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

Question

30^{20} – 20^{20}  is divisible by all of the following values, EXCEPT:

A) 10
B) 25
C) 40
D) 60
E) 64

pushpitkc · Accepted Answer

On simplifying the expression, we get  30^{20} – 20^{20} = (2*3*5)^{20} - (2^2*5)^{20} = 2^{20}*5^{20}(3^{20} - 2^{20})

Prime-factorizing the answer options
A)  10 = 2*5 
B)  25 = 5^2 
C)  40 = 2^3*5 
D)  60 = 2*3*5 
E)  64 = 2^6

Therefore,  Option D(60)  is the only value which does not divide  30^{20} – 20^{20}

VipulGarg80 · Answer

(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.

PKN · Answer

Let's simplify the given exponents:

30^{20} – 20^{20} 
= 10^{20} * 3^{20} - {10}^{20} * 2^{20} 
= 10^{20} ( 3^{20} - 2^{20} )

For divisibility, lets check with the options:-

A . 10^{20}  is divisible by 10
B.  10^{20} = 100 * 10^{18}  is divisible by 25
C.  10^{20} = ( 2^3 *5) * 2^{17} * 5^{19}  is divisible by 40
D.  10^{20} = 2^{20} * 5^{20}  is not divisible by  2^2 *5*  3  
E.  10^{20} =  2^6 *  2^{14} * 5^{20}  is divisible by 64

Hence the correct answer is option (D)

BrentGMATPrepNow · Answer

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)(3^20) = (10^3)(10^17)(3^20) = = (1000)(10^17)(3^20) = ( 40 )(25)(10^17)(3^20), 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

PP777 · Answer

30^20 – 20^20 = 2^20 * 5^20 (3^20- 2^20) = 10^20 (3^20 - 2^20)

Prime factorizing the options:

A) 10 = 2 * 5 
B) 25= 5 * 5
C) 40 = 2^3 * 5
D) 60 = 2^2 * 3 * 5 
E) 64 = 2^6

Option D has 3 as Prime factor. So, 60 does not divide (3^20- 2^20).

Correct Answer- D

SSM700plus · Answer

30^{20} – 20^{20} 
= 10^{20} * 3^{20} - {10}^{20} * 2^{20} 
= 10^{20} ( 3^{20} - 2^{20} )
= 5^{20}  2^{20} ( 3^{20} - 2^{20} )

A) 10 = 2 * 5. divisible by this value.
B) 25 =  5^2  . Clearly divisible by this value.
C) 40 =  2^3 * 5 . divisible by this value.
D) 60 =  2^2 *3 * 5 .  NOT divisible by this value. 
E) 64 =  2^6 . divisible by this value.

Ans - D.

onlymalapink · Answer

i dont understand why is 3 a problem

Bunuel · Answer

If integers  a  and  b  are both multiples of some integer  k > 1  (divisible by  k ), then their sum and difference will also be a multiple of  k  (divisible by  k ):
 
• Example:  a = 6  and  b = 9 , both divisible by 3:  a + b = 15  and  a - b = -3 , again both divisible by 3.
 
If out of integers  a  and  b , one is a multiple of some integer  k > 1  and the other is not, then their sum and difference will NOT be a multiple of  k  (divisible by  k ):
 
• Example:  a = 6 , divisible by 3 and  b = 5 , not divisible by 3:  a + b = 11  and  a - b = 1 , neither is divisible by 3.
 
If integers  a  and  b  both are NOT multiples of some integer  k > 1  (divisible by  k ), then their sum and difference may or may not be a multiple of  k  (divisible by  k ):
 
• Example:  a = 5  and  b = 4 , neither is divisible by 3:  a + b = 9 , is divisible by 3 and  a - b = 1 , is not divisible by 3
 
• OR:  a = 6  and  b = 3 , neither is divisible by 5:  a + b = 9  and  a - b = 3 , neither is divisible by 5
 
• OR:  a = 2  and  b = 2 , neither is divisible by 4:  a + b = 4  and  a - b = 0 , both are divisible by 4.
 
Thus, according to the above, the fact that  30^{20}  is a multiple of 3 and  – 20^{20}  is not, makes  30^{20} – 20^{20}  not a multiple of 3.

Hope it helps.

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

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