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Re: The number 10^30 [#permalink]
15 May 2018, 13:42

1

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Expert's post

Carcass wrote:

The number \(10^{30}\) is divisible for all of the following EXCEPT

A. 250

B. 125

C. 32

D. 16

E 6

When we scan the answer choices, we can see that 6 is the correct answer. In order for a number to be divisible by 6, that number must be divisible by 2 AND divisible by 3

We can see that 10^30 is divisible by 2, and we should also see that 10^30 is NOT divisible by 3. How do we know this? Well, if a number is divisible by 3, then the sum of its digits must also be divisible by 3. For example, we know that 24,501 is divisible by 3, because the sum of its digits is 12 (2 + 4 + 5 + 0 + 1 = 12) and 12 is divisible by 3

What about 10^30? We know that 10^30 = 1 followed by 30 zeros. In other words, 10^30 = 1,000,000,000,000,000,000,000,000,000,000 So, the sum of its digits is 1. Since 1 is NOT divisible by 3, we know that 10^30 is NOT divisible by 3 And, if 10^30 is NOT divisible by 3, then 10^30 is NOT divisible by 6

Answer: E

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Re: The number 10^30 [#permalink]
06 Jun 2018, 12:24

1

This post received KUDOS

Carcass wrote:

The number \(10^{30}\) is divisible for all of the following EXCEPT

A. 250

B. 125

C. 32

D. 16

E 6

Straight Forward question. \(10^{30}\) means that 10 will be ended up with 30 zeros. Option E is 6. this is our answer. why? 6=3*2. any number ends with zero is divisible by 2 but a number is divisible by 3 if the sum of the digits of that number is divisible by 3. if we add up all the values of 100000000000000000000000000000000000000000 it wont be equal to the multiple of 3. Simply the sum of the digits is 1. Thus \(10^{30}\) is not divisible by 6.

Re: The number 10^30 [#permalink]
17 Oct 2018, 07:13

2

This post received KUDOS

Expert's post

Paritrat wrote:

HOW TO ELIMINATE 16? 250 AND 125 HAVE 5 AS UNIT DIDGIT SO THEY WOULD 10 WITH 30 ZERO AS ZERO IN UNIT PLACE FOR 10 AND 5 OR 0 FOR 125 .

WHAT ABOUT 32 AND 16? HOW TO ELIMINATE THEM?

-----ASIDE--------------------- A lot of integer property questions can be solved using prime factorization. For questions involving divisibility, divisors, factors and multiples, we can say: If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples: 24 is divisible by 3 because 24 = (2)(2)(2)(3) Likewise, 70 is divisible by 5 because 70 = (2)(5)(7) And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7) And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7) --------------------------