How many two digit positive integers are there such that they have an

Question

Question. How many two digit positive integers are there such that they have an odd number of total factors, all of which add up to an even number?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6

sumitkrocks · Answer

Condition, positive integers that have an odd number of total factors, qualify our set to a set of perfect squares {16, 25, 36, 49, 64, 81}
For perfect square we know sum of all factors are always odd

So answer must be 0, Option (A)

KarishmaB · Answer

If the number has an odd number of total factors, it means the number is a perfect square. 
The first two-digit perfect square is 16 (square of 4) and the last is 81 (square of 9). So, squares of 4, 5, 6, 7, 8 and 9 will be two digit integers.

We know that perfect squares have an odd number of odd factors and an even number of even factors. If odd number of odd factors are added, the sum will be odd no matter how many even factors are added to them. This means that the perfect squares should have no odd factors. But every number must have 1 as a factor. Hence, there are no such numbers. For every perfect square, the sum of all the factors must be an odd integer.

Answer (A)

Adit_ · Answer

I can see this problem being made into an impossible data sufficiency problem!

egmat · Answer

Great question — this one has a neat  trick  hiding inside it.

The problem asks for two-digit numbers that satisfy TWO conditions simultaneously:
 1.  They have an odd number of total factors.
 2.  The sum of all their factors is even.

Let's tackle each condition.

Condition 1: Odd number of factors 
Factors always come in pairs. For example, for  12 : ( 1 , 12 ), ( 2 , 6 ), ( 3 , 4 ). But for  perfect squares , the square root pairs with itself. For example, for  36 : ( 1 , 36 ), ( 2 , 18 ), ( 3 , 12 ), ( 4 , 9 ), ( 6 , 6 ) — that unpaired  6  gives an odd count.

So only perfect squares have an odd number of factors.   The two-digit perfect squares are:  16 ,  25 ,  36 ,  49 ,  64 ,  81  — that's  6  numbers.

Condition 2: Factor sum must be even 
Now let's check if ANY of these have an even factor sum:
-  16 :  1 + 2 + 4 + 8 + 16  =  31  (odd)
-  25 :  1 + 5 + 25  =  31  (odd)
-  36 :  1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36  =  91  (odd)
-  49 :  1 + 7 + 49  =  57  (odd)
-  64 :  1 + 2 + 4 + 8 + 16 + 32 + 64  =  127  (odd)
-  81 :  1 + 3 + 9 + 27 + 81  =  121  (odd)

Every single one has an odd factor sum!

This is not a coincidence. For any  perfect square , every factor in the sum is odd an odd number of times.   Key Insight:  Since perfect squares have an odd number of factors, and when you look at each factor's parity, the math always forces the total sum to be odd. (You can verify: an odd count of odd numbers always sums to odd.)

So NO number can satisfy both conditions at once.

Answer: A (0)

How many two digit positive integers are there such that they have an

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How many two digit positive integers are there such that they have an

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