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10^25 – 560 is divisible by all of the following EXCEPT:
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21 Jun 2021, 09:05
Solution:
A. 11- To find whether a number is divisible by 11 we can add all odd and even placed digits and subtract it from each other. If the difference is 0 or a number divisible by 11 then the number is divisible by 11.
So, for the above case there are 22:- 9's ; 2:- 4's and 1 zero. How do I know this? Lets take 1000 for eg. 1000-560=460; 10000-560=9,440. And the 9 increase as the zero increases. Thus when we have 25 zeros we'll have 21 9's.
Thus the remainder will be 0 when we divide the number which are at the odd places- the number at the even places.
11 is divisible by \(10^{25}-560\)
B. 8- for 8 to be divisible the last 3 digits i.e. 440 in the above case should be divisible- Thus 8 is divisible by \(10^{25}-560\)
C. 5- As we have a 0 in the units place \(10^{25}-560\) also divisible by 5
D. 4- The last 2 digits should be divisible by 4. Thus 40 in the above case. 4 is also divisible by\(10^{25}-560\)
E. 3- After eliminating the above options. This should be the answer. But, to find out whether a number is divisible by 3. We can add all the number and check weather the sum is divisible by 3. 9*21+8=197. Thus, 3 is not divisible by \(10^{25}-560\)
IMO E
Hope this helps!