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30^20 – 20^20 is divisible by all of the [#permalink]
04 Jun 2018, 05:18
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50% (01:18) wrong based on 40 sessions
30^20 – 20^20 is divisible by all of the following values, EXCEPT: A) 10 B) 25 C) 40 D) 60 E) 64
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Re: 30^20 – 20^20 is divisible by all of the [#permalink]
04 Jun 2018, 05:52
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(30^20  20^20) =10^20 ( 3^20  2^20)
Clearly, divisible by 10.
Can be written as : = 1000 * 10^17 ( 3^20  2^20) Clearly, divisible by 25 & 40 ( 1000 is divisible by 25 & 40 )
Can also be written as : = 2^6 * 5^6 * 10^14 ( 3^20  2^20) Clearly, divisible by 64 ( 2^6 is divisible by 64)
Thus, eliminating options a,b,c & e, option D remains. Hence, D.



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Re: 30^20 – 20^20 is divisible by all of the [#permalink]
05 Jun 2018, 09:02
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GreenlightTestPrep wrote: 30^20 – 20^20 is divisible by all of the following values, EXCEPT:
A) 10 B) 25 C) 40 D) 60 E) 64 Here are some useful divisibility rules: 1. If integers A and B are each divisible by integer k, then (A + B) is divisible by k 2. If integers A and B are each divisible by integer k, then (A  B) is divisible by k 3. If integer A is divisible by integer k, BUT integer B is NOT divisible by integer k, then (A + B) is NOT divisible by k 4. If integer A is divisible by integer k, BUT integer B is NOT divisible by integer k, then (A  B) is NOT divisible by kNow let's check the answer choices.... A) 1030^20 = (10^20)(3^20) = ( 10)(10^19)(3^20), so 30^20 is divisible by 1020^20 = (10^20)(2^20) = ( 10)(10^19)(2^20), so 20^20 is divisible by 10So, by rule #2, 30^20 – 20^20 MUST be divisible by 10ELIMINATE A B) 2530^20 = (5^20)(6^20) = (5^2)(5^18)(6^20) = ( 25)(5^18)(6^20), so 30^20 is divisible by 2520^20 = (5^20)(4^20) = (5^2)(5^18)(4^20) = ( 25)(5^18)(4^20), so 20^20 is divisible by 25So, by rule #2, 30^20 – 20^20 MUST be divisible by 25ELIMINATE B C) 4030^20 = (10^20)(4^20) = (10)(10^19)(4)(4^19) = ( 40)(10^19)(4^19), so 30^20 is divisible by 4020^20 = (10^20)(2^20) = (10)(10^19)(2^2)(2^18) = ( 40)(10^19)(2^18), so 20^20 is divisible by 40So, by rule #2, 30^20 – 20^20 MUST be divisible by 40ELIMINATE C D) 6030^20 = (30^1)(30^19) = (30^1)(2^19)(15^19) = (30)(2)(2^18)(15^19) = ( 60)(2^18)(15^19), so 30^20 is divisible by 6020^20 = (5^20)(4^20) = (5^20)(2^20)(2^20). This tells us that the prime factorization of 20^20 does not have any 3's, which means 20^20 is NOT divisible by 3. And, if 20^20 is not divisible by 3, then 20^20 is NOT divisible by 60So, by rule #4, 30^20 – 20^20 IS NOT divisible by 60Answer: D Cheers, Brent
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Re: 30^20 – 20^20 is divisible by all of the [#permalink]
20 Nov 2018, 19:27
GreenlightTestPrep wrote: GreenlightTestPrep wrote:
C) 40 30^20 = (10^20)(4^20) = (10)(10^19)(4)(4^19) = (40)(10^19)(4^19), so 30^20 is divisible by 40 20^20 = (10^20)(2^20) = (10)(10^19)(2^2)(2^18) = (40)(10^19)(2^18), so 20^20 is divisible by 40 So, by rule #2, 30^20 – 20^20 MUST be divisible by 40 ELIMINATE C
Cheers, Brent
How is it possible 30^20 = (10^20)(4^20) ? I think, 30^20 = (15^20)(2^20) =(15)(15^19) (2^3)(2^17) =(3)(5)(2^3)(15^19)(2^17) =(5)(2^3)(3)(15^19)(2^17) =(3)(5)(2^3)(15^19)(2^17) =(5)(2^3)(3)(15^19)(2^17) = (40)(3)(15^19)(2^17) Any way C is not considered. _____________________________________________________________________________________________ This is my response and may be incorrect. Feel free to rectify any mistakes.



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Re: 30^20 – 20^20 is divisible by all of the [#permalink]
20 Nov 2018, 20:23
30^20 – 20^20 is divisible by all of the following values, EXCEPT: We will modify \(30^{20}20^{20}\) to check for choicesA) 10.......\(30^{20}20^{20}=10^{20}(3^{20}2^{20})\), \(10^{20}\) is divisible by 10, so \(10^{20}(3^{20}2^{20})\) is also divisible by 10...YES B) 25.......\(30^{20}20^{20}=10^{20}(3^{20}2^{20})=(5*2)^{20}(3^{20}2^{20})\), \(5^{20}\) is divisible by 25, so \(10^{20}(3^{20}2^{20})\) is also divisible by 25...YES C) 40.......\(30^{20}20^{20}=10^{20}(3^{20}2^{20})\), \(10^{20}\) is divisible by 40, so \(10^{20}(3^{20}2^{20})\) is also divisible by 40...YES D) 60.......\(30^{20}20^{20}=10^{20}(3^{20}2^{20})\), Now \(3^{20}\) is divisible by 3, but \(2^{20}\) is not divisible by 3, so \(10^{20}(3^{20}2^{20})\) is also not divisible by 3 and hence by all multiples of 3...NO E) 64.......\(30^{20}20^{20}=10^{20}(3^{20}2^{20})=(5*2)^{20}(3^{20}2^{20})\), \(2^{20}\) is divisible by 64, so \(10^{20}(3^{20}2^{20})\) is also divisible by 64...YES D
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Some useful Theory. 1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressionsarithmeticgeometricandharmonic11574.html#p27048 2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effectsofarithmeticoperationsonfractions11573.html?sid=d570445335a783891cd4d48a17db9825 3. Remainders : https://greprepclub.com/forum/remainderswhatyoushouldknow11524.html 4. Number properties : https://greprepclub.com/forum/numberpropertyallyourequire11518.html 5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolutemodulusabetterunderstanding11281.html




Re: 30^20 – 20^20 is divisible by all of the
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20 Nov 2018, 20:23





