So You Know Maths
Can you think of a non prime number so that all its divisors is visible in the number itself ?
For example divisor of 13 is 1 and 13 so it satisfy the first condition but its a prime so not the second. You have to find a non prime. Your time starts now...
For example divisor of 13 is 1 and 13 so it satisfy the first condition but its a prime so not the second. You have to find a non prime. Your time starts now...
Labels: mathschallenge
posted by ToM at
1:31 AM
22 Comments:
125
your example is a prime...
btw first post is correct
125 can't be divided by 2
try to make the questions make sense
1
It is not a prime no. and its divisor is visible in itself.
Why is 1 not considered a prime number? Why doesnt it it go 1,2,3,5,7,11...
and 125 does not work because its not divisible by 12 either. if you thought it would work because of 25x5 then you must also be able to use other multiple digit numbers that are adjacent to one another, like 12 in 125.
125 is a good answer, unless the poster clarifies that all numbers in the answer must be used.
110 = 11 * 10 (yes you must use the 1 twice)
15 can be divided by 1, 5 and 15
125 is right because its divisors are: 5, 5, 5. Also 25 would work. It says nothing about all of the visible numbers being a divisor.
15 wouldn't work because it is also divisible by 3
What's with the title?
Who the heck is Maths?
http://en.wiktionary.org/wiki/maths
No one has answered my original question. Who says that 1 is not a prime number?
The question does specify that ALL of its divisors are visible in the number. But it doesnt specify that all of the numbers must be used. Thus 125 should work...the only divisors are 1, 5, 25, and 125. All of which are visible in the number itself.
1 is not a prime number.
a prime number can only be divided by 2 number, 1 can only be divided by 1 number.
so its a one-of-a-kind number, like 0
I don't get it, why not use 25, it only divides by 5 and 5 is part of 25 meeting the original criteria. Nothing says that all digits from the number must be part of some divisor so 2 in addition to 5 is OK here, isn't it?
I agree with 1, wouldn't 0 work as well though? as nothing can divide 0?
1 is not a prime number by convention. Excluding 1 makes most theorems about the primes simpler to write. If 1 was regarded as a prime, you'd have to keep on saying something like "for all primes greater than 1..." nearly all the time.
As it is, quite a few theorems say "for all odd primes..." because 2 is a prime which would break the theorem.
In short 1 isn't treated as a prime for reasons of convenience.
Having played around with the idea for a bit, I'm almost certain that there is no solution to the posted problem.
Just to give a flavour of how I have tried to tackle the problem:
The number must be divisible by its prime factors (and so it must be at least a 2 digit number), so it must be writable as (e.g.):
p1*p2 = 10*p1 + p2. p1 and p2 must be less that 10 otherwise you'll get a 3 digit number. So p1 and p2 must be one of 1, 2, 3, 5 and 7 (allowing 1 as a prime).
Now p2*(p1-1)=10*p1, so either p2=p1 and p1-1=10 => p1=11 so no good, or p1-1=p1 and p2=10 which is also no good. So there is no 2 digit number that does the trick.
I'm pretty sure that you run into the same sort of difficulties when you try 3 and larger digit numbers. I've only tried a 3 digit number - and that was pretty messy.
I've been a bit lazy as I don't expect that anyone will ever read this.
NB for a 3 digit number, the maximum value is 7*7*7 = 343. This puts a major constraint on what is allowed.
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