\nLet the two-digit number be represented as 10t + u, where t is the tens digit (t \u2208 {1, 2, …, 9}) and u is the units digit (u \u2208 {0, 1, …, 9}).
\nWhen the digits are reversed, the new number is 10u + t.
\nThe problem states that the number increases by 27 when the digits are reversed.
\nSo, (10u + t) – (10t + u) = 27.
\n10u + t – 10t – u = 27
\n9u – 9t = 27
\n9(u – t) = 27
\nu – t = 3.
\nThis condition means the units digit is 3 greater than the tens digit. We need to find all two-digit numbers (10t + u) that satisfy this condition, keeping in mind the constraints on t and u.
\nPossible pairs (t, u) where u – t = 3:
\n– If t = 1, u = 1 + 3 = 4. The number is 14. Reversed: 41 (41 – 14 = 27). Valid.
\n– If t = 2, u = 2 + 3 = 5. The number is 25. Reversed: 52 (52 – 25 = 27). Valid.
\n– If t = 3, u = 3 + 3 = 6. The number is 36. Reversed: 63 (63 – 36 = 27). Valid.
\n– If t = 4, u = 4 + 3 = 7. The number is 47. Reversed: 74 (74 – 47 = 27). Valid.
\n– If t = 5, u = 5 + 3 = 8. The number is 58. Reversed: 85 (85 – 58 = 27). Valid.
\n– If t = 6, u = 6 + 3 = 9. The number is 69. Reversed: 96 (96 – 69 = 27). Valid.
\n– If t = 7, u = 7 + 3 = 10 (not a single digit). So, we stop here.
\nThe two-digit numbers that satisfy the condition are 14, 25, 36, 47, 58, and 69.
\nThe sum of these numbers is 14 + 25 + 36 + 47 + 58 + 69.
\nSum = (10+4) + (20+5) + (30+6) + (40+7) + (50+8) + (60+9)
\nSum = (10+20+30+40+50+60) + (4+5+6+7+8+9)
\nSum = 210 + 39 = 249.
\nAlternatively, Sum = (14 + 69) + (25 + 58) + (36 + 47) = 83 + 83 + 83 = 3 * 83 = 249.
\n<\/section>\n