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# (429)^2 * 237 * (1243)^3 is thrice of ?

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Intern
Joined: 30 Sep 2010
Posts: 17
(429)^2 * 237 * (1243)^3 is thrice of ?  [#permalink]

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Updated on: 02 Aug 2012, 03:25
2
16
00:00

Difficulty:

15% (low)

Question Stats:

78% (01:55) correct 22% (02:04) wrong based on 408 sessions

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(429)^2 * 237 * (1243)^3 is thrice of ?

A. 33605 * 31960 * (1243)^2
B. 33654 * 538219 * (1243)^2
C. 33891 * 533247 * (1243)^2
D. 34122 * 532004 * (1243)^2
E. 34606 * 534572 * (1243)^2

Originally posted by surendar26 on 20 Dec 2010, 09:35.
Last edited by Bunuel on 02 Aug 2012, 03:25, edited 2 times in total.
Edited the question and added the OA
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Posts: 64240

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20 Dec 2010, 12:27
22
3
surendar26 wrote:
(429)^2 * 237 * (1243)^3 is thrice of ?

33605 * 31960 * (1243)^2
33654 * 538219 * (1243)^2
33891 * 533247 * (1243)^2
34122 * 532004 * (1243)^2
34606 * 534572 * (1243)^2

(429)^2*237*(1243)^3 is an odd number.

Among answer choices only C is an odd number, (so thrice C also will be an odd number).

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02 May 2013, 10:59
6
arpanpatnaik wrote:
Bunuel wrote:
surendar26 wrote:
(429)^2 * 237 * (1243)^3 is thrice of ?

33605 * 31960 * (1243)^2
33654 * 538219 * (1243)^2
33891 * 533247 * (1243)^2
34122 * 532004 * (1243)^2
34606 * 534572 * (1243)^2

(429)^2*237*(1243)^3 is an odd number.

Among answer choices only C is an odd number, (so thrice C also will be an odd number).

If I were to approach the problem in this way...

429^2 * 237 * 1243 * 3 * 1243^2 must be equal to one of the choices...hence 429^2 * 237 * 1243 * 3 must be equal to the part after cancelling^2.
Considering the multiplication and its unit place...the value unit place 429^2 * 237 * 1243 * 3 must be 3. None of the choices seem to satisfy that.

Unit place Choice 1 : 0 Choice 2: 6 Choice 3: 7 Choice 4: 8 and Choice 5: 2....So if I were to guess, none of them might be the answer

Am I doing something wrong? Please correct me if either my process or values are off

Thanks!

$$(429)^2$$ * 237 * $$(1243)^3$$
So we have $$(429)^2$$ * 237 * $$(1243)$$ * $$(1243)^2$$
the units digit, will be 9^2 = 81 , 237 , 1243 and $$(3)^2$$ = 9
units digit will be, 1 * 7 * 3 * 9 = 189

Now test the units digit of the answer choices
As $$(429)^2$$ * 237 * $$(1243)^3$$ = 3 times of (X),
we are looking for an answer choice that gives the unit digit of 3.

33605 * 31960 * (1243)^2 = 5 * 0 * 9 = 0 (ruled out)
33654 * 538219 * (1243)^2 = 4 * 9 * 9 = 324 (ruled out)
33891 * 533247 * (1243)^2 = 1 * 7 * 9 = 63 ---> thats our answer
34122 * 532004 * (1243)^2 = 2 * 4 * 9 = 72 (ruled out)
34606 * 534572 * (1243)^2 = 6 * 2 * 9 = 108 (ruled out)
##### General Discussion
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14 Jan 2011, 16:49
1
Bunuel wrote:
surendar26 wrote:
(429)^2 * 237 * (1243)^3 is thrice of ?

33605 * 31960 * (1243)^2
33654 * 538219 * (1243)^2
33891 * 533247 * (1243)^2
34122 * 532004 * (1243)^2
34606 * 534572 * (1243)^2

(429)^2*237*(1243)^3 is an odd number.

Among answer choices only C is an odd number, (so thrice C also will be an odd number).

Wow !!! the best approach.... toooo good...
+1 kudos...
Intern
Joined: 20 Jun 2011
Posts: 42
Re: (429)^2 * 237 * (1243)^3 is thrice of ?  [#permalink]

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03 Jul 2012, 13:22
Is there a factor rule or something which allows it to be possible to solve this by taking the numbers digit of 429^2 and see that it will be 1 which suits only answer choice C ?
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02 May 2013, 10:15
Bunuel wrote:
surendar26 wrote:
(429)^2 * 237 * (1243)^3 is thrice of ?

33605 * 31960 * (1243)^2
33654 * 538219 * (1243)^2
33891 * 533247 * (1243)^2
34122 * 532004 * (1243)^2
34606 * 534572 * (1243)^2

(429)^2*237*(1243)^3 is an odd number.

Among answer choices only C is an odd number, (so thrice C also will be an odd number).

If I were to approach the problem in this way...

429^2 * 237 * 1243 * 3 * 1243^2 must be equal to one of the choices...hence 429^2 * 237 * 1243 * 3 must be equal to the part after cancelling^2.
Considering the multiplication and its unit place...the value unit place 429^2 * 237 * 1243 * 3 must be 3. None of the choices seem to satisfy that.

Unit place Choice 1 : 0 Choice 2: 6 Choice 3: 7 Choice 4: 8 and Choice 5: 2....So if I were to guess, none of them might be the answer

Am I doing something wrong? Please correct me if either my process or values are off

Thanks!
Intern
Joined: 14 Feb 2013
Posts: 25
Schools: Duke '16

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02 May 2013, 11:03
arpanpatnaik wrote:

If I were to approach the problem in this way...

429^2 * 237 * 1243 * 3 * 1243^2 must be equal to one of the choices...hence 429^2 * 237 * 1243 * 3 must be equal to the part after cancelling^2.
Considering the multiplication and its unit place...the value unit place 429^2 * 237 * 1243 * 3 must be 3. None of the choices seem to satisfy that.

Unit place Choice 1 : 0 Choice 2: 6 Choice 3: 7 Choice 4: 8 and Choice 5: 2....So if I were to guess, none of them might be the answer

Am I doing something wrong? Please correct me if either my process or values are off

Thanks!

the mistake you did was multiplying (429^2 * 237 * 1243^3) by 3, when the question states
(429^2 * 237 * 1243^3) = 3 times of (answer)
which is, (429^2 * 237 * 1243^3) / 3 = (answer)
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02 May 2013, 11:08
karishmatandon wrote:
arpanpatnaik wrote:

If I were to approach the problem in this way...

429^2 * 237 * 1243 * 3 * 1243^2 must be equal to one of the choices...hence 429^2 * 237 * 1243 * 3 must be equal to the part after cancelling^2.
Considering the multiplication and its unit place...the value unit place 429^2 * 237 * 1243 * 3 must be 3. None of the choices seem to satisfy that.

Unit place Choice 1 : 0 Choice 2: 6 Choice 3: 7 Choice 4: 8 and Choice 5: 2....So if I were to guess, none of them might be the answer

Am I doing something wrong? Please correct me if either my process or values are off

Thanks!

the mistake you did was multiplying (429^2 * 237 * 1243^3) by 3, when the question states
(429^2 * 237 * 1243^3) = 3 times of (answer)
which is, (429^2 * 237 * 1243^3) / 3 = (answer)

Ahh! Didnt read the question properly Sorryy!! and thanks for the help Karishma!
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Re: (429)^2 * 237 * (1243)^3 is thrice of ?  [#permalink]

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15 Sep 2013, 23:23
Square of 429: As the last digit is "9", its square should give "1" at the end.
Only option C best fits in, so answer = C
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Re: (429)^2 * 237 * (1243)^3 is thrice of ?  [#permalink]

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17 Jan 2014, 03:51
somalwar wrote:
Square of 429: As the last digit is "9", its square should give "1" at the end.
Only option C best fits in, so answer = C

did it just like you. quite risky but took 30 seconds :D
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17 Jan 2014, 04:13
Bunuel wrote:
surendar26 wrote:
(429)^2 * 237 * (1243)^3 is thrice of ?

33605 * 31960 * (1243)^2
33654 * 538219 * (1243)^2
33891 * 533247 * (1243)^2
34122 * 532004 * (1243)^2
34606 * 534572 * (1243)^2

(429)^2*237*(1243)^3 is an odd number.

Among answer choices only C is an odd number, (so thrice C also will be an odd number).

I wish Amitabh Bacchan (pardon my spellings: a famous Bollywood movie star) was planning to take GMAT...and if he would have seen this...he would have said ADBUDH (incredible)

Once again Bunuel sama...why r u so awesome

We need to have tab for +10 kudos as well

Respect

Posted from my mobile device
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“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
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Re: (429)^2 * 237 * (1243)^3 is thrice of ?  [#permalink]

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26 Feb 2014, 18:04
unceldolan wrote:
somalwar wrote:
Square of 429: As the last digit is "9", its square should give "1" at the end.
Only option C best fits in, so answer = C

did it just like you. quite risky but took 30 seconds :D

One more check, divide 237 by 3 comes up 79

79 x 1237 = Last digit is 3, so option C works fine

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Re: (429)^2 * 237 * (1243)^3 is thrice of ?  [#permalink]

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11 Dec 2018, 05:56
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Re: (429)^2 * 237 * (1243)^3 is thrice of ?   [#permalink] 11 Dec 2018, 05:56