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

Question

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?

Bunuel · Accepted Answer

Yes, you were on a right track. 10^(25) is a 26-digit number: 1 with 25 zeros. 10^(25) - 560 will be a 25-digit number: 22 9's followed by 440 at 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 the 9's add up to some multiple of 3, naturally, and 440 adds 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 eleven 99's, as well as 440, leave no remainder upon division by 11 (or by applying the divisibility rule for 11). Check Divisibility Rules chapter of Number Theory: https://gmatclub.com/forum/math-number- ... 88376.html

Bunuel · Answer

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.

Hope it's clear.

AzWildcat1 · Answer

+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.

enigma123 · Answer

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.

enigma123 · Answer

As I keep saying - you are a true genius. Many thanks and + 1.

MHIKER · Answer

Will the sum be : 22 x 9 +4+4+0 =

Bunuel · Answer

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.

premnath · Answer

Mod: Can you please edit the math function to correct the powers.

I am following below approach to solve this problem:

10^25 - 560 
 =2^25 * 5^25 - 2^4 *5*7

we'll check divisibility with the options one by one
a.11

10^25-560  when divided by 11 remainder will be
 (-1)^25 - 560 
 = -1-560 
 =-561  this is divisible by 11

b.8
 =2^25 * 5^25 - 2^4*5*7 
 =2^3 *2^22*5-2^3*2*5*7 
as both part have factor  2^3  , it is divisible by 8

c.5
5 is appearing in both the parts. So, divisible by 5.

d.4
4 is appearing in both the parts. So, divisible by 4.

e.3
As all other options are out, so the answer is E.

If required this can also be checked by remainder method

10 ^25-560 
 =(1)^25 - 560 
 =1-560 
 =-559

Clearly this is not divisible by 3

EvaJager · Answer

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

alex1233 · Answer

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?

Thanks!

IanStewart · Answer

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.

mejia401 · Answer

Here's a trick to calculate whether the number is a mulitple of 11 (and 7 or 13):

1. Seperate the number from right to left in groups of 3 digits
2. Add the groups in the even and odd positions separately. 
3. If the difference between the sums is divisible by 7, 11, or 13, then the entire number is divisible by 7, 11, or 13.

Since the all the digits from the 7th digit through the 22nd digit are nine and the difference is zero, the first two groups of three will be sufficient. So, 999 - 440 = 550 and 550/11 = integer, and therefore is divisible by 11.

mikeyg51 · Answer

Thank you Bunuel. I used this tactic after seeing you reference it in other posts. I learned this before but had not committed it to memory until now. Great explanation as always

Anonymous · Answer

My method for the almost identical PS problem from mgmat   was to use the prime box of 560= 2^4 * 5 *7 and 10^5 = (2*5)^5 = 2^5 * 5^5

From this point, I began to observe which numbers from the answer choices were present in the prime boxes, eliminating them, and I ended up with 3 and 11. This was were I stuck and I'd like your opinion on how can I continue and secondary, whether my way of thinking is accurate and effective  IanStewart   Bunuel   KarishmaB   GMATCoachBen

KarishmaB · Answer

I am not sure how you are solving the question using prime factorization. If you are taking common factors, some options may get eliminated but you will still be left with a large number whose value you do not know. Hence you do not know whether it is a multiple of 11 or 3. Look at the way Bunuel has solved above.

MoeKo · Answer

For those who are thinking whether 9,999........440 is divisible by 11 or not, like me, here's my solution for that.

9/11 = 0.smth ( not divisible )
99/11 = 9 ( divisible )
999/11 = 90.smth ( not divisible )
9999/11 = 9 divisible ). .... so on.

That means  even number of 9 will be divisible by 11 .
That's why 10^25 – 560 is NOT divisible by 3.

bumpbot · Answer

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

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

Prep Toolkit

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