What is the smallest number equally divisible by every whole number from 1 to 100?

Question

I calclated: 69,720,375,229,712,477,164,533,808,935,312,303,556,800
I want to see if I am right (and share the answer with everybody).

that should be 808,935,312,303,556,800 after the 533... (It got cut off due to length)

Answer 1

You are absolutely right. Well done.

das Nerd: you are almost right, but the number you suggest is not divisible by 4,8,16,32,64,9,27,81,25, or 49. To find the correct result, you must multiply all the prime numbers to the largest power they appear in any number from 1-100. In the case of prime factors greater than 7, this is 1 (since any number in which they appeared twice would be greater than 100), but the factor 2 can appear up to six times, 3 four times, and both 5 and 7 can appear twice. Therefore the correct number is 2^6 * 3^4 * 5^2 * 7^2 * 11 *13 * 17... * 97, which is the number the questioner gave.

Answer 2

You can't just multiply the primes from 1 to 100, because the resulting number would be divisible by 4, 9, or any squares of primes.

You need to multiply the largest powers of primes less than 100. So this is:

64 * 81 * 25 * 49 * 11 * 13 * .... * 97

where the primes more than 10 only need a single power.

I may have typoed, but the result I get is:

69,720,375,229,712,477,164,533,808,935,312,303,556,800

which appears to agree with your answer.

Answer 3

multiply all the prime numbers between 1 to 100 and call the calculated number x.

but x is not the answer.

x is divisible by 5 but it is not divisible by 25 because there is only one 5 among its factors but we need two.

or

x is divisible by 2 but it is not divisible by 64 because there is only one 2 among its factors but we need 6.

we need max. six 2's no more because 2^7=128 which is greater than 100.

as well, we need max. two 5's no more because 5^3 is 125

the same thing for other prime numbers

so the answer is

2^6 * 3^4 * 5^2 * 7^2 * 11 * 13 * 17 * the product of all prime numbers between 18 and 100

Answer 4

I make it the product of all the prime numbers below 100 multiplied by 64,81,25 and 49. That comes to about 3.95710x10^41. My arithmetic is correct but my reasoning may be wrong. Correction. Divide the above number by 2,3,5 and 7 to get 1.88x10^39.

Answer 5

multiply all the prime numbers between 2 and 97, make sure number ends in 00.

Since I was too lazy to do it on paper... my calculator punched up some answer around 1.195898102898540406919434083821...x10^40 (i think the actual answer is x10^41)