Challenge

I’m doing my homework and have hit an extra credit problem that I’m struggling with. I will give away half of my current tokens (for a total of 6,645 tokens) to the first person who can answer the following question. (I searched google, and could not find it.)

Find three consecutive integers the first of which is divisible by the square of a prime; the second, by the cube of a prime; and the third, by the fourth power of a prime.

I have an idea of how to do it, so if you ask the right questions, I’ll give hints. PLEASE help me get this extra credit! I didn’t do as well on my midterm as I’d have liked!

one dollar bob…

Yeah, I suck at teh math. Sorry.

Is this something you must show your work on?

The numbers are:

363, 364, 365

:smiley:

… or maybe not… Keep trying!

The same prime number?

1,2,3

Can I buy a vowel!.. E!
crowd cries: AAAAAAWWWWWWWWWWWWWW

I do have to show my work on this. Once I get the answer, I can figure out how to work it backwards, I think,

1 is not a prime. I have tested all the prime numbers between 2 and 71 by hand, no luck.

Yes, it is the same prime number.

inserts evil laugh

This can’t work because integers are numbers that have no fractions or decimals.

2,3,4.

2 is divisible by the square of a prime (1). 1^2 = 1. 2 is divisible by 1.
3 is divisible by the cube of a prime (1). 1^3=1. 3 is divisible by 1.
4 is divisible by the fourth power of a prime (1). 1^4=1. 4 is divisible by 1.

1 is not prime. 2 is the first prime. I have a list of all primes between 73 and 10001 if anyone is interested…

I thought 1 was considered a prime number.
I believe her question is asked improperly because it must be 3 different prime numbers.

I quoted it from my text book. I’ve tried using the technique that I’ve used on similar problems, but it doesn’t work for this one.

Another way of stating the question:

x = ppk
x+1 = pppl
x+2 = p
ppp*m, where k,l, m are just integers

Although, now that I’m thinking about it, Fred might be right. I’m going to try to work it with different primes for each.

I used primes 2 thru 73 and found in no case did I find a sequence when I squared, cubed & quadrupled.

[quote=“HottyToddyChick, post: 1083146”]I quoted it from my text book. I’ve tried using the technique that I’ve used on similar problems, but it doesn’t work for this one.

Another way of stating the question:

x = ppk
x+1 = pppl
x+2 = p
ppp*m, where k,l, m are just integers

Although, now that I’m thinking about it, Fred might be right. I’m going to try to work it with different primes for each.[/quote]

You’re on to something, of course by the original logic, Sophie is dead on;)

It’s not necessarily a direct sequence. When you quadruple the prime, subtract one and then divide by the cube of the prime. I’ve done 2 through 73 also, so don’t redo those!

I’m not so sure it’s the same prime anymore, but it would be nice if the problem said “by the same prime” or “by a different prime”, you know?

I have toyed with this long enough to discover a pattern. All 4th power primes beyond 5 are ending with 1. Therefore the cube would have to end in 0. I do not believe any primes cubed will end in 0. And the square would have to end in 9. If 1 is not considered a prime number then I cannot come up with a solution.

Ask Leah… she’s a math major.