Small is greater than large!

Computer Science Level 4

Find the sum of all integers n ≤ 999 n \leq999 such that sum of digits of 2 n 2^{n} > > sum of digits of 3 n + 1 3^{n+1} .


The answer is 75.

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Harsh Shrivastava by Harsh Shrivastava , 20, India

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5 solutions

Chew-Seong Cheong
May 14, 2015

I used the following Python coding.

There are only two values, n = 20 , 55 n = 20, 55\quad , therefore the required answer is 75 \boxed{75} .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution sir! I also used a similar code!

Harsh Shrivastava - 6 years, 1 month ago 
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def sumDigits( n ):
    sum1 = 0
    while n:
        sum1 += n%10
        n /= 10
    return sum1

ans = 0    
for i in range(0,1000):
    sumDigits2 = sumDigits( 2**i )
    sumDigits3 = sumDigits( 3**(i+1) )
    if sumDigits2 > sumDigits3:
        ans += i

print ans

3 Helpful 0 Interesting 0 Brilliant 0 Confused

ahahaha.. idol!!

Keil Cerbito - 6 years ago
Brock Brown
May 20, 2015

Python 2.7: 

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def digit_sum(n):
    return sum([int(i) for i in str(n)])
def goal(n):
    return digit_sum(2**n) > digit_sum(3**(n+1))
print "Answer:", sum([n for n in xrange(1000) if goal(n)])

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Bill Bell
Jul 2, 2015

Exponentiation over and over again is expensive of computer cycles, and therefore to be avoided.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Arulx Z
Jul 2, 2015 
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def summ(n):
    return sum(int(x) for x in str(n))
tot = 0
for n in xrange(1, 1000):
    tot += n if summ(2**n) > summ(3**(n+1)) else 0
print tot

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Good, standard solution.

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