \frac{1}{3}  - already simplified, denominator = 3
-  \frac{41}{99}  - already simplified (41 is prime), denominator = 99
-  \frac{2}{3}  - already simplified, denominator = 3
-  \frac{23}{37}  - already simplified (37 is prime), denominator = 37

Now, let's find the period length for each denominator:

For denominator 3: 
 10^1 - 1 = 9 , and  9 \div 3 = 3  ✓
Period length = 1

For denominator 11: 
 10^1 - 1 = 9  (not divisible by 11)
 10^2 - 1 = 99 , and  99 \div 11 = 9  ✓
Period length = 2

For denominator 99: 
 10^2 - 1 = 99 , and  99 \div 99 = 1  ✓
Period length = 2

For denominator 37: 
 10^1 - 1 = 9  (not divisible by 37)
 10^2 - 1 = 99  (not divisible by 37)
 10^3 - 1 = 999 , and  999 \div 37 = 27  ✓
Period length = 3

Notice how we systematically test powers of 10 until we find one that works!

The Answer: 
Comparing all period lengths:
- Options A, C: period 2
- Options B, D: period 1 
- Option E: period 3

Therefore,  \frac{23}{37}  has the longest sequence of different digits.

Answer: E

---

You can check out the  step-by-step solution on Neuron by e-GMAT  to master the systematic framework for finding period lengths and discover a time-saving pattern that works for all repeating decimal problems. You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice  here .

anushree01 · Answer

Could you please let me know which all fractions shall I learn, that are frequently used in gmat.
Thanks & I appreciate!

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