## Problem A: Deficient, Perfect, and Abundant

Write a program that repeatedly reads a positive integer, determines if the integer is deficient, perfect, or abundant, and outputs the number along with its classification.

A positive integer, n, is said to be perfect if the sum of its proper divisors equals the number itself. (Proper divisors include 1 but not the number itself.) If this sum is less that n, the number is deficient, and if the sum is greater than n, the number is abundant.

The input starts with the number of integers that follow. For each of the following integers, your program should output the classification, as given below. You may assume that the input integers are greater than 1 and less than 32500.

```3
4
6
12```

### Sample Output

```4 is a deficient number.
6 is a perfect number.
12 is an abundant number.
```

Point Value: 5
Time Limit: 2.00s
Memory Limit: 16M

Problem Types: [Show]

Languages Allowed:
C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, PHP, SCM, CAML, PERL, C#, C++11, PYTH3

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• (3/0)
Not to be overly 'answer checking' but i tried this multiple times on TP and they work fine but i dont get the problem here...I dont think its a typo either...

• (2/1)
Didn't you see this?
Pascal users: make sure you're using longints, if necessary.
In general you should always use longint instead of integer.