2007 iTest Problems/Problem 38

Problem

Find the largest positive integer that is equal to the cube of the sum of its digits.

Solution

Clearly this number is the cube of a positive integer $n$ so let $n^3$ be the number we seek. The sum of the digits of $n$ is equal to $n\mod{9}$ , so $n^3\equiv n\mod{9}$ or $n(n+1)(n-1)=9k$ for some natural $k$ . Because there is only one factor of 3 in a set of 3 consecutive numbers, one of $n$ , $n-1$ , or $n+1$ must be divisible by 9. Now note that $n$ has maximum 5 digits because else the prompt is always false (solution to $10^n>(9(n+1))^3$ ), so $n\le\sqrt[3]{100000}\approx46$ . Thus $n$ could be $46,45,44,37,36,35,28,27,26\dots$ . The first three numbers are clearly ineligible because they would require too many high digits. Then we test each number. $37^3=50653$ , $36^3=46656$ , $35^3=42875$ , $28^3=21952$ , and finally $27^3=19683\rightarrow1+9+6+8+3=27$ , so the answer is $\boxed{19683}$ .

2007 iTest (Problems, Answer Key)
Preceded by: Problem 37	Followed by: Problem 39
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4

Page

Toolbox

Search

2007 iTest Problems/Problem 38

Problem

Solution

See Also