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27 Oct 2020, 06:08
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
55% (02:47) correct
45%
(02:24)
wrong
based on 205
sessions
History
Date
Time
Result
Not Attempted Yet
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
A. 582
B. 588
C. 594
D. 606
E. 612
28 Oct 2020, 01:43
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Given
To Find
Approach and Working Out
Correct Answer: Option C
- • 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
General Discussion
27 Oct 2020, 06:36
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Bunuel
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 \(99*6=594\)
C
