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Bunuel
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first find last three digits of the big number =1000-120=880

(A) 2 [divisible by 2 because the number has 0 in unit digit
(B) 3[not divisible by 3]
(C) 4 [divisible by 4 because last two digit of the number 80 is divisible by 4]
(D) 8 [divisible by 8 because last three digit of the number 880 is divisible by 8]
(E) 10 [divisible because last digit has 0]

so B is the right answer :-)
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Bunuel
The expression 10^14 - 120 is divisible by all of the following integers EXCEPT


(A) 2

(B) 3

(C) 4

(D) 8

(E) 10

Since 120 is divisible by 3 but 10^14 is not, their difference is not divisible by 3.

Answer: B
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I found prime factors of 10 and 120 here

then prime factors were 10 = 2,5 and 120 = 2^3, 3, 5

Then I chose B because 3 was not commonly present in both numbers.

Is this a lucky coincident?
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awchoi1213
I found prime factors of 10 and 120 here

then prime factors were 10 = 2,5 and 120 = 2^3, 3, 5

Then I chose B because 3 was not commonly present in both numbers.

Is this a lucky coincident?
Yes, it was a lucky co-incidence.

3 properties--->

1) if both integers a and b are multiple of some integer k (k>1), their sum and difference will be divisible by k

2) if only one integer out of a and b is multiple of integer k (k>1), their sum and difference will not be a multiple of k

3) if none of integer a and b are multiple of integer k (k>1), their sum and difference may be or may not be divisible by k

Eg: a=5, b=4
K= 3

Hope this helps


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Bunuel
The expression 10^14 - 120 is divisible by all of the following integers EXCEPT


(A) 2

(B) 3

(C) 4

(D) 8

(E) 10

We see that 120 is divisible by 3, but 10^14 is not. Thus, the difference 10^14 - 120 is not divisible by 3, either.

Answer: B
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if someone is not mathematically inclined or does not understand number theory then he can just simply do the following:
pick a random value for 10^x. For example, 1000. Then 1000-120=880. Therefore not divisible by 3. Same will happen for any value of 10^x>100
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Bunuel
The expression 10^14 - 120 is divisible by all of the following integers EXCEPT

Hopefully I can help someone understand this question better. Testing values is the wrong way to approach this problem. Not learning the fundamentals for an easy question will hurt you.

For this questions, let's simply test divisibility rules - that's all it's asking. But instead, let's not actually solve anything here. Let's just try to take it at face value and test divisibility of the terms of the equation, not necessarily the difference of the equation.

Let's leave the terms separated as they are.

10^14 - 120


(A) 2
Is 10^14 divisible by 2? Yes, it's even
Is 120 divisible by 2? Yes, it's even

(B) 3
Is 10^14 divisible by 3? Hard to tell, let's not waste time here and move on
Is 120 divisible by 3? Yes (3 * 40)

(C) 4
Is 10^14 divisible by 4? Yes, 4 evenly goes into 100. (25 * 4)
Is 120 divisible by 4? Yes (4 * 30)

(D) 8
Is 10^14 divisible by 8? Yes, because it's divisible by both 2 and 4.
Is 120 divisible by 8? Yes (8 * 15)

(E) 10
Is 10^14 divisible by 10? Of course
Is 120 divisible by 10? Yes

We see that easily, 3 must be the answer
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