Too gruesome for me

Computer Science Level 3

( n + 3 ) n = ∑ k = 3 n + 2 k n \Large \displaystyle (n+3)^n = \sum_{k=3}^{n+2} k^n

How many natural numbers of n < 500 n < 500 satisfy the equation above?


The answer is 2.

Bill Bell by Bill Bell , 74, Canada

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4 solutions

Brock Brown
Jun 27, 2015

Let r r be the number of roots that exist.

( n + 3 ) n = ∑ k = 3 n + 2 k n ⟹ 0 = ( n + 3 ) n − ∑ k = 3 n + 2 k n (n + 3)^{n} = \sum^{n+2}_{k=3} k^{n} \implies 0 = (n + 3)^{n} - \sum^{n+2}_{k=3} k^{n}

Let f ( n ) = ( n + 3 ) n − ∑ k = 3 n + 2 k n f(n) = (n + 3)^{n} - \sum^{n+2}_{k=3} k^{n} 

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def f(n):
    return (n + 3)**n - sum((k**n for k in range(3,n+3)))
roots = 0
for n in range(501):
    if f(n) == 0:
        roots += 1
print("r =", roots)

⟹ r = 2 \implies r = 2

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Note: range(501) includes 0 0 in the search too but n ∈ N n\in N .

Arulx Z - 5 years, 9 months ago
Bill Bell
Jun 3, 2015

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

A no-frills solution. Clear and to the point.

Note, you can format code directly using the markup system, check out code markup

Thanks for the tip. I gave up on formatting code at one point. However, there is one advantage of using images and that is that it's a simple matter to display output along with the code.

Bill Bell - 5 years, 11 months ago
Dheeman Kuaner
Mar 20, 2018

Here is a C program code for the problem.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Arulx Z
Aug 23, 2015 
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>>> count = 0
>>> for n in xrange (1, 500):
        count += (n + 3) ** n == sum (k ** n for k in xrange (3, n + 3))
>>> count
2

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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