Answer: (1) 1859, (2) 3876.
The winner is Don Bradford of Belconnen, ACT, Australia. There were 313 entries.
Worked answer
The difference between the 4-digit integer abcd and its reverse is:
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999(d – a) + 90(c – b)] = 9[111(d – a) + 10(c – b)]
111(d – a) can be 0, 111, 222, 333, 444, 555, 666, 777, or 888.
10(c – b) can be 0, 10, 20, 30, 40, 50, 60, 70, 80 or 90, or the equivalent negatives since b can be greater than c.
11 x 13 = 143; [111(d – a) + 10(c – b)] can only be 0 if the integer is palindromic and cannot come to 143, 286, 429, 572 or 715; but 858 is possible if d = 9, a = 1, (c – b) = -3.
So the only possibilities are 1309, 1419, 1529, 1639, 1749, 1859 and 1969. These are all divisible by 11, but only 1859 is divisible by 13. (1859 = 11 x 132; and 9581 = 11 x 13 x 67)
The ban on leading zeros eliminates 0858; and the word “larger” eliminates the palindromes 1001 and its multiples.
17 x 19 = 323; so either d – a = 3, c – b = -1 or d – a = 6, or c – b = -2.
So the smaller integer must be 1bc4, 2bc5, 3bc6, 4bc7, 5bc8 or 6bc9 (with b – c = 1) or 1bc7, 2bc8 or 3bc9 (with b – c = 2).
- 1bc4: 323 x 8 = 2584
- 2bc5: 323 x 5 = 1615
- 3bc6: 323 x 12 = 3876
- 1bc7 or 4bc7: 323 x 9 = 2907; 323 x 19 = 6137
- 2bc8 or 5bc8: 323 x 6 = 1938; 323 x 16 = 5168, but (b – c) ≠ 1
- 3bc9 or 6bc9: 323 x 13 = 4199; 323 x 23 = 7429
3876 = 17 x 19 x 12; 6783 = 17 x 19 x 21
The ban on leading zeros eliminates 0323, 0646 and 0969; and the word “larger” excludes the palindrome 3553 = 17 x 19 x 11.
