Not Vieta's

Computer Science Level 3

$N$ is a natural number between 10 and 1000. $P$ denotes the product of its digits and $S$ denotes the sum of its digits. If $(6 \times P ) + (4 \times S ) = 4 \times N$ , how many values can N take ?

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The answer is 11.

5 solutions

Brock Brown
Dec 29, 2014

def s(x):
    total = 0
    for digit in str(x):
        total += int(digit)
    return total
def p(x):
    product = 1
    for digit in str(x):
        product *= int(digit)
    return product
n = 10
count = 0
while n <= 1000:
    if 6*p(n) + 4*s(n) == 4*n:
        count += 1
    n += 1
print "Answer:", count

The values that N can take are 16, 26, 36, 46, 56, 66, 76, 86, 96, 889, and 998.

just like i did it.

you could use

def s(x):
        return sum ([int(c) for c in str(x)])

Ori Ashual - 6 years, 2 months ago

Yeah, you're right, that's the way better version of the digit sum function. I wrote that code before I was familiar with list comprehensions .

Brock Brown - 6 years, 2 months ago

Also, for product of digits, you can do reduce(operator.mul, (int(t) for t in str(x)), 1) . By the way, you need to import operator .

Arulx Z - 5 years, 11 months ago

Bill Bell
Jul 2, 2015

Number can be split into its integers using Python's map and list functions.
reduce can be used in conjunction with a lambda function to compute the product needed.
Conditional incrementation of counter can be done with predicate based on required condition.

Aryan Gaikwad
Apr 3, 2015

public static void main(String args[]){
    int val = 0;
    for(int n = 11; n < 1000; n++)
        if(mul(Integer.toString(n)) + add(Integer.toString(n)) == 4 * n)
            val++;
    System.out.println(val);
}

private static int add(String n){
    int sum = 0;
    for(int i = 0; i < n.length(); i++)
        sum += Character.getNumericValue(n.charAt(i));
    return sum * 4;
}

private static int mul(String n){
    int prod = 1;
    for(int i = 0; i < n.length(); i++)
        prod *= Character.getNumericValue(n.charAt(i));
    return prod * 6;
}

Chew-Seong Cheong
Apr 1, 2015

I use Python coding as follows:

n = 0
for N in range(10,1001):
  T = str(N)
  P = 1
  S = 0
  for i in range(len(T)):
      P *= int(T[i])
      S += int(T[i])
  if 6*P + 4*S == 4*N:
      n += 1
      print n, P, S, N

Arulx Z
Jul 18, 2015

import operator
n = 0
for x in xrange(11, 1000):
    n += (6 * reduce(operator.mul, (int(t) for t in str(x)), 1)) + (4 * sum(int(z) for z in str(x))) == 4 * x

Moderator note:

A good solution. Consider breaking the statement inside the for loop up into a few more statements. It's good coding practice to have one major operation take place per line.

Not Vieta's

You can try more of my Questions here .

The answer is 11.

5 solutions

Moderator note:

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