Basic number problem

• May 14th 2010, 11:27 PM
BBAmp
Basic number problem
Hi everyone,

I couldn't figure out what to call this. All I know is that it is in a Unique Factorization chapter and I only know the very basics at the moment. I'm just trying to get a head start on number theory before I head into the class this Fall. Here is the problem:

a) Verify that (2^5)(9^2) = 2592 (I can do this but I included this to set the context)
b) Is (2^5)(a^b) = 25ab possible for other a, b? (Here 25ab denotes the digits of (2^5)(a^b) and not a product).

So my problem is I came up with the right solution which is yes and no respectively for a and b but I found both answers by inspection. Part a was easy enough but for part b I noticed there are 81 combinations of numbers and cut that combination down to 2^6, 8^2, 5^2, and 4^3 since they are the closest to 9^2 and so the closest to a number between 2500 and 2599. I checked each number and there was no other combination of a and b to make (2^5)(a^b) = 25ab possible.

Is there a way to do this problem aside from inspection? If anyone can give me a small hint I can try to work it out from there.

Thank you.
• May 14th 2010, 11:42 PM
undefined
Quote:

Originally Posted by BBAmp
Hi everyone,

I couldn't figure out what to call this. All I know is that it is in a Unique Factorization chapter and I only know the very basics at the moment. I'm just trying to get a head start on number theory before I head into the class this Fall. Here is the problem:

a) Verify that (2^5)(9^2) = 2592 (I can do this but I included this to set the context)
b) Is (2^5)(a^b) = 25ab possible for other a, b? (Here 25ab denotes the digits of (2^5)(a^b) and not a product).

So my problem is I came up with the right solution which is yes and no respectively for a and b but I found both answers by inspection. Part a was easy enough but for part b I noticed there are 81 combinations of numbers and cut that combination down to 2^6, 8^2, 5^2, and 4^3 since they are the closest to 9^2 and so the closest to a number between 2500 and 2599. I checked each number and there was no other combination of a and b to make (2^5)(a^b) = 25ab possible.

Is there a way to do this problem aside from inspection? If anyone can give me a small hint I can try to work it out from there.

Thank you.

I'm not sure how the writers of the problem intended this to be solved, but there is an easier solution by inspection that what you wrote.

2^5 = 32.

Write down all the multiples of 32 that are in the range {2500...2599}.
• May 14th 2010, 11:43 PM
simplependulum
Quote:

Originally Posted by BBAmp
Hi everyone,

I couldn't figure out what to call this. All I know is that it is in a Unique Factorization chapter and I only know the very basics at the moment. I'm just trying to get a head start on number theory before I head into the class this Fall. Here is the problem:

a) Verify that (2^5)(9^2) = 2592 (I can do this but I included this to set the context)
b) Is (2^5)(a^b) = 25ab possible for other a, b? (Here 25ab denotes the digits of (2^5)(a^b) and not a product).

So my problem is I came up with the right solution which is yes and no respectively for a and b but I found both answers by inspection. Part a was easy enough but for part b I noticed there are 81 combinations of numbers and cut that combination down to 2^6, 8^2, 5^2, and 4^3 since they are the closest to 9^2 and so the closest to a number between 2500 and 2599. I checked each number and there was no other combination of a and b to make (2^5)(a^b) = 25ab possible.

Is there a way to do this problem aside from inspection? If anyone can give me a small hint I can try to work it out from there.

Thank you.

$2^5 a^b = 25ab$

$25ab$ is divisible by $32$

so the possible numbers $25ab$ are

$2528 , 2560 , 2592$

$2^5 2^8 = 2^{13} = 8192$

$2^5 6^0 = 32$

$2^5 9^2 =2592$

only $2592$ satisfies it .