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If k is an integer, and 35^21/k is an integer, then k could
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Updated on: 10 Apr 2012, 10:50
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If k is an integer, and (35^21)/k is an integer, then k could be each of the following, EXCEPT A. 8 B. 9 C. 12 D. 16 E. 17
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Originally posted by dimitri92 on 19 May 2010, 01:15.
Last edited by Bunuel on 10 Apr 2012, 10:50, edited 2 times in total.
Edited the question and added the OA




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Re: 35^21/k
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19 May 2010, 01:21




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Re: 35^21/k
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19 May 2010, 02:34
Bunuel wrote: dimitri92 wrote: If k is an integer, and 35^21/k is an integer, then k could be each of the following, EXCEPT
(A) 8(B) 9(C) 12(D) 16(E) 17 \(\frac{35^21}{k}=\frac{(351)(35+1)}{k}=\frac{34*36}{k}=\frac{2^3*3^2*17}{k}\) From the answer choices only \(16=2^4\) is not a factor of numerator, hence in case \(k=16\), \(\frac{35^21}{k}\) won't be an integer, hence k can not be 16. Answer: D. thanks man ..you make it seem too simple ...kudos to you ....i will make sure i remember this x^2  1 formula so i can use it in these situations



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Re: 35^21/k
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19 May 2010, 06:20
This threw me off. I thought this was asking for 35 to be raised to the 21/k power and this number was somehow an integer.
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Re: 35^21/k
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20 May 2010, 23:32
nicely explained Bunuel.. made it very simple..
Thanks



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Re: 35^21/k
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13 Jul 2010, 19:40
My "messy" way: I just plug in the number and see if it will give a remainder of 1 or 1 ex: 35^2/8 > remainder is 1 >11=0 35^2/12 >remainder is 1 ... and so on the only choice that doesn't give remainder of 1 or 1 is 16
it works..just not as pretty as Bunuel's



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Re: 35^21/k
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26 Aug 2010, 18:43
i picked the same way .i.e by putting in numbers . Bunuel's way has to be the best



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Re: Help on q...
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01 Jan 2011, 23:33
m990540 wrote: I've seen a few of these; how do I go about solving them? Thanks so much in advance!
If K is an integer, and (35^21)/k is an integer, then k could be each of the following, EXCEPT:
A 8 B 9 C 12 D 16 E 17 \(35^21= 1224\). So for 1224/k to be an integer, k should be a factor of this 1224. 1224 can be written as \(2^3*3^2*17^1\). From this factorization, we see that all answer choices except 16 have prime factors that are fully included in the prime factorization of 1224. 16 has four 2s in it whereas 1224 has only 3. So this extra 2 means that 16 is not a factor of 1224 therefore, when 1224 is divided by 16, there will be some remainder left, making the whole term a non integer. Answer D.
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18 Aug 2011, 16:06
You can use prime factorization in this problem: (34*36)/k The prime factors are 2,2,2,3,3,17A 8 = 2,2,2 B 9 = 3,3 C 12 = 2,2,3 D 16 = 2,2,2,2 (Not divisible as there is not enough 2's)E 17 = 17
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Re: Question
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18 Aug 2011, 17:35
IanSmith wrote: Any help with this question?
If k is an integer, and (35^2  1)/k is an integer, then k could be each of the following, except:
A 8 B 9 C 12 D 16 E 17 = (35^2  1)/k = [(351) (35+1)]/k = (34) (36)/k = (2x17) (2x2x3x3)/k = (2x17) (2x2x3x3)/k a) Is there 8? yes, there are three 2's i.e. 2x2x2. b) Is there 9? yes, there are two 3's i.e. 3x3. c) Is there 12? yes, there are 2x2x3. d) Is there 16? No, there are not four 2's 2x2x2x2. e) Is there 17? yes, there is 17. So the answer should be D. 16.



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Re: Question
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18 Aug 2011, 19:19
= 36*34/k
B,C,E are ruled out as they yield non integer value.
36 has two 2's and 24 has one 2. so their product has three 2's . so divisible by 8. So A is ruled out too.
Answer is D. ( as 36*34 is not divisible by 16)



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Re: Question
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10 Apr 2012, 10:47
Hello guys, How do I get \(\frac{(35^21)}{k} = \frac{(35)(35)1}{k} = \frac{(35+1)(351)}{k}\)? I calculated the two terms and, yeas they are equal, but I can't imagine that I have to calculate that within 2 min?! Thanks for your help!



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10 Apr 2012, 10:54



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23 Sep 2013, 11:43
Pipp wrote: My "messy" way: I just plug in the number and see if it will give a remainder of 1 or 1 ex: 35^2/8 > remainder is 1 >11=0 35^2/12 >remainder is 1 ... and so on the only choice that doesn't give remainder of 1 or 1 is 16
it works..just not as pretty as Bunuel's Hi, Why are you looking for remainder of 1 or 1? You are dividing the whole numerator by k, so if you were to pick numbers, (35^2/k)(1/k)..... 1/k = 1/8 (if you are replacing k with 8)



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Re: If k is an integer, and 35^21/k is an integer, then k could
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08 Mar 2018, 10:42
dimitri92 wrote: If k is an integer, and (35^21)/k is an integer, then k could be each of the following, EXCEPT
A. 8 B. 9 C. 12 D. 16 E. 17 35^2  1 = (35 + 1)(35  1) = (36)(34) = 4 x 9 x 2 x 17 = 2^3 x 3^2 x 17 Since 16 = 2^4, we see that 16 doesn’t divide 35^2  1, so k can’t be 16. Answer: D
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Re: If k is an integer, and 35^21/k is an integer, then k could
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11 Jun 2018, 12:27
Bunuel wrote: dimitri92 wrote: If k is an integer, and 35^21/k is an integer, then k could be each of the following, EXCEPT
(A) 8(B) 9(C) 12(D) 16(E) 17 \(\frac{35^21}{k}=\frac{(351)(35+1)}{k}=\frac{34*36}{k}=\frac{2^3*3^2*17}{k}\) From the answer choices only \(16=2^4\) is not a factor of the numerator, hence in case \(k=16\), \(\frac{35^21}{k}\) won't be an integer, hence k can not be 16. Answer: D. hi pushpitkc, can you explain why Bunuel expressed this \(35^21\) as \((351)(35+1)\) why is he using this formula \(a^2b^2\) as you see in this formula both numbers \(a\) and \(b\) are raised to power two but in the above question we only have 35 raised to power of 2 and another number is 1 which is not raised to power of 2 thank you



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Re: If k is an integer, and 35^21/k is an integer, then k could
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11 Jun 2018, 12:33
dave13 wrote: Bunuel wrote: dimitri92 wrote: If k is an integer, and 35^21/k is an integer, then k could be each of the following, EXCEPT
(A) 8(B) 9(C) 12(D) 16(E) 17 \(\frac{35^21}{k}=\frac{(351)(35+1)}{k}=\frac{34*36}{k}=\frac{2^3*3^2*17}{k}\) From the answer choices only \(16=2^4\) is not a factor of the numerator, hence in case \(k=16\), \(\frac{35^21}{k}\) won't be an integer, hence k can not be 16. Answer: D. hi pushpitkc, can you explain why Bunuel expressed this \(35^21\) as \((351)(35+1)\) why is he using this formula \(a^2b^2\) as you see in this formula both numbers \(a\) and \(b\) are raised to power two but in the above question we only have 35 raised to power of 2 and another number is 1 which is not raised to power of 2 thank you Hi dave13This is a thumb rule when it comes to exponents: \(1^{anything} = 1\). So, the reverse also holds true! In this case, 1 has been written as \(1^2\) and that is the reason \(35^2  1 = 35^2  1^2 = 36 * 34\) Hope this helps you!
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Re: If k is an integer, and 35^21/k is an integer, then k could
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12 Jun 2018, 05:02
pushpitkc thank you what if it were \(35^2  2\) would it be correct t write like this \((352)(35+2)\) ?



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If k is an integer, and 35^21/k is an integer, then k could
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12 Jun 2018, 07:58
dave13 wrote: pushpitkc thank you what if it were \(35^2  2\) would it be correct t write like this \((352)(35+2)\) ? Hey dave13If it was \(35^2  2\), we could write it as \(35^2  (\sqrt{2})^2\). Now, \(35^2  (\sqrt{2})^2 = (35 + \sqrt{2})(35  \sqrt{2})\) Whenever you come across such expressions, you need to bring it to the form \(a^2b^2\) and then it is equal to \((a+b)(ab)\) Hope this helps you!
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If k is an integer, and 35^21/k is an integer, then k could &nbs
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