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Bunuel
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Bunuel
\(\frac{49^3−147}{49} =\)

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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Bunuel
\(\frac{49^3−147}{49} =\)

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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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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