MGAMT divisibility exception

[fang6]

this is a question from MGMAT - I have seen how this is resolved but the method offered by MGAT to resolve this does not seem that logic or straightforward to me and I want to see if someone else has another way to solve this

here is the question

10^25 - 560 is divisible by all of the following EXCEPT which one of the following :

11
8
5
4
3

thanks for your help

[Guest]

I am not sure MGMAT solved it but i used the following method.

I quickly eliminated 8,5,4 since both 10^25 and 560 are divisible by these 3 numbers. So that left me 11 and 3.

I picked 3, since any factor of 10 divided by 3 will give me 0.3333... and so all i had to do was to prove 560/3 wasnt 0.333333. Sure enough its 0.66666..

I quickly check 11 and found a similar pattern after the decimal 909090.. so that confirmed that 3 was the answer.

this is a question from MGMAT - I have seen how this is resolved but the method offered by MGAT to resolve this does not seem that logic or straightforward to me and I want to see if someone else has another way to solve this

here is the question

10^25 - 560 is divisible by all of the following EXCEPT which one of the following :

11
8
5
4
3

thanks for your help

[Guest]

10^25 is just 1 followed by 25 zeros. So 10^25 - 560 is just 22 9's followed by 440, i.e. 9999999999999999999999440. So as the previous poster pointed out you can eliminate 5, 4, and 8 since both 10^25 and 560 are both multiples is 5, 4 and 8. If you don't see that trick, you can just apply the usual tests for divisibility by 5,4, and 8 to eliminate them as answer choices. In order to test 3, you could sum up all the digits and you'd get 206, which is not divisible by 3,so the answer is 3.

However, you might also get there by process of elimination by seeing that the number above is really just a bunch of "99"'s with a "44" tacked on the end so it is going to be divisible by 11.

[RonPurewal]

yeah.

you could also apply divisibility tests for all of the answer choices. the number has 25 digits (9,999,999,999,999,999,999,999,440) - which, although perhaps a bit intimidating, is nowhere close to unmanageable.

11: add and subtract alternate digits, giving 0 (all the digits cancel in pairs, since there are twenty-two 9's and two 4's) --> divisible
8: 8 goes into 440 --> divisible
5: ends in 0 --> divisible
4: 4 goes into 40 --> divisible
3: add all the digits: 22x9 + 2x4 = 206, which is not divisible by 3 --> not divisible

there you go. quite do-able.

lesson: even if a method seems pretty labor-intensive, just go with it. do not stare at problems.

[vietst]

The remainder = 1-2 = -1 or 2 when 10^25 - 560 is divided by 3. Then 3

[rahul]

I agree with Ron, that's how i solved
the last three digits ( 1000 - 560 ) 440

440 is divisible by 11,4,8,5

440 is not evenly divisible by 3 so the answer is 3

[fang6]

thanks to all for the above replies this was very helpful ! great community on MGMAT forums ! Thanks !

[StaceyKoprince]

Glad you're finding this useful! Happy studying!

[Guest]

Happily, you don't have to sum all of the digits...

The number is a bunch of 9's followed by 440.

We know the rule that the a number is divisible by three if the sum of the digits is divisble by three.

However, 3 goes evenly into 9, so no matter how many 9's you add together, 3 will go into that number.

But, if you add 8 (the two 4's at the end) to a number that 3 already goes into (the bunch of 9's), the number is no longer divisible by 3, since 3 doesn't go into 8.

[RonPurewal]

Happily, you don't have to sum all of the digits...

The number is a bunch of 9's followed by 440.

We know the rule that the a number is divisible by three if the sum of the digits is divisble by three.

However, 3 goes evenly into 9, so no matter how many 9's you add together, 3 will go into that number.

But, if you add 8 (the two 4's at the end) to a number that 3 already goes into (the bunch of 9's), the number is no longer divisible by 3, since 3 doesn't go into 8.

very true. the more solutions, the merrier.

this is clearly more efficient than the div-by-3 test that i propounded above, although if you don't think of it right away - and that means in a couple of seconds tops - then you should start summing digits. remember that time management, not finding the most elegant solution, is the principal issue.