Date: Fri, 04 Oct 1996 23:27:17 +0800
Sender: Rita 
Subject: Mathematics

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?

Thanx!

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
6106556447863882000035605638833716703554207400894540191395023621436050639970523
1203021164366069389563701733455174652493802096528279659381259483508916176782516
8926163221548818705965056545777743298081872565023704682568753763162781359379857
8816088851880913783787318008632718379275774870294646072072077043617747737722978
4500022657580657233628383930137914619684009220791267089768552182903618603146950
0842192427800725780716480012657266798737517723023431143584285521349919380564468
0391721696262026736880627308986765963917721348896015521169814921103068177978857
8141054359274289556411400436598704927821275214881488970218576557325551889577507
340928956338410400961096026352642413831783448576

Thanks for the question.

Cheers

Penny