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30^20 – 20^20 is divisible by all of the

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30^20 – 20^20 is divisible by all of the [#permalink] New post 04 Jun 2018, 05:18
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45% (01:24) correct 54% (01:16) wrong based on 11 sessions
30^20 – 20^20 is divisible by all of the following values, EXCEPT:

A) 10
B) 25
C) 40
D) 60
E) 64
[Reveal] Spoiler: OA

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Re: 30^20 – 20^20 is divisible by all of the [#permalink] New post 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] New post 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 k


Now let's check the answer choices....

A) 10
30^20 = (10^20)(3^20) = (10)(10^19)(3^20), so 30^20 is divisible by 10
20^20 = (10^20)(2^20) = (10)(10^19)(2^20), so 20^20 is divisible by 10
So, by rule #2, 30^20 – 20^20 MUST be divisible by 10
ELIMINATE A

B) 25
30^20 = (5^20)(6^20) = (5^2)(5^18)(6^20) = (25)(5^18)(6^20), so 30^20 is divisible by 25
20^20 = (5^20)(4^20) = (5^2)(5^18)(4^20) = (25)(5^18)(4^20), so 20^20 is divisible by 25
So, by rule #2, 30^20 – 20^20 MUST be divisible by 25
ELIMINATE B


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

D) 60
30^20 = (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 60
20^20 = (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 60
So, by rule #4, 30^20 – 20^20 IS NOT divisible by 60

Answer: D

Cheers,
Brent
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Re: 30^20 – 20^20 is divisible by all of the [#permalink] New post 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.
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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] New post 20 Nov 2018, 20:23
Expert's post
30^20 – 20^20 is divisible by all of the following values, EXCEPT:

We will modify \(30^{20}-20^{20}\) to check for choices

A) 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/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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

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