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Sub 505 (Easy)|   Exponents|      
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Bunuel
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(49^3−147)/49 = 31 Jul 2018, 21:56
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

5% (low)

Question Stats:

85% (01:17) correct 15% (01:23) wrong based on 145 sessions

History

Date
Time
Result
Not Attempted Yet
\(\frac{49^3−147}{49} =\)

A. 1654
B. 1977
C. 2045
D. 2398
E. 2401
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Official Answer
D
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sudarshan22
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(49^3−147)/49 = 31 Jul 2018, 22:03
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Bunuel
\(\frac{49^3−147}{49} =\)
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= (49^3−147) / 49
= ( 49 ( 49^2 - 3)) / 49
= 49^2 - 3
= 2401 - 3
= 2398

Hence, D.
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NandishSS
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(49^3−147)/49 = 31 Jul 2018, 22:06
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Bunuel
\(\frac{49^3−147}{49} =\)

A. 1654
B. 1977
C. 2045
D. 2398
E. 2401
Show more

\(49^3−147\)

Unit digit of 9^3 =9

Unit digit of 147 = 7

So by the logic \(49^3−147\) will end with 2 if you are going to divide it by 49 then it will be 8 only option is D
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(49^3−147)/49 = 31 Jul 2018, 22:11
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Bunuel
\(\frac{49^3−147}{49} =\)

A. 1654
B. 1977
C. 2045
D. 2398
E. 2401
Show more

OA: D
\(\frac{49^3−147}{49} =\frac{{49^3−49*3}}{49}=\frac{{49(49^2−3)}}{49}=49^2−3\)
Unit digit of \(49^2\) will be \(1\).
The correct option will have \(11-3=8\) as unit's digit.
Only Option D has \(8\) as unit's digit
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Akash720
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(49^3−147)/49 = 31 Jul 2018, 22:14
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Bunuel
\(\frac{49^3−147}{49} =\)

A. 1654
B. 1977
C. 2045
D. 2398
E. 2401
Show more

=(49^3-49*3)/49
=49(49^2-3)/49
=49^2-3
Here 9^2 gives 1 as unit digit. Therefore, x1-3 will give 8 as unit digit

Hence D
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Nijssen
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(49^3−147)/49 = 02 Dec 2018, 06:18
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So for this 'difficult' mental math I just have to compute the final digit if the answers all have different final digits?
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