the cube of a digit is equal to that digit what is the digit
Answers
Answer:
digit is the and the cube also called the equal to digit
Let integer n＞−1 and n3 got k+1 digits {ak,ak−1,…,a1,a0} . Then
n3=10kak+10k−1ak−1+…+10a1+a0
n=ak+ak−1+…+a1+a0
n3−n=(10k−1)ak+(10k−1−1)ak−1+…+(10−1)a1
Therefore n3−n mod 9=0
(n−1)n(n+1)=9m
One and only one of three consecutive integers is divisible by 3 . Thus in this case it is also divisible by 9
0, 1, 8, 17,18, 26,27 all are solutions. Are there more?
n≦9k+9⟹
[0]n3≦93(k+1)3
How big is k ? n3 got k+1 digits. Thus
10k+10k−1+…+10+1≦n3
10k+10k−1+…+10+1=10k+1−110−1=10k+1−19
10k+1−19≦n3
With [0]
10k+1−1≦94(k+1)3
Let y=k+1
10y≦94y3+1＜104y3
10y−4＜y3
[1]10y−43＜y
f(y)=10y−43 is a continuous and increasing differentiable function. So is g(y)=y .
Minimum y=ymin=1 and [1] holds for y=ymin=1 . At y＜7 [1] holds. With y≧7 [1] does not hold, and as f(y) grows exponentially fast, y=6 is last one where solutions for n are still possible.
For y=6 k=5 , and n≦6∗9=54 because n is sum of k+1=6 digits. All need to be done is to check all integers n less than 55 within (n−1)n(n+1) mod 9=0 range (starting where left of, n＞27 ). Check for
28, 35,36,37, 44,45,46, 53,54
Cubes of these 9 numbers above do not have property required by the question [manual verification with calculator]. Therefore only solutions are
0,1,8,17,18,26,27.