A problem by Rama Devi

Computer Science Level 3

I am thinking of a 6-digit number. The sum of the digits is 43.

And only two of the following three statements about the number are true:

(1) it's a square number,

(2) it's a cube number, and

(3) the number is under 500000.

Find the number.


The answer is 499849.

Rama Devi by Rama Devi , 43, India

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3 solutions

Arulx Z
Jun 21, 2015 
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>>> def sum(n):
    s = 0
    while n:
        s += n % 10
        n /= 10
    return s

>>> def sqrt(n):
    for x in xrange(317, 1000):
        z = x ** 2
        if(z == n):
            return True
        elif(z > n):
            return False


>>> def cbrt(n):
    for x in xrange(47, 100):
        z = x ** 3
        if(z == n):
            return True
        elif(z > n):
            return False

>>> for n in xrange(100000,1000000):
    if(sum(n) == 43):
        if(sqrt(n) and cbrt(n)):
            print n
        elif(cbrt(n) and n < 500000):
            print n
        elif(n < 500000 and sqrt(n)):
            print n


499849

Yay! I can finally code in python!

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Congrats on being able to code in Python!

Here 's a C++ solution. Adding it here for the sake of variety.

Prasun Biswas - 5 years, 9 months ago

Did you like the problem ?

Rama Devi - 5 years, 11 months ago

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Yes. I do.

Arulx Z - 5 years, 11 months ago

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Then like and re share it for others.

Rama Devi - 5 years, 11 months ago

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@Rama Devi – Please up vote my solution if you like it.

Rama Devi - 5 years, 11 months ago
Rama Devi
Jun 16, 2015

We can deduce that the answer is 499849 by the following steps.

Max sum of numbers is 4+9+9+9+9+9, which equals 49, from this You can deduce that the number is fairly close to the 499999.

It may be a square number. So the square is below square of max sqrt(499999)~707.10607. By trying squares below 707.1.... You will find the right answer that is 707^2.

Only statements 1 and 3 are true because there exists no cube number,whose digits' sum is 43.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I got this wrong, because I failed to read the part that says "only two of the following three statements about this number are true".

But here's some Python anyway:

Python 3.4: 

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def goal(n):
    count = 0
    if digit_sum(n) == 43:
        if has_root(n,2):
            count += 1
        if has_root(n,3):
            count += 1
        if n < 500000:
            count += 1
    return count == 2
def digit_sum(n):
    return sum((int(i) for i in str(n)))
def has_root(n, root):
    test = 0
    while test**root <= n:
        if test**root == n:
            return True
        test += 1
    return False
n = 100000
while not goal(n):
    n += 1
print("Answer:", n)

Brock Brown - 5 years, 12 months ago

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So , you got trapped.Then why don't you up vote my solution.

Rama Devi - 5 years, 12 months ago

Your solution is incomplete. You didn't show that there is no other number with only satisfies 2 of the 3 statements.

Can you come up with a one line argument why condition 2 cannot be true if the sum of the digits is 43?

Calvin Lin Staff - 5 years, 11 months ago

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I did not mean that 2 cannot be true.

Rama Devi - 5 years, 11 months ago

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I know you did not mean that, nor did you say that. That is why your solution is incomplete, because you didn't show that we could have 1 and 2 being satisfied.

I am saying that

There is no perfect cube whose digit sum is 43.

Calvin Lin Staff - 5 years, 11 months ago

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@Calvin Lin – Exactly.That is what I mean.

Any way , thanks, I will edit my solution right now.

Rama Devi - 5 years, 11 months ago

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@Rama Devi – Note that you should still explain why "there is no cube number whose digit sum is 43". That is not immediately obvious.

Calvin Lin Staff - 5 years, 11 months ago

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@Calvin Lin – How it that possible?

Rama Devi - 5 years, 11 months ago

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@Rama Devi – Show that the "cubic residues mod 9" are only 0, 1, 8.

We know that n 3 ≡ digit sum of n 3 ≡ 43 ( m o d 9 ) n^3 \equiv \text{digit sum of } n^3 \equiv 43 \pmod{9} .

Hence, there is no solution.

Calvin Lin Staff - 5 years, 11 months ago
Bill Bell
Sep 17, 2015

Since there aren't many six-digit squares and cubes it's reasonably inexpensive to use a lookup list for them. 

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cubes=[c**3 for c in range(47,100)]
squares=[s**2 for s in range(317,1000)]

def sum_digits(n):
    r = 0
    while n:
        r, n = r + n % 10, n / 10
    return r

for n in range(10**5,10**6):
    if sum_digits(n)==43:
        if int(n in squares) + int(n in cubes) + int(n<500000)==2:
            print n

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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