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M15-29

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Bunuel
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M15-29 [#permalink] New post   16 Sep 2014, 00:56
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41% (02:10) correct 59% (02:18) wrong based on 134 sessions
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If in a six-digit integer \(n\), \(f(k)\) represents the value of the \(k_{th}\) digit from the leftmost digit, is \(n\) is divisible by 7? (For example, \(f(4)\) corresponds to the value of the hundreds digit of \(n\).)


(1) \(f(1) = f(4), f(2) = f(5), f(3) = f(6)\)

(2) \(f(1) = f(2) = f(3) = f(4) = f(5) = f(6)\)
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Bunuel
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Re: M15-29 [#permalink] New post   21 Oct 2014, 07:35
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martinnotceoyet wrote:
Quote:
(For example, F(4) is the value of the hundreds digit of N)?

Sorry if I am wrong, but I believe the example should be F(4) is the value of the thousands digit of N .


123,456
1 - HUNDRED THOUSANDS
2 - TEN THOUSANDS
3 - THOUSANDS
4 - HUNDREDS
5 - TENS
6 - UNITS
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Re: M15-29 [#permalink] New post   17 Dec 2016, 11:51
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In my opinion, in actual GMAT, we are not allowed to use calculator, so every calculation is done by ourselves.

In case that we face this question, I think we could still complete it under 2 mins.

In statement (1), we simply check whether 1001 is divisible by 7 or not.

In statement (2), we simply check whether 111,111 is divisible by 7 or not. And the division by 7 is not complex. I believe that anyone could make it in about 10 secs.

Here is how I make the calculation
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Bunuel
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Re M15-29 [#permalink] New post   16 Sep 2014, 00:56
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Official Solution:


If in a six-digit integer \(n\), \(f(k)\) represents the value of the \(k_{th}\) digit from the leftmost digit, is \(n\) is divisible by 7? (For example, \(f(4)\) corresponds to the value of the hundreds digit of \(n\).)

(1) \(f(1) = f(4), f(2) = f(5), f(3) = f(6)\)

The above condition implies that the last three digits of \(n\) are the same as the first three digits of \(n\). Therefore, \(n = abc,abc = abc*1000 + abc = abc*(1000 + 1) = abc*1001\). Since 1001 is divisible by 7 (1,001 = 143*7), then \(n = abc*1001\) must also be divisible by 7. Sufficient.

(2) \(f(1) = f(2) = f(3) = f(4) = f(5) = f(6)\)

The above condition implies that all digits of \(n\) are the same. Therefore, \(n = aaa,aaa = aaa*1000 + aaa = aaa*(1000 + 1) = aaa*1001\). Since 1001 is divisible by 7 (1,001 = 143*7), then \(n = aaa*1001\) must also be divisible by 7. Sufficient.


Answer: D
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Re: M15-29 [#permalink] New post   21 Oct 2014, 07:32
Quote:
(For example, F(4) is the value of the hundreds digit of N)?

Sorry if I am wrong, but I believe the example should be F(4) is the value of the thousands digit of N .
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Re: M15-29 [#permalink] New post   22 Aug 2016, 04:13
I think this is a high-quality question and I agree with explanation.
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Re: M15-29 [#permalink] New post   15 Oct 2017, 03:00
Can you please explain how did you arrive at N=abc∗1000+abc
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Re: M15-29 [#permalink] New post   15 Oct 2017, 03:06
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GeetikaR wrote:
Can you please explain how did you arrive at N=abc∗1000+abc


We can write 6-digit number abc,abc as abc∗1000+abc because abc,abc = abc,000 + abc = abc∗1000+abc. For example, 123,123 = 123,000 + 123 = 123*1000 + 123.
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Re: M15-29 [#permalink] New post   21 Jun 2020, 11:17
I think this is a high-quality question and I agree with explanation.
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Re: M15-29 [#permalink] New post   31 May 2021, 17:58
I think this is a high-quality question and I agree with explanation.
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Re: M15-29 [#permalink] New post   22 Apr 2023, 13:52
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M15-29 [#permalink] New post   21 Sep 2023, 02:56
I think this is a high-quality question and I agree with explanation. I think this is a high-quality question and I agree with the explanation.

Great question Bunuel
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Re: M15-29 [#permalink] New post   14 Nov 2023, 21:04
Bunuel wrote:
If in a six-digit integer \(n\), \(f(k)\) represents the value of the \(k_{th}\) digit from the leftmost digit, is \(n\) is divisible by 7? (For example, \(f(4)\) corresponds to the value of the hundreds digit of \(n\).)


(1) \(f(1) = f(4), f(2) = f(5), f(3) = f(6)\)

(2) \(f(1) = f(2) = f(3) = f(4) = f(5) = f(6)\)



I took great amount of time in understanding formation of six digit integer.

I incorrectly calculated 'k'th digit from leftmost digit.

Example:
Six digit number be 123123. So I felt f(4) should be 4th digit after leftmost digit "1"- "2"
Though, I should have read example mentioned in the question correctly.
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Re: M15-29 [#permalink]
14 Nov 2023, 21:04
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Bunuel
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