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Three digits have been removed from each of the following numbers. If
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23 May 2018, 06:21
23 May 2018, 06:21
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Difficulty:
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Question Stats:
70% (01:28) correct
30%
(01:34)
wrong
based on 131
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Three digits have been removed from each of the following numbers. If n = 25, which of the numbers is equal to \(3 * 2^{n - 1}\) ?
A. 47, __ __6, __23
B. 47, __ __6, __32
C. 49, __ __2, __64
D. 49, __ __2, __36
E. 50, __ __1, __48
A. 47, __ __6, __23
B. 47, __ __6, __32
C. 49, __ __2, __64
D. 49, __ __2, __36
E. 50, __ __1, __48
Re: Three digits have been removed from each of the following numbers. If
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23 May 2018, 11:46
23 May 2018, 11:46
Bunuel wrote:
Three digits have been removed from each of the following numbers. If n = 25, which of the numbers is equal to \(3 * 2^{n - 1}\) ?
A. 47, __ __6, __23
B. 47, __ __6, __32
C. 49, __ __2, __64
D. 49, __ __2, __36
E. 50, __ __1, __48
A. 47, __ __6, __23
B. 47, __ __6, __32
C. 49, __ __2, __64
D. 49, __ __2, __36
E. 50, __ __1, __48
\(3 * 2^{n - 1}=3*2^{25-1}=3*2^{24}\)
\(2^{24}\) will have a units digit of \(6\) and when it is multiplied by \(3\), the resulting number's unit's digit will be \(8\)
Only Option E has a units digit 8. Answer E
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Re: Three digits have been removed from each of the following numbers. If
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23 May 2018, 12:01
23 May 2018, 12:01
Expert Reply
Solution
Given:
• Three digits have been removed from each of the given options
• The value of n = 25
To find:
• Which of the given numbers is equal to \(3 * 2^{n-1}\)
Approach and Working:
As it is given that n =25,
• The value of \(2^{n-1}\) = \(2^{24}\)
If we observe the unit digit cycle of the powers of 2, we can see the pattern of 2, 4, 8, and 6 are getting repeated in that order
- o \(2^1\) = 2
o \(2^2\) = 4
o \(2^3\) = 8
o \(2^4\) = 16
o \(2^5\) = 32
o \(2^6\) = 64 and so on.
If we multiply \(2^{24}\) by 3, then the resultant number will end with 6 * 3 = 8
Only option E has a number ending with 8
Hence, the correct answer is option E.
Answer: E
