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Let m be the 1,003rd digit to the right of the decimal point
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13 Oct 2020, 09:19
13 Oct 2020, 09:19
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83% (01:04) correct
16% (01:36) wrong based on 43 sessions
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Let \(m\) be the \(1,003rd\) digit to the right of the decimal point in the repeated decimal expansion \(0.\overline{13579}\), and let n equal the \(78th\) digit to the right of the decimal point in the repeated decimal expansion \(0.\overline{2468}\). What is the value of \(m^n\)?
Kudos for the right answer and explanation
Show: :: OA
625
Kudos for the right answer and explanation
Re: Let m be the 1,003rd digit to the right of the decimal point
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19 Oct 2020, 08:35
19 Oct 2020, 08:35
1
from the first decimal
0.13579
The digit repeat itself after five digit and since 1000 is divisible by 5 then the 1000th digit will be 9
Then the 1003 digit will be 5
From the second digit as well
76 is divisible by 4 then the 78th digit will be 4(we starts counting from the beginning 0.2478)
M^n is 5^4=625
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0.13579
The digit repeat itself after five digit and since 1000 is divisible by 5 then the 1000th digit will be 9
Then the 1003 digit will be 5
From the second digit as well
76 is divisible by 4 then the 78th digit will be 4(we starts counting from the beginning 0.2478)
M^n is 5^4=625
Posted from my mobile device
Re: Let m be the 1,003rd digit to the right of the decimal point
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08 Nov 2020, 07:26
08 Nov 2020, 07:26
1
(1000+3) digit of the first number, m=5
(76+2) th digit of the second number, n= 4
therefore, m^n=5^4=625
(76+2) th digit of the second number, n= 4
therefore, m^n=5^4=625
