For each integer n 1, let Fn be the set containing all the positive

Question

For each integer n > 1, let F(n) be the set containing all the positive factors of n that are less than n. For example, F(20) = {1, 2, 4, 5, 10}. If m is an integer greater than 1 such that the sum of the elements of F(m) is less than m, which of the following could be the largest element in F(m)?

A. 18
B. 24
C. 30
D. 32
E. 36

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Show Answer

Official Answer

D

A. 18
B. 24
C. 30
D. 32
E. 36

Are You Up For the Challenge: 700 Level Questions

Raben · Accepted Answer

I'll give this one a try:

I'm not an expert by any means, but here was my approach:

I started by thinking which one intuitively had as little factors as possible:

So I started with 18, which considering m>1 implies that the number being examined is 36

18 = F(36) = {2,3,6,8,18} --> Doesn't work since 2+3+6+8+18 = 37

I applied the same logic to all the options and eventually found the correct answer for 32:

32 = F(64) = {2,4,8,16,32} --> Works since 2+4+8+16+32 = 63

venkycs98 · Answer

Small mistake -> 8 isn't a factor of 36. 
The factors are - {2,3,4,6,9,12,18} = 54

asingh91 · Answer

1 would also be factor per the question sample mentioned, hence sum would = 64 right?

onlymalapink · Answer

started with 18, which considering m>1 implies that the number being examined is 36
 huh don't get this logic

Aashimabhatia · Answer

We need to find the largest element in the set of proper factors, F(m), of a number m, given that the sum of F(m) is less than m. This means m is a deficient number, where the sum of all proper divisors is less than the number itself.

Option A: 18
If 18 is the largest element in F(m), then m = 36 since 18 is the largest proper divisor.
F(36) = {1, 2, 3, 4, 6, 9, 12, 18}
Sum = 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 = 55, which is greater than 36, so this does not satisfy the condition.

Option B: 24
If 24 is the largest element, then m = 48.
F(48) = {1, 2, 3, 4, 6, 8, 12, 16, 24}
Sum = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 16 + 24 = 56, which is greater than 48, so this does not satisfy the condition.

Option C: 30
If 30 is the largest element, then m = 60.
F(60) = {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30}
Sum = 1 + 2 + 3 + 4 + 5 + 6 + 10 + 12 + 15 + 20 + 30 = 88, which is greater than 60, so this does not satisfy the condition.

Option D: 32
If 32 is the largest element, then m = 64.
F(64) = {1, 2, 4, 8, 16, 32}
Sum = 1 + 2 + 4 + 8 + 16 + 32 = 63, which is less than 64, so this satisfies the condition.

Option E: 36
If 36 is the largest element, then m = 72.
F(72) = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}
Sum = 1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 36 = 123, which is greater than 72, so this does not satisfy the condition.

The only valid answer is 32, which is the largest element in F(64), making option D the correct answer.

Kishoreadi · Answer

The question asks for the largest integer. So start from testing the largest option which is 36. As stated by others, 36 is not the right answer. Move on to 32, and you see sum of factors of 32(except 32) is less than 32. Hence that's your answer. You don't have to test other options as they are anyway smaller than 32

Akarsh97 · Answer

In a trial to understand the question, i started with smaller no. like 2,3,4,6,8. With this it will become clear that only prime and thier power satisfies the condition i.e. 2, 3, 4(2^2), 5, 7, 8(2^3), 9(3^2), 11.........

In the given option of only 32 is 2^5 (prime no)

In the given option of only 32 is 2^5 (prime no)

egmat · Answer

Hi bpn303,

Great question — here's a shortcut that takes under a minute.

Key principle:  Powers of  2  are always "deficient" — the sum of their proper divisors is always less than the number itself.

Why? Take  2 ^k. Its proper divisors are  1 ,  2 ,  4 , ...,  2 ^(k- 1 ). That sum equals  2 ^k -  1 , which is always exactly  1  less than the number. So the condition (sum < m) is always satisfied.

Now, the largest proper divisor of any power of  2  is half of it — which is also a power of  2 .

So the shortcut becomes: just scan the answer choices for a power of  2 !

-  18  — not a power of  2 
-  24  — not a power of  2 
-  30  — not a power of  2 
-  32  =  2 ^ 5  — YES!
-  36  — not a power of  2

If  32  is the largest proper divisor, then m =  64  =  2 ^ 6 . The proper divisors are { 1 ,  2 ,  4 ,  8 ,  16 ,  32 }, and their sum is  63 , which is less than  64 . Done.

Why don't the others work? Each of  18 ,  24 ,  30 , and  36  would force m to be a number with many divisors (like  36 ,  48 ,  60 , or  72 ), and  numbers with lots of divisors tend to have divisor sums much larger than themselves — they're "abundant," not deficient.

General principle to remember:  When a problem asks about numbers whose divisor sum is small relative to the number, think  powers of primes  — especially powers of  2 . They have the fewest divisors for their size.

Answer: D

For each integer n > 1, let Fn be the set containing all the positive

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