Homepage Solution manuals Ivan Niven An Introduction to the Theory of Numbers Exercise 1.3.51* (Largest integer divisible by all integers less than its square root.)

Exercise 1.3.51* (Largest integer divisible by all integers less than its square root.)

Show that 24 is the largest integer divisible by all integers less than its square root.

(Hint: Consider the highest power of 2 , of 3 , of 5 , of 7 less that the square root.)

Answers

Proof. The positive integers less than the square root of 24 are 1 , 2 , 3 , 4 , and 24 is divisible by all these integers.

Let n 25 . Consider the highest powers of 2 , of 3 , of 5 , of 7 less that the square root:

2 s < n 2 s + 1 , 3 t < n 3 t + 1 , 5 u < n 5 u + 1 , 7 v < n 7 v + 1 .

Suppose that n is divisible by all integers less than its square root. Then n is divisible by 2 s , 3 t , 5 u , 7 v , therefore

2 s 3 t 5 u 7 v n ,

a fortiori 2 s 3 t 5 u 7 v n . Therefore

n 2 = ( n ) 4 2 s + 1 3 t + 1 5 u + 1 7 v + 1 = 210 × 2 s 3 t 5 u 7 v 210 n .

This implies n 210 , so every integer greater than 210 is not divisible by all integers less than its square root.

Its remains the integers n , where 25 n 210 , for which we verify the same property with a computer program, which returns all integers n up to 210 divisible by all integers less than n .

def verif(maxi):
    l = []
    for n in range(1,maxi + 1):
        test = True
        i = 1
        while test and i * i < n:
            if n % i != 0:
                test = False
            i = i + 1
        if test:
            l.append(n)
    return l

verif(210)
                          [1, 2, 3, 4, 6, 8, 12, 24]

So 24 is the largest integer divisible by all integers less than its square root. □

Note: there is another solution using the fact that n is divisible by n 1 , n 2 , n 3 , but not shorter.

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2024-07-18 17:18
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