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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

Guys any idea what concept has been Tested over here and what will be the answer?

I have started doing it this way but got stuck. So can someone please help?

I have started from

10^5 - 560 = 99,440 i.e. it has two 9s followed by 440. . . 10^10 - 560 = 99,99,440 ------------------------------> Am I doing it right this way?

Yes, you were on a right track.

10^(25) is a 26 digit number: 1 with 25 zeros. 10^(25)-560 will be 25 digit number: 22 9's and 440 in the end: 9,999,999,999,999,999,999,999,440 (you don't really need to write down the number to get the final answer). From this point you can spot that all 9's add up to some multiple of 3 (naturally) and 440 add up to 8 which is not a multiple of 3. So, the sum of all the digits is not divisible by 3 which means that the number itself is not divisible by 3.

Answer: E.

You can also quickly spot that the given number is definitely divisible: By 2 as the last digit is even; By 4 as the last two digits are divisible by 4; By 8 as the last three digits are divisible by 8; By 11 as 11 99's as well as 440 have no reminder upon division by 11 (or by applying divisibility by 11 rule).

Re: 10^25 – 560 is divisible by all of the following EXCEPT [#permalink]

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23 Jan 2012, 16:19

+1 E as well

I know that 10^x will always end in a zero so i just took a sample ( did 1000 - 560 = 440) and just checked the answers to see if 440 is divisible by any of them. Only one it is not divisible is by E.

Bunuel - really struggling to understand how did you arrive at

10^(25)-560 will be 25 digit number: 22 9's and 440 in the end. Can you please explain a bit? I may be missing a trick over here.

\(10^{25}\) is a 26 digit number: 1 and 25 zeros. For example: \(10^4=10,000\) --> 1 and 4 zeros;

\(10^{25}-560\) will be 25 digit number (so 1 less than above number): some number of 9's and 440 in the end. For example: \(10^4-560=10,000-560=9,440\) --> 9 and 440 in the end. So, out of 25 digits of \(10^{25}-560\), 3 digits in the end will be 440 and first 25-3=22 digits will be 9's.

Yes. 10^25-560 will have 25 digits: 22 9's and 440 in the end --> sum of the digits=22*9+4+4+0=22*9+8, which is obviously not a multiple of 3.
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Re: 10^25 – 560 is divisible by all of the following EXCEPT [#permalink]

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30 Jan 2012, 09:56

answer is E, its simple, 10^25 -560 will give u 21 times 9, 2times 4 and a 0, there will be total of 25 digits, so if look at the face of the number - 9999...9999 44 0.. is divisible by 11, 4, 5, & 8.. except by rule of divisibility - sum of all digits is 197, which is not divisible by 3 since it is not a multiple of 3.. so u get you answer E

Re: 10^25 – 560 is divisible by all of the following EXCEPT [#permalink]

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07 Feb 2012, 13:59

devinawilliam83 wrote:

10^25-560 is divisible by all of the following except a.11 b.8 c.5 d.4 e.3

10^25-560 will give us 9999......99440 a. Leave 11 for now b. divisibility by 8 - last 3 digits should be divisible by 8 i.e. 440/8 = 55 c. since last 2 digits are 40 no will be divisible by 5 d. since its divisible by 8 it will be obviously divisible by 4 e. Sum of nos should be divisible by 3: 9x + 4+4+0 = 17

Re: 10^25 – 560 is divisible by all of the following EXCEPT [#permalink]

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30 Aug 2012, 09:36

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enigma123 wrote:

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

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

Guys any idea what concept has been Tested over here and what will be the answer?

I have started doing it this way but got stuck. So can someone please help?

I have started from

10^5 - 560 = 99,440 i.e. it has two 9s followed by 440. . . 10^10 - 560 = 99,99,440 ------------------------------> Am I doing it right this way?

\(10^{25}\) has the sum of its digits 1 (being 1 followed by 25 zero's). Therefore, this number is a multiple of 3 plus 1. 560 is a multiple of 3 minus 1, because the sum of its digits is \(11 = 12 - 1.\) Reminder: the number and the sum of its digits leave the same remainder when divided by 3 (also true when dividing by 9).

When you subtract from a \(M3 + 1\) (multiple of 3 plus 1) an integer which is a \(M3 - 1\), the result is a \(M3 + 2.\) Like \(3a + 1 - (3b - 1) = 3(a-b)+2.\) So, our number is not divisible by 3.

Answer E
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Re: 10^25 – 560 is divisible by all of the following EXCEPT [#permalink]

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10 Feb 2013, 09:45

Bunuel wrote:

enigma123 wrote:

10^25 – 560 is divisible by all of the following EXCEPT: a)11 b)8 c)5 d)4 e) 3

Guys any idea what concept has been Tested over here and what will be the answer?

I have started doing it this way but got stuck. So can someone please help?

I have started from

10^5 - 560 = 99,440 i.e. it has two 9s followed by 440. . . 10^10 - 560 = 99,99,440 ------------------------------> Am I doing it right this way?

Yes, you were on a right track.

10^(25) is a 26 digit number: 1 with 25 zeros. 10^(25)-560 will be 25 digit number: 22 9's and 440 in the end: 9,999,999,999,999,999,999,999,440 (you don't really need to write down the number to get the final answer). From this point you can spot that all 9's add up to some multiple of 3 (naturally) and 440 add up to 8 which is not a multiple of 3. So, the sum of all the digits is not divisible by 3 which means that the number itself is not divisible by 3.

Answer: E.

You can also quickly spot that the given number is definitely divisible: By 2 as the last digit is even; By 4 as the last two digits are divisible by 4; By 8 as the last three digits are divisible by 8; By 11 as 11 99's as well as 440 have no reminder upon division by 11 (or by applying divisibility by 11 rule).

When solving these type of questions do you always try and do the subtraction and then test divisibility or do you ever try and see by what is 10^25 divisible by and what os 560 divisible by and see if there is anything in common?

When solving these type of questions do you always try and do the subtraction and then test divisibility or do you ever try and see by what is 10^25 divisible by and what os 560 divisible by and see if there is anything in common?

It really depends what the question asks. The rules are these (I wouldn't recommend memorizing them; you'll find it much easier to use them if you understand why they're true). Let's use 11 as an example, but we could use any number at all:

if you add or subtract two multiples of 11, you will always get a multiple of 11. This is because we'd be able to factor out 11: 11m + 11n = 11(m+n). So 121 + 77, and 11^7 + 11^3, and 13! + 12! must all be divisible by 11, because each is the sum of two multiples of 11.

If you add or subtract one number which is a multiple of 11 and one number which is not a multiple of 11, you will never get a multiple of 11. So 121 + 76, and 11^7 + 3^7, and 13! + 10! are not multiples of 11, because in each case we are adding a multiple of 11 and a number which is not a multiple of 11.

If you add or subtract two numbers neither of which is a multiple of 11... anything can happen. So 13+9 is a multiple of 11, but 13+10 is not. Or to take a more interesting example, 18^2 - 15^2 is a multiple of 11 (as you can see using the difference of squares factorization), but 18^2 - 13^2 is not.

Just which principle you should use really depends on the nature of the question being asked. In the question in the OP, we are subtracting two multiples of 4, 5, and 8, so the result must be divisible by 4, 5 and 8. But when it comes to checking divisibility by 3 or by 11, we need to look at the expression in a different way.
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Re: 10^25 – 560 is divisible by all of the following EXCEPT [#permalink]

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26 Feb 2013, 09:04

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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

Ans : e 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 -> (-1)^25 - 10 -> -11 -> -11 + 11 -> 0 so the given no is divisible by 11 also.Eliminate option a. Hence,we are now left with only option e which is the answer.
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