N is a number made up of 999 digits of 1

AlphaBlaziken

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Feb 12, 2015
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What is N?
a. Prime number
b. Composite number
c. Even number
d. Number divisible by 9 and 18
e. Number divisible by 9 but not 18
So basically N is 111,111,111,...until 999 digits. Sorry for the English, I'm not native and this question is originally a test question in my language. I know for sure it's not c nor e and I don't think it's a prime either. Is there a trick to this? I don't think we should try proving 999 digits number using the primality test because we can't use calculator and we have limited time. The name of the test is "Arithmetic and Logic" so is this the right sub-forum to post this? If not please move my thread. Thank You.
 
This problem requires you to know the divisibility test for 9. To determine if a number, n, is divisible by 9, just add up the digits from n. If that sum is divisible by 9 then the n is divisible by 9, otherwise n is not divisible by 9.
The number is clearly not prime since 111 clearly goes into it.
 
In the end it will become 9. 9+9+9=27, 2+7=9.
But you still have not said how to use this result. Since you insist on not doing so I will. The sum of the digits is 999 and 9 goes into this number then 9 goes into the original number, 999,999,...,999. Since the number is not even (it ends in 9) is is not divisible by 18 (since multiples of 18 are all even). In the end the answer is divisible by 9 but not 18. Can you now state the divisibility rule for 18?
 
It can be divided by 18, if it can be divided by 9 and 2. From adding the digits I learned that it's divisible by 9. And because it's not even, it can't be divided by 18.
 
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