Date: Fri, 04 Oct 1996 23:27:17 +0800
Sender: Rita 
Subject: Mathematics
I am a secondary student. I have tried many many times to figure out this problem, but still no answer! Can you help me?

The question is divided into 2 parts.

A)The number 64 is a sq., a cube, and a sixth power because 64=8*8, 4*4*4, 2*2*2*2*2*2. Find the smallest integer greater than 1 that is a sq., a cube, a 4th, 5th, 6th, 7th, 8th, 9th and 10th power. I have found that the answer is 2 to the power 2520th.

B) How many digits are in the correct answer to part A? The answer is uncertain. It is either 758 or 759. Can you give me any ideas?


Hi Rita

This is a nice question. Before I answer your question in part (B) I want to give my solution to part (A)

To find the smallest number (other than 1) which is a first power, a second power, a third power, ... , a tenth power, involves finding the least common multiple, the lcm, of the set of numbers {1, 2, 3, ... , 10} and raising 2 to this power.

To see this let's look at a simpler problem -- find the smallest number that is a sixth, an eight and ninth power. Note that 6 = 2*3, 8 = 2^3 and 9 = 3^2. Suppose k^m is the smallest 6th, 8th and 9th power. To make k^m as small as possible we take k = 2. To have 2^m a sixth power we must have m/6 is an integer; to be an 8th power we must have m/8 is an integer; similarly m/9 is an integer. The smallest such integer m is the lcm{6, 8, 9} = lcm{2*3, 2^3, 3^2} = 2^3*3^2 = 72; here, by expressing each number in our set in terms of its building blocks, the primes, we can easily find out how often each prime must appear in the lcm. In our original problem we must find the lcm{1, 2, 3, 2^2, 5, 2*3, 7, 2^3, 3^2, 2*5} = 2^3*5*7*3^2 = 2520. That is the number we are after is 2^2520.

To find the smallest integer (other than 1) that is a 1st, 2nd, 3rd, .... , nth power one would have to find the lcm{1, 2, 3, ... , n} and raise 2 to this power. These number are enormous once we take n to be anything at all large.

Now suppose that d is the number of digits in 2^m then 2^m = r*10^d where 0.1 <= r < 1. Taking the log base 10 of both sides we get m*log(2) = log(r) + d or d = m*log(2) - r where -1 <= log(r) < 0. Thus, since d is an integer, to find d calculate m*log(2), truncate to an integer and add 1. In your example m = 2520 so m*log(2) = 758.596 and d = 759.

Of course the other way do do this problem is to calculate 2^2520 and count the digits. 2^2520= 3940842455221416269534854318363891517281917224975164265532215418234933676588009

Thanks for the question.



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