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x and n are positive integers, such that 7x = 10^n – 1. What is the sm 27 Oct 2020, 06:08
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C
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57% (02:41) correct 43% (02:29) wrong based on 260 sessions

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x and n are positive integers, such that \(7x = 10^n – 1\). What is the 99th smallest possible value of n?

A. 582
B. 588
C. 594
D. 606
E. 612
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C
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x and n are positive integers, such that 7x = 10^n – 1. What is the sm 28 Oct 2020, 01:43
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Given

    • x and n are positive integers, such that 7x = \(10^n\) – 1.


To Find

    • What is the 99th smallest value of n.

Approach and Working Out


    • \(10^n\) – 1 = 999 … n digits.
      o Hence n is the number of digits and that is the answer.

    • Any number written 6 times is always divisible by 7.
      o This is because aaaaaa = aaa × 1001 = aaa × 7 × 11 × 13
      o Hence 999999 will be divisible by 7.
      o 999999… 12 digits will also be divisible by 7.

    • Hence the number of digits for that number will be 6*99 = 594 and that will be the value of n.


Correct Answer: Option C
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x and n are positive integers, such that 7x = 10^n – 1. What is the sm 27 Oct 2020, 06:36
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Bunuel
x and n are positive integers, such that \(7x = 10^n – 1\). What is the 99th smallest possible value of n?

A. 582
B. 588
C. 594
D. 606
E. 612
Show more


There has to be a pattern and that is what one has to find...

\(7x = 10^n – 1.....10^n=7x+1\)
So 7x has to be of type 9, 99, 999, and so on.
Let us check which is the first multiple of 7 in the series..
We can save time by knowing some multiples of 7..
9=7+2...NO
99=70+29...NO, 70+28 is.
999 = 777+210+12...NO
9999=7777+2121+101=7777+2121+70+31...NO
99999=77777+21210+1019=77777+21210+700+280+39....NO
999999=777777+212121+10101=777777+212121+7000+2800+280+21...YES
That is 6 times 9s...Now 999999=10^6-1
The next multiple in the series will be twice the number of 9s, so 2*6 times 9s or 10^12
Thus 99th smallest value of n will be \(99*6=594\)


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x and n are positive integers, such that 7x = 10^n – 1. What is the sm 01 Nov 2020, 18:16
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I got the correct option C but using the cyclicity method. As 7x+1 = 10^n , I plugged in values for x until I noticed the last digit to be repeated which was at x=11 (digit 8, 7+1=8, 7(11)+1=78) . Then I divided 99 by the cyclicity factor 11 which equals 9 and the 9th value had a 4 as its unit digit same as choice C. Is this method valid or a complete fluke? Would appreciate your feedback
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x and n are positive integers, such that 7x = 10^n – 1. What is the sm 03 Nov 2020, 05:52
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Numairrr
I got the correct option C but using the cyclicity method. As 7x+1 = 10^n , I plugged in values for x until I noticed the last digit to be repeated which was at x=11 (digit 8, 7+1=8, 7(11)+1=78) . Then I divided 99 by the cyclicity factor 11 which equals 9 and the 9th value had a 4 as its unit digit same as choice C. Is this method valid or a complete fluke? Would appreciate your feedback
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How much time did it take?
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x and n are positive integers, such that 7x = 10^n – 1. What is the sm 03 Nov 2020, 05:53
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Bunuel
x and n are positive integers, such that \(7x = 10^n – 1\). What is the 99th smallest possible value of n?

A. 582
B. 588
C. 594
D. 606
E. 612
Show more



The way we can approach this is through only trial and error and putting values in.

Are we sure about that this a GMAT question?
Caus it takes me far more than two minutes to solve.
Thankyou in advance.
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x and n are positive integers, such that 7x = 10^n – 1. What is the sm 10 Jun 2022, 17:14
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The faster we get to the value of 'n', the lesser time it will take to solve this. Note: 10^3/7, gives a remainder of - 1. So, 10^3/7 X 10^3/7, give a remainder of 1 {(-1)(-1)}. So, 10^6 - 1 is divisible by 7.
Ans. 6x99 = 594
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x and n are positive integers, such that 7x = 10^n – 1. What is the sm 23 Aug 2026, 00:02
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Here's a simple way. Use cyclicity of remainders method. Take 7 to RHS. As you can see there is -1 also. Since the solution should be an real number and not decimals. -1 has to cancel out with something. How will that happen. Something remainder has to come from 10^n which will cancel out -1 or make it a number which is divisible by 7.

Let's find out. We will see what all remainders will get upon dividing 10^n with 7. So, the remainders came out as 3,2,6,4,5,1. This is the complete cycle of 6 numbers post that it will repeat again. We found out number. It is 1. If 1 is there it will cancel out -1. So, the 'n' should be a multiple of 6.

'n' could be 6,12,18, etc... Question asked to find 'n' of 99th smallest number. It will be 99x6=594
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