A number theory problem by Akshat Sharda

Number Theory Level 5

Given a positive integer $n$ , let $P(n)$ be the product of the nonzero digits of $n$ .

If $n$ has only one digit, then $P(n)$ is equal to that digit.

$S = P(1)+ P(2)+\ldots+P(999)$

Find the largest prime factor of $S$ ?

The answer is 103.

2 solutions

Rajdeep Brahma
Mar 25, 2018

P (1) to P (10) summation is 46. P (11) to P (20) is 46×1. P (21) to P (30) is 46×2....See a pattern? So P (1) to P (99) will be 46×(1+1+2....+9)=46×46-1 P(100) to P (199)=46×46 too P (200) to P (299)=46×46×2 (Since a factor 2 comes in hundred place) Similarly,P (900) to P (999) is 46×46×9 The reqd sum will be 46×46×46-1....have 103 as highest factor... hence the answer.

Arkajyoti Banerjee
Jun 5, 2017

Computer science solution xD

#function to find no. of factors
def nf(n):
    i=1
    s=0
    while i <= n:
        if n % i == 0:
            s += 1
        i += 1
    return s
# function to check primality
def isprime(n):
    if(n%2==0) & (n!=2):
        return 0
    else:
        if nf(n) == 2:
            return 1
        else:
            return 0
#function to evaluate product of the non-zero digits of a number
def p(n):
    p=1
    while n > 0:
        if n % 10 != 0:
            p *= n % 10
            n -= n % 10
        n /= 10
    return p

i=1
s=0
while i <= 999:
    s += p(i)
    i += 1
maxprime=1
i=1
while i <= s:
    if (s % i == 0) & (isprime(i)==1) & (i > maxprime):
        maxprime = i
    i += 1
print(maxprime)

Takes some time to show the output though.

A number theory problem by Akshat Sharda

The answer is 103.

2 solutions

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