A three digit number can be divided by 11, and the two digit number obtained after removing the last digit can be divided by 9. What are the three digit numbers?

Using (ABC) to express three digits: the multiple of 11 must satisfy the following conditions: a + C-B is the multiple of 11, a + C-B = 11 or 0, a + B is the multiple of 9. AB can be divisible by 9, which may be: 18, 27, 36, 45, 54, 63, 72, 81, 90, 99; ① when AB = 18, 1 + C-8 = 11 or 0, so C = 17 (rounding off) or 7; ② when AB = 27, 2 + C-7 = 11 or 0, so C = 16 (rounding off) or 5; ③ when AB = 36, 3 + C-6 = 11 or 0, so When C = 14 (rounding off) or 3, ④ AB = 45, 4 + C-5 = 11 or 0, so, C = 12 (rounding off) or 1, ⑤ AB = 54, 5 + C-4 = 11 or 0, so, C = 10 (rounding off) or - 1 (rounding off), ⑥ AB = 63, 6 + C-3 = 11 or 0, so, C = - 3 (rounding off) or 7, ⑦ AB = 72, 7 + C-2 = 11 or 0, so, C = - 5 (rounding off) or 6, Ⅷ AB = 81, 8 + C-1 = 11 or 0, so, C = - 7 (rounding off) or 4, Ⅸ ab =At 90, 9 + c-0 = 11 or 0, so C = - 9 (rounding off) or 2, and at 10 AB = 99, 9 + C-9 = 11 or 0, so C = 11 (rounding off) or 0, so there are 1872753651 638726814902 990, a total of 9

If a four digit number can be divided by 9, and the three digit number obtained after removing the last digit is just a multiple of 4, then the last digit of the largest of the four digits is______ ．

The maximum number of four digits should be 9. Therefore, the maximum number of four digits satisfying the condition can be set as 99ab. Because 99A can be divided by 4, the maximum value of a is 6. Because 99ab can be divided by 9, B = 3, so the answer is 3

A three digit number can be divided by 11, and the two digit number obtained after removing the last digit can be divided by 9. What are the three digit numbers?

Using (ABC) to express three digits: the multiple of 11 must satisfy: a + C-B is the multiple of 11, a + C-B = 11 or 0, a + B is the multiple of 9. AB can be divided by 9, which may be: 18, 27, 36, 45, 54, 63, 72, 81, 90, 99; ① when AB = 18, 1 + C-8 = 11 or 0, so C = 17 (rounding off) or 7, ② when AB = 27

Proof: a three digit minus the sum of its digits, will be divisible by 9
Write the derivation

Let three digits be (ABC)
=100A+10B+C
(ABC)-A-B-C=99A+9B
Because a and B are integers
So 99A and 9b are multiples of 9
So 99A + 9b is a multiple of 9
Where (ABC) is the three digit tree

A three digit number can be divided by 11, and the two digit number obtained after removing the last digit can be divided by 9. What are the three digit numbers?