A doubt in a question.........

Calvin edit: Discussion about mistakes made in problems should occur on the solution discussion itself, so that others can also easily learn from it. I have locked this discussion.

This is in response to We have so much in common by brian charlesworth.

Let S be the set of all positive integers such that each of and have exactly four divisors and have their divisors add to the same value.

Let m be the number of elements in and b be the sum of these elements.

Find m+b.

Here the question says that the two consecutive numbers n and n+1 have exactly four divisors and have the divisors add up to the same value. I observed that the number which is derived by the product of two numbers always has 4 divisors(1 , the number itself and the two other primes). So let the two numbers satisfying the above condition be n and n+1.

Let n = ab and n+1=cd where a,b,c,d are prime numbers.

Let us now transfer the given condition from english to mathematics.

ab=cd+1 (1)

and

1+a+b+ab=1+c+d+cd

a+b+cd+1=c+d+cd (from 1)

So a+b = c+d+1

(a+b) - (c+d) = 1

So the sum of two different primes should be 1 more than the sum of the other two primes.(condition 1)

Now let us list down a few prime numbers.....

2,3,5,7,11,13,17,19,23,29............

Consider 2,3 to be a set .Now if we pick twin primes and group it in such a way that the larger number goes with two and the smaller number goes with three,then we will have a pair which satisfies condition 1. Which implies that the product of these two numbers in a pair will yield two numbers which satisfy the given condition in the problem.

For example

5,7 are a twin prime pair.(7+2) - (5+3) = 1.

7x2=14 5x3=15.

14,15 have four factors and the sum of their divisors is equal to 24.

Similarly there can be many numbers possible.(Since there are infinite twin primes) So we cannot determine the number of elements in set S.

Am I wrong??? If I am can you correct me.....

Note by Hari prasad Varadarajan
6 years, 3 months ago

1 vote

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Comments

I have added a response to your comment to my posted solution, so if you would like to have a look that would be great. :)

Brian Charlesworth - 6 years, 3 months ago

Can you point out where I went wrong. I started my proof by taking them to be consecutive.But the result does not turn out so............

Hari prasad Varadarajan - 6 years, 3 months ago

You have three statements, namely

(i) ab=cd+1ab = cd + 1,

(ii) 1+a+b+ab=1+c+d+cd1 + a + b + ab = 1 + c + d + cd, and

(iii) (a+b)−(c+d)=1(a + b) - (c + d) = 1.

Now it is true that from (i) and (ii) we can get (iii). But this only goes in one direction, i.e., it is not the case that if statement (iii) is true then (i) or (ii) must be true. For example, if a=2,b=13,c=3a = 2, b = 13 , c = 3 and d=11d = 11 then (2+13)−(3+11)=1(2 + 13) - (3 + 11) = 1 but 2∗13=26,3∗11=332*13 = 26, 3*11 =33, i.e., statement (iii) is true but (i) is not, (nor is (ii)).

Brian Charlesworth - 6 years, 3 months ago

You are much more likely to receive a response if you comment on the problem itself (or file a report), as Brian would be able to address your concerns.

Calvin Lin Staff - 6 years, 3 months ago
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