There are three 6 digit numbers with the following properties applicable to each:

1. All digits are unique.

2. The first three digits

**ABC** form a triangular number as do the latter three,

**DEF**; both are multiples of 3.

3. The

digital root/sum of the first triangular number is greater than that of the second.

4. Three consecutive digits form the difference of the triangular numbers, either being ascending or descending.

Identify the three 6 digit numbers.

At the outset, we observe that each of ABC and DEF are triangular numbers divisible by 3. Since their absolute difference consists of consecutive ordered digits, we must have:

|T(x) - T(y)| = 100*a + 10*(a+1) + (a+2), or 100*a + 10*(a-1) + (a-2), where T(x) and T(y) are triangular numbers.

or, |T(x) - T(y)| = 111*a +/- 12

or, |T(x) - T(y)|(mod 111) = +/-12 ......(i)

Again, for a triangular number T(n), the given conditions stipulate that:

100 <= T(n) <= 999

or, 200 <= n(n+1) <= 1998

or, 14 <= n <= 44 .........(ii)

But 3 divides T(n) = n(n+1)/2, and so:

n must possess the form 3t, or 3t-1 .......(iii)

We now construct a table in conformity with (i), (ii) and (iii) as

follows:

n T(n) residue of T(n)

in mod 111

14 105 105

15 120 9

17 153 42

18 171 60

20 210 99

21 231 9

23 276 54

24 300 78

26 351 18

27 378 45

29 435 102

30 465 21

32 528 84

33 561 6

35 630 75

36 666 0

38 741 75

39 780 3

41 861 84

42 903 15

44 990 102

From the above table, we observe that the valid pairs of triangular numbers (T(m),T(n)) differing by 12 in their absolute values of residues (mod 111) are:

(210, 666), (231, 465), (120, 465), (153, 276), (351, 561),

(435, 780), (780, 903), (780, 990).

Of these, the respective magnitudes of the absolute differences between the elements of each of the of the pairs (351, 561), (780, 990) do not consist of consecutive ordered digits.

Also, each of the pairs (210, 666),(780, 903) contain duplicate digits.

We denote d(x) as the digital root of x, and observe that by conditions of the problem d(ABC) > d(DEF). Thus, for the remaining pairs:

d(231) = 6 = d(465), a contradiction

d(120) = 3 < 6 = d(465), and so: ABC = 465, DEF = 120

d(153) = 9 < 6 = d(276), and so: ABC = 153, DEF = 276

d(435) = 3 < 6 = d(780), and so: ABC = 780, DEF = 435

Consequently, the required 6 digit numbers are *153276, 465120* and* 780435*.

*Edited on ***August 18, 2008, 1:34 pm**