Answer:
Answer c is correct. Both statements arent sufficient separately, but together they are sufficient.
Step-by-step explanation:
Neither of the statments are sufficient alone. Lets first analyze why statment (1) alone is not sufficient.
If we suppose that our list is scattered, so our values are pretty distant one to the other, one value increasing by one wont change this. The list will still be scattered and our values will remain distant one to the other.
For a concrete example, consider a list with values multiples of a thousand. Our list can look this way {1000,2000,3000, ....., 19000, 20000}. So we have 20 distinct values. Adding one to any of this 20 values wont change the fact that we have 20 different values. However, our list doesnt have 2 consecutive integers.
Proving (2) not being sufficient is pretty straightfoward. If our list contains 20 equal values it wont have 2 consecutive values, because all values are equal. For example the list with values {0, 0, 0, 0, ....., 0} will have the value 0 occuring more than once but it doesnt have 2 consecutive values.
Now, lets assume both statements (1) and (2) are valid.
Since (2) is valid there is an integer, lets call it <em>k</em>, such that <em>k</em> appears on the list at least twice. Because (1) is True, then if we take one number with value <em>k </em>and we increase its value by one, then the number of distinct values shoudnt change. We can observe a few things:
- There are 19 numbers untouched
- The only touched value is <em>k</em>, so our list could only lost <em>k </em>as value after adding 1 to it.
- Since <em>k</em> appears at least twice on the list, modifying the value of one<em> number</em> with value <em>k</em> wont change the fact that the rest of the numbers with value <em>k</em> will <em>preserve</em> its value. This means that k is still on the list, because there still exist numbers with value k.
- The only number that <em>could</em> be new to the list is k+1, obtained from adding 1 to k
By combining points 2 and 3, we deduce that the lists doesnt lose values, because point 2 tells us that the only possible value to be lost is k, and point 3 says that the value k will be preserved!
Since the list doesnt lose values and the number of different values is the same, we can conclude that the list shoudnt gain values either, because the only possibility for the list to adquire a new value after adding one to a number is to lost a previous value because the number of distinct numbers does not vary!
Point 4 tells us that value k+1 was obtained on the new list after adding 1 to k. We reach the conclusion that the new list doesnt have new values from the original one, that means that k+1 was alredy on the original list.
Thus, the original list contains both the values k (at least twice) and k+1 (at least once), therefore, the list contains at least two consectutive values.
