A 4-digit number has the same two digits in the front and the same two digits in the back (like 1122), and it is divisible by 33. What is the smallest such number?

视频信息

答案文本 复制

视频字幕 复制

We need to find the smallest 4-digit number with a special pattern: the same two digits in front and the same two digits in back, like 1122, 3344, or 5566. This number must also be divisible by 33. We can express this in general form as 1000A plus 100A plus 10B plus B, which simplifies to 1100A plus 11B, where A is from 1 to 9 and B is from 0 to 9. To check divisibility by 33, we need to check divisibility by both 3 and 11, since 33 equals 3 times 11. For divisibility by 3, the sum of digits must be divisible by 3. For divisibility by 11, the alternating sum of digits must be divisible by 11. For our AABB format, the sum of digits is A plus A plus B plus B, which equals 2A plus 2B, or 2 times A plus B. The alternating sum is A minus A plus B minus B, which equals zero. Since the alternating sum is always zero and therefore divisible by 11, we only need to ensure that 2 times A plus B is divisible by 3. Since the alternating sum is always zero and divisible by 11, we only need 2 times A plus B to be divisible by 3. Since 2 and 3 are coprime, this is equivalent to A plus B being divisible by 3. Given that A ranges from 1 to 9 and B ranges from 0 to 9, the possible values for A plus B that are divisible by 3 are: 3, 6, 9, 12, 15, and 18. We can create a table showing all valid pairs. For A plus B equals 3, we have pairs like 1,2 and 2,1 and 3,0. For larger sums, we get more combinations, with the constraint that A cannot be zero. To find the smallest number, we start systematically with A equals 1. For A equals 1, we need 1 plus B to be divisible by 3, so B can be 2, 5, or 8. This gives us three candidates: 1122, 1155, and 1188. Since we want the smallest, we check 1122 first. Let's verify: 1122 divided by 33 equals 34 with no remainder. We can double-check: 34 times 33 equals 1122. For divisibility rules: the sum of digits is 1 plus 1 plus 2 plus 2 equals 6, which is divisible by 3. The alternating sum is 1 minus 1 plus 2 minus 2 equals 0, which is divisible by 11. Therefore, 1122 is indeed divisible by 33 and is our answer. Let's perform a complete verification of our answer 1122. First, the division check: 1122 divided by 33 equals 34 with remainder 0. We can verify this by calculating 34 times 33: that's 34 times 30 plus 34 times 3, which equals 1020 plus 102, giving us 1122. Next, let's check the divisibility rules. For divisibility by 3, the sum of digits is 1 plus 1 plus 2 plus 2 equals 6, and 6 divided by 3 equals 2, so it's divisible by 3. For divisibility by 11, the alternating sum is 1 minus 1 plus 2 minus 2 equals 0, and 0 divided by 11 equals 0, so it's divisible by 11. Therefore, our complete solution process confirms that 1122 is the smallest 4-digit number of the form AABB that is divisible by 33.