Positive Integers

       What are the three positive integers whose sum is 43 and the sum of the cubes of the three integers is 17299².

Answer:    25,11,1 ─ First of all we must consider even- odd possibilities for A +B+C = 43

A, B and C must all be odd or two of the three numbers must be even and the other odd. Let, us, to start with, investigate the case where all three are odd. By using a table of cubes, we can see that the unit digit of A, B and C must be odd ─ 1,3,5,7, or 9. Also we can see the following pattern:

Unit Digit        A	Unit Digit       B	Unit Digit     C	Unit Digit of     A+B+C  =43
1	1	1	3*
3	3	3	9
5	5	5	5
7	7	7	1
9	9	9	7
5	7	9	1
3	7	9	9
3	5	9	7
3	5	7	5
1	7	9	7
1	5	9	5
1	5	7	3*
1	3	9	3*
1	3	7	1
1	3	5	9

Now it is clear that we only need to consider the three cases where 3 appears in the last column i.e. (1, 1, 1), (1, 5, 7) and (1, 3 9),

Unit Digit    A	Unit Digit    B	Unit Digit    C	Unit Digit of    A³ + B ³ + C³
1	1	1	3
1	5	7	9x
1	3	9	7

 Now we need consider only one or two digit numbers with unit digits 1, 5 and 7. And the possible value for the ten’s digits of A, B and C are 0, 0, 3 or 0, 1, 2.

31³ > 17299 therefore 35³ and 37³ are too large.

27³ > 17299

25³   = 15625

11³ = 1331

07³ = 343                    sum total = 17299

Now I leave it to the reader the investigation of the case where two of A, B, C are even and the other odd.