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10^25 – 560 is divisible by all of the following EXCEPT:

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by BTGmoderatorDC » Sat Jun 15, 2019 4:47 pm
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10^25 – 560 is divisible by all of the following EXCEPT:

A. 11
B. 8
C. 5
D. 4
E. 3

OA E

Source: Manhattan Prep
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Source: — Problem Solving |

by GMATGuruNY » Sun Jun 16, 2019 2:09 am
BTGmoderatorDC wrote:10^25 – 560 is divisible by all of the following EXCEPT:

A. 11
B. 8
C. 5
D. 4
E. 3
RULE:
MULTIPLE OF X + MULTIPLE OF X = MULTIPLE OF X
MULTIPLE OF X + NON-MULTIPLE OF X = NON-MULTIPLE OF X

For an integer to be divisible by 3, its digit sum must be a multiple of 3.

10²� - 560 = (100000...) - (560) = 99999...440
Sum of the digits = (9+9+9+9+9...) + (4+4+0) = (multiple of 3) + (non-multiple of 3) = non-multiple of 3.
Since its digit sum is not a multiple of 3, 10²� - 560 is not divisible by 3.

The correct answer is E.
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by swerve » Sun Jun 16, 2019 2:55 pm
1. From the answer choices, we can easily eliminate options B, C, and D in the first shot. We are now left with options A and E.

2. For option A :
Remainder = 10^25/11 - 560/11 \(\Rightarrow\) (-1)^25 - 10 \(\Rightarrow\) -11 \(\Rightarrow\) -11 + 11 \(\Rightarrow\) 0
so the given number is divisible by 11 also. Eliminate option A.

Hence, we are now left with option E only, which is the answer.
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by Scott@TargetTestPrep » Fri Jun 21, 2019 6:48 pm
BTGmoderatorDC wrote:10^25 – 560 is divisible by all of the following EXCEPT:

A. 11
B. 8
C. 5
D. 4
E. 3

OA E

Source: Manhattan Prep

We see that 10^25 is divisible by 8, 5 and 4 and so is 560, therefore, the difference, 10^25 - 560, is also divisible by 8, 5 and 4. This leaves us to check 11 and 3.

Notice that when a number is subtracted from 10^n and suppose that number is at least 2 digits fewer than 10^n, then the difference will have a sequence of leading 9’s. For example, if 56 is subtracted from 10^5, the difference is 100,000 - 56 = 99,944. Furthermore, we see that when subtracting 56 from 10^5, we just need to use 100 to subtract 56 (since 100 is greater than 56 already). Notice that 100 - 56 = 44 and the remaining digits (to the left of 44) will be a sequence of leading 9’s. Let’s extend this concept to 10^25 - 560.

We see that we need to use 1000 to subtract 560 (since 1000 is greater than 560 already). Notice that 1000 - 560 = 440 and the remaining digits (to the left of 440) will be a sequence of leading 9’s. That sequence of 9’s is divisible by 3; however, 440 isn’t. Therefore, the difference, 10^25 - 440 = 99...9440 (we don’t need to care how many 9’s there are), will not be divisible by 3.

(Note: We will leave as an exercise for readers to show that 10^25 - 560 is divisible by 11. Hint: Argue that 440 in 99...9440 is divisible by 11 and that the sequence of 9’s has an even number of 9’s.)

Answer: E

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