Each of X and Y is a positive integer such that:

X⁴ and Y⁴ share identical last four digits in the same order, and

X-Y = 2016

(A) Find the smallest solution satisfying the given conditions.

(B) Derive the general form of X and Y satisfying the given conditions.

Step 1:
Find all 1 digit numbers where X⁴ and (X+6)⁴ end in the same digit.
{1,2,3,6,7,8}

Step 2:
Extend this list to all two digits numbers where X⁴ and (X+16)⁴ end in the same two digits.
{11,42,23,36,17,48,+50 to each of these}

Step 3:
Extend this list to all three digit numbers where X⁴ and (X+016)⁴ end in the same three digits.
{411, +11 others, +500 to each of these}

Step 4:
Extend this list to all four digit numbers where X⁴ and (X+2016)⁴ end in the same four digits.
{2411,
4911,
1161,
3661,
0242, <---smallest
2742,
1492,
3992,
4323,
1823,
3073,
0573,
3036,
0536,
4286,
1786,
2117,
4617,
3367,
0867,
2448,
4948,
1198,
3698,
+5000 to each of these}

A general form would be any number with one of these numbers at its final four digits.

Posted by Jer on 2016-06-23 21:17:49