Be Quiet Children!
Math brain teasers require computations to solve.
One day, a frustrated math teacher lost his patience with his students' non-stop chatting. Thus, he decided to give the ultimate hard problem:
Find the only positive integer number less than 20,000 that is also a sum of three positive integers, all containing exactly 7 factors.
There was a steady silence. Can you break the silence by figuring it out?
HintFor the prime factorization of positive integer x, we get
x=[y(sub 1)^K (sub 1)][y(sub 2)^ K (sub 2)]..., where y sub(n) is a prime number, and K sub(n) is the power to which y sub(n) is raised, we can figure out the number of prime factors by:
[K(sub 1) + 1][K(sub 2) + 1]...
For example, the prime factorization of 228 is (2^2)(3^1)(19^1). 228 has 12 factors, since (2+1)(1+1)(1+1) = 12.
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Answer
16,418.
First, you need to figure out the separate integers. If you have read the hint, you would know how to use prime factorization to your advantage. The only way to get 7 as the number of factors is 1 x 7. So the prime factorization of any number with 7 factors is n^6 times x^0, where n and x are both primes. But if you look, x^0, for all nonzero values of x, is 1! So, any number with exactly 7 factors must be a prime raised to the sixth power. This yields 2^6 +3^6+5^6 = 16418. Any prime 7 or greater to the sixth power is greater than 20,000.
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Comments
McBobby1212  
Mar 09, 2007
| alright! first comment!
that was pretty hard, but i think i get the jist of it. |
vlerma  
Mar 09, 2007
| So sorry, this is out of my data base. Thanks for submitting tho.  |
gooseash
Mar 10, 2007
| got the ans in 30 secs... Not that difficult. If one knows the concept of finding the num of factors of a given NUM. This is simple. Any number N can be written as
N= P^a * Q^b * R^c, where P,Q,R are prime. Then the number of factors is given by (a+1)(b+1)(c+1). Now for your question, since the num of factors is 7 an odd number, the number has to be an even number. the questions says three nums are involved and hence 2,3 and 5 can be taken. |
gooseash
Mar 10, 2007
| got the ans in 30 secs... Not that difficult. If one knows the concept of finding the num of factors of a given NUM. This is simple. Any number N can be written as
N= P^a * Q^b * R^c, where P,Q,R are prime. Then the number of factors is given by (a+1)(b+1)(c+1). Now for your question, since the num of factors is 7 an odd number, the number has to be a square number. the questions says three nums are involved and hence 2,3 and 5 can be taken. |
Mon05 
Mar 10, 2007
| tried and failed, Math is not my forte, getting too old, but great teaser and I got to remember my math.  |
Kellington 
Mar 19, 2007
| Good one. A little difficult, but I figured it out with a bit of figuring.  |
Jimbo 
Mar 24, 2007
| Personally I am astounded at the number of patterns that can be established using the properties of special numbers such as primes! Thanks for that excellent puzzle! |
colawiamo 
Apr 21, 2007
| that was intresting, only took me a minute to find out but good! |
frozenrose_885
May 12, 2007
| Lol, I'm not going to even try. |
Pissfer
Jun 18, 2007
| What if the factors in the number weren't all the same. In other words, what if you used the numbers 2^6, 2^5* 3 and 2^4 * 3^2? P.S. It would be 304! |
pating
Jun 18, 2007
| i dont get this teaser. i agree with Pissfer.
and what do you mean when you say a number containing exactly 7 factors? is it the same as prime factorization?  |
t4mt 
Jul 20, 2007
| "What if the factors in the number weren't all the same. In other words, what if you used the numbers 2^6, 2^5* 3 and 2^4 * 3^2? P.S. It would be 304!"
Did you read the hint and teaser correctly? 2^4, and 2^6 simplify to 2^10, so you're already off there. Remember, prime factorization is multiplication. Actually, all that would be 884736. Plus, none of them have exactly 7 factors. |
pating
Jul 22, 2007
| Please enlighten us, t4mt.
2^6 = 64, 2^5* 3 = 96 and 2^4 * 3^2 = 144. (64 + 96 + 144 = 304) I think that's what Pissfer meant. So how many factors do 96 and 144 have? Getting confused here.
|
t4mt
Aug 28, 2007
| Use your brain! 
96=2^5 * 3
3=3^1, obviously.
so (5+1)*(1+1) = 12. Despite what
you think, 12 does not equal 7. Hope I have enlightened you. |
bfriedfischtom
Apr 02, 2008
| Found a loophole.
It sais
"three positive integers"
It does not say they have to be different.
Thus 192 for instance is also a solution |
pzakielarz
Sep 23, 2008
| I agree with bfriedfishtom...I actually thought I was doing something wrong because I found a bunch of solutions... |
javaguru
Jan 04, 2009
| OK, I have some problems with this. In fact, the teaser either has many answers (thus not unique) or no answer depending on how you interpret the ambiguous question.
First, 2^6 has five factors, not seven. Factors are the NON-TRIVIAL divisors, which do not include 1 and the number itself. This means that the even positive numbers with 7 factors are 2^8 and those numbers which are the square of 2xp where p is an odd prime number. This gives:
2^8 = 256 (2,4,8,16,32,64,12
(2x3)^2 = 36 (2,3,4,6,9,12,1
(2x5)^2 = 100 (2,4,5,10,20,25,50)
(2x7)^2 = 196
. . .
Second, the wording of the teaser describes four positive integers, all of which have seven factors. It doesn't delineate that only the three smaller positive integers have seven factors. 16,418 has two factors: 2, 8209
In attempting to solve this, first I identified the 19 numbers less than 20000 that have 7 factors:
36, 100, 196, 256, 484, 676, 1156, 1444, 2116, 3364, 3844, 5476, 6724, 7396, 8836, 11236, 13924, 14884 and 17956
Other than 100, they all end in 4 or 6, so if the largest ends in a 6 it must be the sum of one that ends in a 4 and two that end in a 6; if the largest ends in a 4 it must be the sum of one that ends in a 6 and two that end in a 4.
I was rather disappointed after I eliminated all possibilities, because of course there is no answer to this teaser.
Even if I interpret the question as was apparently intended, that only the three smaller integers need to have 7 factors, the teaser is obviously wrong as there are well over 100 positive integers that meet the criteria.
What I find particularly curious is that all of the people who supposedly solved this teaser before looking at the answer would make the same error as t4mt who wrote it. |
javaguru
Jan 04, 2009
| OK, a correction to the last part of my last post giving my attempt to solve the problem. The teaser is still incorrect as I stated.
I mistakenly was only using EVEN positive integers. Any number which is either a prime number to the eighth power or which is the square of the product of two different prime numbers will have seven factors.
There are many solutions even when all four numbers must have 7 factors. The smallest number with 7 factors that is the sum of three other integers with 7 factors is 1225.
1225 = 1089 + 100 + 36
1225 (5, 7, 25, 35, 49, 175, 245)
1089 (3, 9, 11, 33, 99, 121, 363)
100 (2, 4, 5, 10, 20, 25, 50)
36 (2, 3, 4, 6, 9, 12, 1 |
dalfamnest 
Sep 30, 2009
| Perhaps this problem should be removed - let's see!
I am not sure what definition you use for the word 'factor' in USA. I have been teaching maths (that's New Zealand for 'math') for over 30 years now and will always count the factors of 36 as {1, 2, 3, 4, 6, 9, 12, 18, 36}. All these, and only these, can be grouped for a product of 36. How can you think 36 has seven factors, javaguru? Perhaps by omitting 36 and 1 ... but 1 x 36 = 36. There is nothing about prime or even factors in the question.
My solution is as follows:
If the number has an odd number of factors, it must be a perfect square. Hence, we check the square numbers and find that 64 is the first. {1, 2, 4, 8, 16, 32, 64}
Continuing with the patterns, and making a few deductions, I found the next numbers had to be 3^6 and 5^6.
The sum? 64 + 729 + 15625 = 16418.
Amazing! That's the answer given by the author. Great puzzle - well done!
As for the idea that the numbers could all be the same, technically that is correct - like i have three wives but they are all the same person. Let's use common sense here - it's bad enough that mathematicians say a quadratic might have two roots, both equal! |
javaguru
Nov 02, 2009
| By definition, a factor is a non-trival divisor (i.e. excludes 1 and the number itself). |
dalfamnest
Nov 02, 2009
| So a prime number has NO factors by your definition! I would define a prime as having exactly two distinct factors. Wolfram's Mathworld Resource gives the term 'proper factor' as having your definition, javaguru. |
javaguru
Nov 03, 2009
| OK, I'll go with that distinction between "factor" and "proper factor".
My comment and solution above use the definition I gave which apparently is a "proper factor". I blame it on my source for the definition of "factor", which is not as authoritative as Wolfram. |
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