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(49^3−147)/49 =

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Bunuel
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(49^3−147)/49 = [#permalink] New post   31 Jul 2018, 21:56
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A
B
C
D
E

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15% (low)

Question Stats:

83% (01:12) correct 17% (01:25) wrong based on 111 sessions
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\(\frac{49^3−147}{49} =\)

A. 1654
B. 1977
C. 2045
D. 2398
E. 2401
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D
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sudarshan22
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Re: (49^3−147)/49 = [#permalink] New post   31 Jul 2018, 22:03
Bunuel wrote:
\(\frac{49^3−147}{49} =\)


= (49^3−147) / 49
= ( 49 ( 49^2 - 3)) / 49
= 49^2 - 3
= 2401 - 3
= 2398

Hence, D.
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Re: (49^3−147)/49 = [#permalink] New post   31 Jul 2018, 22:06
Bunuel wrote:
\(\frac{49^3−147}{49} =\)

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


\(49^3−147\)

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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Re: (49^3−147)/49 = [#permalink] New post   31 Jul 2018, 22:11
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Bunuel wrote:
\(\frac{49^3−147}{49} =\)

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


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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Re: (49^3−147)/49 = [#permalink] New post   31 Jul 2018, 22:14
Bunuel wrote:
\(\frac{49^3−147}{49} =\)

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


=(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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Re: (49^3−147)/49 = [#permalink] New post   02 Dec 2018, 06:18
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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Re: (49^3−147)/49 = [#permalink]
02 Dec 2018, 06:18
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