Digits

[cinelli]

This puzzle is to find a number whose square and cube, when taken together, contain all ten digits precisely once.

Cinelli

[psychodom]

69

For the square and the cube to have exactly 10 digits the original number must be greater than 46 (46^2 has 4 digits, 46^3 has 5 digits) and, similarly, less than 100.
We can discount all numbers ending in 0,1,5,6 as these have repeated last digits.

At this point I loaded up Excel and inspected the remainder to find 69, I'm not sure if there's a more refined sieve that could be applied.

-Dom

[cinelli]

69 is the right answer. As Dom has pointed out, you can discount numbers ending in 0, 1, 5 and 6. You can also observe that if the combination of square and cube include all ten digits, the digital root of this combination is 9. This further restricts the trial number to having a digital root of 3, 6, 8 and 9. So the list of trial numbers is 48, 53, 54, 57, 62, 63, 69, 72, 67, 84, 87, 93, 98, 99, which is not too long. Then I think you just have to work through them. For the record 69^2 is 4761 and 69^3 is 328509.

Cinelli

[psychodom]

Thanks Cinelli,
I must admit I struggled with the logic of

You can also observe that if the combination of square and cube include all ten digits, the digital root of this combination is 9. This further restricts the trial number to having a digital root of 3, 6, 8 and 9

so I did some digging on the properties of digital roots:

The digital root (also repeated digital sum) of a non-negative integer is the (single digit) value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.

properties of digital roots
dr(a+b) = dr(dr(a) + dr(b))
dr(a*b) = dr(dr(a) * dr(b))
The digital root of a square is 1, 4, 7, or 9. Digital roots of square numbers progress in the sequence 1, 4, 9, 7, 7, 9, 4, 1, 9 (ie periodicity of 9)
The digital root of a perfect cube is 1, 8 or 9, in that strict sequence (ie periodicity of 3)

so for a root (R) the digital roots for the respective squares and cubes have the following repeating pattern
R|123456789...
dr(S)|149779419...
dr(C)|189189189...

we know that digital root of combination of our square (S) and cube (C) is 9
therefore
dr(S) = 1 and dr(C) = 8 (ie R is a multiple of 9 less one; 8, 17, 26, 35, ...)
or
dr(S) = 9 and dr(C) = 9 (ie R is a multiple of 3)

this (along with the original sieve) gives Cinelli's refined candidate list of
48, 53, 54, 57, 62, 63, 69, 72, 78, 84, 87, 93, 98, 99 as required
(98 and 98 can both be discounted too by observing both their squares and cubes have leading 9s)

Cheers,

-Dom

[Gengulphus]

The digital root (also repeated digital sum) of a non-negative integer is the (single digit) value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.

properties of digital roots
dr(a+b) = dr(dr(a) + dr(b))
dr(a*b) = dr(dr(a) * dr(b))

Just to add some further detail: the digital root is just the remainder of the number when divided by 9, except that when the number is divisible by 9, the digital root is 9 rather than 0 (except of course when the original number is 0). This follows mainly from the fact that each of 1, 10, 100, 1000, etc, has remainder 1 when divided by 9, which implies that any non-negative number and the sum of its digits have the same remainder when divided by 9 (because it implies for instance that 123 = 100+20+3 and 1+2+3 have the same remainder when divided by 9). The argument needs to be completed by observing that the digits sum of any multi-digit number is smaller than the original number, so that the process of repeatedly summing the digits cannot terminate with more than a single-digit number, and that 0 is only the digits sum of 0 itself.

The remainder when divided by any other number N has corresponding properties if one writes the numbers in base N+1. Using N=99 and base 100, for instance, that means e.g. that 1234567890, 12+34+56+78+90 = 270 and 2+70 = 72 have the same remainder when divided by 99. They therefore have the same remainder when divided by 11, so the remainder of 1234567890 when divided by 11 is the same as that of 72, which is 6. With a bit more mathematical work along the same lines, that leads to the divisibility-by-11 test, that the sum of the even digits differs from the sum of the odd digits by a multiple of 11 - e.g. for 12345657890, the sum of the odd digits is 1+3+5+7+9 = 25 and the sum of the even digits is 2+4+6+8+0 = 20, which differ by 5, a non-multiple of 11, and so 1234567890 is not divisible by 11. (The test is sometimes wrongly presented as the sum of the odd digits being the same as the sum of the even digits, by the way - which works for enough of the low multiples of 11 that people tend to believe it: it first goes wrong for 209 = 11*19.)

Gengulphus