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Is Sum divisible by 11 ?

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Is Sum divisible by 11 ?

by anubhavgulati » Tue Sep 27, 2011 10:07 am
Question.

The digits of n digit number are interchanged and the number so formed is added to the original number. Is the sum divisible by 11 ?

1) n = pqr , where p,q,r are three consecutive natural numbers.
2) The original number had 7 odd and 5 even digits.

Answer choices:
a) Stmt. 1 is sufficient
b) Stmt. 2 is sufficient
c) Both Stmt 1 and 2
d) Either of the statements
e) 1,2 together not sufficient.

Can someone provide me the solution for this question.
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by shankar.ashwin » Tue Sep 27, 2011 10:15 am
1) 3 digit No with consecutive integers can be of the form

Odd-Even-Odd (123)(or) Even-Odd-Even(234)

In both cases when you add the No. with the reverse, you get a even No. in the units place.

So you can say for sure the sum will not be divisible by 11.

2) I really cant make out much from this statement. IMO it could have cases where Nos are div by 11 and in some cases not.

A IMO. OA?

anubhavgulati wrote:Question.

The digits of n digit number are interchanged and the number so formed is added to the original number. Is the sum divisible by 11 ?

1) n = pqr , where p,q,r are three consecutive natural numbers.
2) The original number had 7 odd and 5 even digits.


Can someone provide me the solution for this question.
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by knight247 » Tue Sep 27, 2011 10:29 am
@ashwin

Since u've inferred that u you'll get Odd-Even-Odd or Even-Odd-Even and when added with Odd-Even-Odd or Even-Odd-Even respectively you will get Even always in the units digit i.e. the resultant number will be even. But the even number could still be divisible by 11 example 22 or 44 or 66 etc
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by shankar.ashwin » Tue Sep 27, 2011 10:40 am
Oops yes..

But again but divisibility test for 11 is (sum of odd places)-(sum of even places) should be div by 11

Take 123, 234 or any 3 digit No (with consecutive int)and its reverse you will find its difference is not div by 11.

123&321 = 444
234&432 = 666
345&543 = 888
456&654= 1110
567+765= 1332
and so on;

Since there are only 9 cases guess you can count them here.
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by anubhavgulati » Tue Sep 27, 2011 10:44 am
As u said, you get a even no. in the units place, so how can u r sure the sum will not divisible by 11. Any specific rule is this ?

Second statement is sufficient, the number is divisible by 11.
shankar.ashwin wrote:1) 3 digit No with consecutive integers can be of the form

Odd-Even-Odd (123)(or) Even-Odd-Even(234)

In both cases when you add the No. with the reverse, you get a even No. in the units place.

So you can say for sure the sum will not be divisible by 11.

2) I really cant make out much from this statement. IMO it could have cases where Nos are div by 11 and in some cases not.

A IMO. OA?

anubhavgulati wrote:Question.

The digits of n digit number are interchanged and the number so formed is added to the original number. Is the sum divisible by 11 ?

1) n = pqr , where p,q,r are three consecutive natural numbers.
2) The original number had 7 odd and 5 even digits.


Can someone provide me the solution for this question.
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by anubhavgulati » Tue Sep 27, 2011 10:47 am
Ok...

So it means the answer should be D, both are sufficient.

Thanks.
shankar.ashwin wrote:Oops yes..

But again but divisibility test for 11 is (sum of odd places)-(sum of even places) should be div by 11

Take 123, 234 or any 3 digit No (with consecutive int)and its reverse you will find its difference is not div by 11.

123&321 = 444
234&432 = 666
345&543 = 888
456&654= 1110
567+765= 1332
and so on;

Since there are only 9 cases guess you can count them here.
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by gmatclubmember » Tue Sep 27, 2011 10:52 am
To me statement 1 looks sufficient.
cant say nething about statement 2 though.

1. 1 says that the no. of digits in the no. can be 6 or 24 or 60.. multiples of 6.
lets say n=6.
the no. can be 100000a+10000b+1000c+100d+10e+f when we reverse its digits:100000f+10000e+1000d+100c+10b+a.
Add up both:
100001a+10010b+1100c+1100d+10010e+100001f.
Now a no. is divisible by 11 if the difference in the summation of its odd and even digits is 0, which is satisfied by the no. above.
so this no. is divisible by 11.
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by rishimaharaj » Tue Sep 27, 2011 10:55 am
Anubhav,
Here's my attempt:

If you have an n digit number, that means that:
If n=1, it can be any number between 1-9.
If n=2, it can be any number between 10-99.
If n=3, it can be any number between 100-999.
etc.

For n=1:
  • Digits interchanged and added will never be a multiple of 11.
  • Example: 1 + 1 = 2 or 9 + 9 = 18.
  • This will give an answer of NO to the question.
For n=2:
  • Digits interchanged and added seem to always give a multiple of 11.
  • Exmaple: 99 + 99 = 198. 27 + 72 = 99. 12 + 21 = 33.
  • This will give an answer of YES to the question.
For n=3:
  • Interchanged digits added can sometimes give a multiple of 11.
  • Negative Example: 113 + 131 = 244 --> Not a multiple of 11.
  • Positive Example: 111 + 111 = 222 --> Multiple of 11.
We can keep going on, but I doubt it will help much.

STATEMENT 1.
n = product of p, q, and r --> three consecutive natural numbers.
-n could be 1*2*3 = 6; 6 + 6 = 12; not a multiple of 11, which is an answer NO to the question.
-n could be 2*3*4 = 24, 24 + 42 = 66; a multiple of 11, which is an answer YES to the question.
STATEMENT 1 is INSUFFICIENT.

STATEMENT 2.
n is a 12 digit number with 7 odd digits and 5 even digits.
I would assume this is insufficient without doing much work, but to double check, let the number be:
111111122222. Rearrange it to be: 222221111111. Add them together to get 333332233333.
Using the divisibility by 11 rule, add every other number and subtract the remaining numbers:
We have (3+3+3+2+3+3) - (3+3+2+3+3+3)= 17 - 17 = 0 --> Multiple of 11.

Now let the number be: 12121212121213 and rearrange it to be 212113211211. Add them together to get 333325332424.
We have (3+3+2+3+2+2) - (3+3+5+3+4+4) = 15 - 22 = -7 --> Not a multiple of 11.
STATEMENT 2 is INSUFFICIENT.

STATEMENTS 1 & 2 Combined.
By this point, I'm fed up of the question and would guess between either C or E. I mean, really...Who is going to sit there during to test to try and find 3 consecutive numbers whose product contains 12 digits, 5 of which are even and 7 of which are odd. Not me, for sure!

I'd make an educated guess that there are some possibilities that will give a multiple of 11 and some which don't (referring to the work in STATEMENT 2). I'd then markanswer choice E, and move on to the next question.

Sorry for the rushed ending -- I'm interested in seeing the correct answer, and possibly a faster, more efficient/elegant solution that what I've proposed above, hehehe :-).

--Rishi
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by shankar.ashwin » Tue Sep 27, 2011 10:57 am
Ami
Statement 1 sounded like it was a 3 digit integer, why consider n=6?
I listed down n=3 and consecutive (see above) and none of the Nos were divisible.

gmatclubmember wrote:To me statement 1 looks sufficient.
cant say nething about statement 2 though.

1. 1 says that the no. of digits in the no. can be 6 or 24 or 60.. multiples of 6.
lets say n=6.
the no. can be 100000a+10000b+1000c+100d+10e+f when we reverse its digits:100000f+10000e+1000d+100c+10b+a.
Add up both:
100001a+10010b+1100c+1100d+10010e+100001f.
Now a no. is divisible by 11 if the difference in the summation of its odd and even digits is 0, which is satisfied by the no. above.
so this no. is divisible by 11.
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by gmatclubmember » Tue Sep 27, 2011 11:05 am
shankar.ashwin wrote:Ami
Statement 1 sounded like it was a 3 digit integer, why consider n=6?
I listed down n=3 and consecutive (see above) and none of the Nos were divisible.

gmatclubmember wrote:To me statement 1 looks sufficient.
cant say nething about statement 2 though.

1. 1 says that the no. of digits in the no. can be 6 or 24 or 60.. multiples of 6.
lets say n=6.
the no. can be 100000a+10000b+1000c+100d+10e+f when we reverse its digits:100000f+10000e+1000d+100c+10b+a.
Add up both:
100001a+10010b+1100c+1100d+10010e+100001f.
Now a no. is divisible by 11 if the difference in the summation of its odd and even digits is 0, which is satisfied by the no. above.
so this no. is divisible by 11.
n is the no. of digits. and n= (a-1)a(a+1) so the least value of n (no. of digits) is 6.
I hope I make it clear now :)
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by shankar.ashwin » Tue Sep 27, 2011 11:07 am
Ooh! Completely misinterpreted the problem.. Time to catch up sleep I guess :)
Last edited by shankar.ashwin on Tue Sep 27, 2011 11:09 am, edited 1 time in total.
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by rishimaharaj » Tue Sep 27, 2011 11:08 am
Wait a second...
Does n = pqr mean that
-p is the Hundreds digit of n
-q is the Tens digit of n, and
-r is the Units digit of n?

Or does n = pqr mean that n is the product of p, q, and r?

If it is the first meaning, where each letter represents a digit, then how could statement 2 be possible? How could a 3 digit number have 7 odd digits and 5 even digits?
It's not possible.
The statements must be true, and not contradict each other, right?

If that's the case, then n = pqr must mean that n is the product of p, q, and r...
Right?
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by gmatclubmember » Tue Sep 27, 2011 11:11 am
shankar.ashwin wrote:Ooh! Completely misinterpreted the problem.. Time to catch up sleep I guess :)
To me the answer should be either A or D. first statement is sufficient (check with any 6 digit no.: 231453 + 354132 = 585585 divisible by 11).
Second statement is still elusive :(
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by rishimaharaj » Tue Sep 27, 2011 11:15 am
I agree with gmatclubmember...I screwed up with statement STATEMENT 1.
He's correct by saying STATEMENT 1 is sufficient.

But STATEMENT 2 is definitely insufficient.

I'll change my answer to answer choice A now! :-P
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by anubhavgulati » Tue Sep 27, 2011 11:18 am
Statement no 2 is also sufficient, its 12 digit number as mentioned in the question. In your previous post for this question you have taken the 14 digit number instead of 12 digit number which results in wrong answer.
rishimaharaj wrote:I agree with gmatclubmember...I screwed up with statement STATEMENT 1.
He's correct by saying STATEMENT 1 is sufficient.

But STATEMENT 2 is definitely insufficient.

I'll change my answer to answer choice A now! :-P
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