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# (49^3−147)/49 =

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Math Expert
Joined: 02 Sep 2009
Posts: 64226
(49^3−147)/49 =  [#permalink]

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31 Jul 2018, 20:56
00:00

Difficulty:

15% (low)

Question Stats:

83% (01:12) correct 17% (01:32) wrong based on 89 sessions

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$$\frac{49^3−147}{49} =$$

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

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Re: (49^3−147)/49 =  [#permalink]

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31 Jul 2018, 21:03
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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Joined: 06 Jan 2015
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Re: (49^3−147)/49 =  [#permalink]

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31 Jul 2018, 21:06
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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Joined: 22 Feb 2018
Posts: 404
Re: (49^3−147)/49 =  [#permalink]

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31 Jul 2018, 21:11
1
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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Re: (49^3−147)/49 =  [#permalink]

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31 Jul 2018, 21:14
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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Joined: 03 Nov 2018
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Re: (49^3−147)/49 =  [#permalink]

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02 Dec 2018, 05:18
So for this 'difficult' mental math I just have to compute the final digit if the answers all have different final digits?
Re: (49^3−147)/49 =   [#permalink] 02 Dec 2018, 05:18

# (49^3−147)/49 =

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