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D01-11

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D01-11 [#permalink]

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New post 16 Sep 2014, 00:11
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If \(S\) is the sum of the digits of a given number, \(T\) is the sum of the digit of \(S\), and \(G\) is the sum of digits in \(T\). For example \(S\) of 987 is \(9+8+7 = 24\), \(T\) of \(S\) is \(2+4 = 6\) and \(G\) of 6 is 6. Therefore \(G\) of 987 is 6. Which of the following has the greatest \(G\)?

A. 94123
B. 91964
C. 64678
D. 62355
E. 45689
[Reveal] Spoiler: OA

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Re D01-11 [#permalink]

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New post 16 Sep 2014, 00:11
Official Solution:

If \(S\) is the sum of the digits of a given number, \(T\) is the sum of the digit of \(S\), and \(G\) is the sum of digits in \(T\). For example \(S\) of 987 is \(9+8+7 = 24\), \(T\) of \(S\) is \(2+4 = 6\) and \(G\) of 6 is 6. Therefore \(G\) of 987 is 6. Which of the following has the greatest \(G\)?

A. 94123
B. 91964
C. 64678
D. 62355
E. 45689


The explanation follows as under: \(G\) of 45689 is 5 whereas the rest have \(G\) smaller than 5.

A. \(S = 9+4+1+2+3 = 19\), \(T = 1+9 = 10\) and \(G = 1+0 = 1\).

B. \(S = 29\), \(T = 11\) and \(G = 2\).

C. \(S = 31\), \(T = 4\), and \(G = 4\).

D. \(S = 21\), \(T = 3\), and \(G = 3\).

E. \(S = 32\), \(T = 5\), and \(G = 5\).

Therefore E is the highest.


Answer: E
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Collection of Questions:
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Re: D01-11 [#permalink]

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New post 04 Dec 2014, 11:05
Is there a shortcut to do this? It's not hard to add up all the options manually..

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Re: D01-11 [#permalink]

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New post 05 Dec 2014, 06:41
E
But took 2.15sec
Solved by adding each Digits in mind and noted down
Finding Out just S is Enough
Greater S value >>Mean Greater T >> Greater G!
:-D
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Re: D01-11 [#permalink]

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New post 05 Dec 2014, 15:56
kanusha wrote:
E
Greater S value >>Mean Greater T >> Greater G!
:-D


it doesn't work
look
S in B) is greater than S in D), however T is greater in D) than in B)

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Re: D01-11 [#permalink]

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New post 20 Feb 2016, 14:39
this requires just the ability to process basic arithmetic quickly in your head.

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Re: D01-11 [#permalink]

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New post 05 Oct 2016, 05:50
another way is to cancel out the number in your head first cut 9 then 5 etc.. then u will be left with e last value with maximum weightage

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Re D01-11 [#permalink]

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New post 16 Jul 2017, 12:28
I think this is a poor-quality question and I agree with explanation. Because there is no shorter way to solve this question.
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Re: D01-11 [#permalink]

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New post 18 Jul 2017, 09:53
seeing the question stem I understand that its simply adding the individual digits of each number. so option E yields 32 which is the highest.

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Re: D01-11 [#permalink]

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New post 21 Aug 2017, 15:28
rayzha wrote:
Is there a shortcut to do this? It's not hard to add up all the options manually..


Yes, these type of problems are called digit sum. The shortest way is to find out the remainder when divided by 9.
This should give the answer.

Cheers!

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Re: D01-11   [#permalink] 21 Aug 2017, 15:28
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