hvelas93 wrote:
I thought I came across this somewhere but when a number is divisible by 11 its hundreds digit and ones digit add up to the tens digit?
In this case, 13,333-3=13,330. ---> 3+0=3 which would make this divisible by 11.
It's obviously not right but can someone tell me what rule I may be confused with? (I know I saw it somewhere)
I know the alternating odd-even rule for divisibility by 11 so no need to explain that.
Hello hvelas93,
If you knew the alternating odd-even rule, you SHOULD have applied that rule to solve this question. Because, that IS THE divisibility test for 11.
Whatever you have mentioned in your first sentence is essentially a corollary of this rule and NOT the rule itself. It works ONLY in the case of 3-digit numbers.
Think about it, will you generalize a divisibility test based on a certain number of digits? Not really, right?
Think about the number 1001. The sum of the ones and the hundreds digit here is 1 and the tens digit is 0. Both are not equal. Does that mean that 1001 is not divisible by 11? A resounding NO. 1001 IS divisible by 11.
So my advice to you would be to learn a rule in its entirety. Do not try to extrapolate specific cases and expect it to work under all conditions. This is essentially the reason why you got 3, which is the wrong answer. Also, do not attempt to make a short-cut out of a short-cut; divisibility tests are short-cuts themselves. They help you ascertain whether a number is divisible without you having to actually divide it.
Apply the divisibility test for 11 (the alternating odd-even rule), which you said you know, and you will see that n has to be 1 so that the difference between the respective sums is ZERO.
The correct answer option is A.
Hope that helps!
_________________
Crackverbal Prep Team
www.crackverbal.com