2016 is absolutely awesome 3

Logic Level 2

For any positive integer $k$ let $f_1(k)$ denote the sum of the squares of the digits of $k$ (when written in decimal), and for $n \ge 2$ define $f_n(k)$ iteratively by $f_n(k)=f_1(f_{n-1}(k))$ .

Find $f_{2017}(2016)$ .

The answer is 42.

2 solutions

Khang Nguyen Thanh
Dec 14, 2015

Starting from $k = 2016$ and iterating the map “sum of squares of the digits” we obtain the chain $2016 \rightarrow 41 \rightarrow 17 \rightarrow 50 \rightarrow 25 \rightarrow 29 \rightarrow 85 \rightarrow 89 \rightarrow 145 \rightarrow 42 \rightarrow 20 \rightarrow 4 \rightarrow 16 \rightarrow 37 \rightarrow 58 \rightarrow 89$ , after which the sequence repeats itself, with period $8$ .

Thus, $f_1(2016) = 41, f_2(2016) = 17$ , etc., and $f_{n+8}(2016) = f_n(2016)$ for all integers $n \ge 7$ .

Since $2017 = 8 . 251 + 9$ , it follows that follows that $f_{2017}(2016) = f_9(2016) = 42$ .

Yes Same Way! Is there any other method?

Kushagra Sahni - 5 years, 6 months ago

I solved it this way as well - if you are good at carrying out simple calculations it is a very easy problem.

Thomas Sutcliffe - 3 years, 7 months ago

@Khang Nguyen Thanh I apologise for this strange question, but is there a reason why such a cycle would exist?

John Frank - 3 years, 2 months ago

One reason (probably not the best) to thik about a cycle is that Any number with 4 digits have f(a) with less than 4 digits, therefore you know that when you apply f a certain number of times you will find a same value again at somepoint and therefore enter into a cycle. As many numbers have the same “f” you can hope to fall into a cycle very quickly

Omar Rekhis - 2 years, 7 months ago

Masbahul Islam
Jan 4, 2016

def a(n):

s=0

x=str(n)

for i in range(len(x)):

    y=int(x[i])

    s=s+y*y

return s

f1=a(2016)

for i in range(1,2017):

f2=a(f1)

f1=f2

print f2

2016 is absolutely awesome 3

The answer is 42.

2 solutions

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