answer：Okay, so I have this inequality: ( x^2 - 2x + 5 geq a^2 - 3a ). The problem says this inequality holds for any real number ( x ). I need to find the range of the real number ( a ). Hmm, let me think about how to approach this.First, I remember that quadratic expressions can be analyzed by completing the square. Maybe that will help me find the minimum or maximum value of the quadratic in terms of ( x ). Let me try completing the square for ( x^2 - 2x + 5 ).So, starting with ( x^2 - 2x ). To complete the square, I take half of the coefficient of ( x ), which is ( -2 ), so half of that is ( -1 ), and then square it, getting ( 1 ). So, I can write:( x^2 - 2x + 1 = (x - 1)^2 ).But wait, the original expression is ( x^2 - 2x + 5 ). So, I added 1 to complete the square, but I also have the +5. So, I need to adjust for that. Let me write it as:( x^2 - 2x + 5 = (x - 1)^2 + 4 ).Yes, because ( (x - 1)^2 = x^2 - 2x + 1 ), so adding 4 gives ( x^2 - 2x + 5 ). That makes sense.So, now the inequality becomes:( (x - 1)^2 + 4 geq a^2 - 3a ).Since ( (x - 1)^2 ) is a square term, it's always non-negative for any real ( x ). The smallest value it can take is 0, which happens when ( x = 1 ). Therefore, the minimum value of the left-hand side ( (x - 1)^2 + 4 ) is 4.So, for the inequality ( (x - 1)^2 + 4 geq a^2 - 3a ) to hold for all real ( x ), the right-hand side ( a^2 - 3a ) must be less than or equal to the minimum value of the left-hand side, which is 4.Therefore, we have:( a^2 - 3a leq 4 ).Now, I need to solve this quadratic inequality for ( a ). Let me rewrite it as:( a^2 - 3a - 4 leq 0 ).To solve ( a^2 - 3a - 4 leq 0 ), I can factor the quadratic or use the quadratic formula. Let me try factoring first.Looking for two numbers that multiply to -4 and add up to -3. Hmm, let's see. The factors of -4 could be 1 and -4, or -1 and 4. Let me check:1 * (-4) = -4, and 1 + (-4) = -3. Perfect.So, the quadratic factors as:( (a - 4)(a + 1) leq 0 ).So, ( (a - 4)(a + 1) leq 0 ).To find the solution to this inequality, I can analyze the sign of each factor.The critical points are ( a = -1 ) and ( a = 4 ). These divide the real number line into three intervals:1. ( (-infty, -1) )2. ( (-1, 4) )3. ( (4, infty) )I need to test each interval to see where the product ( (a - 4)(a + 1) ) is less than or equal to zero.Let's pick test points:1. For ( a < -1 ), say ( a = -2 ): ( (-2 - 4)(-2 + 1) = (-6)(-1) = 6 ), which is positive. So, this interval does not satisfy the inequality.2. For ( -1 < a < 4 ), say ( a = 0 ): ( (0 - 4)(0 + 1) = (-4)(1) = -4 ), which is negative. So, this interval satisfies the inequality.3. For ( a > 4 ), say ( a = 5 ): ( (5 - 4)(5 + 1) = (1)(6) = 6 ), which is positive. So, this interval does not satisfy the inequality.Now, we also need to check the critical points:- At ( a = -1 ): ( (-1 - 4)(-1 + 1) = (-5)(0) = 0 ), which satisfies the inequality.- At ( a = 4 ): ( (4 - 4)(4 + 1) = (0)(5) = 0 ), which also satisfies the inequality.Therefore, the solution to the inequality ( (a - 4)(a + 1) leq 0 ) is ( a in [-1, 4] ).So, the range of the real number ( a ) is from -1 to 4, inclusive.Let me double-check my work to make sure I didn't make any mistakes.1. Completing the square for ( x^2 - 2x + 5 ) correctly gave me ( (x - 1)^2 + 4 ). That seems right.2. Recognizing that the minimum value of ( (x - 1)^2 + 4 ) is 4 is correct because the square term is always non-negative.3. Setting ( a^2 - 3a leq 4 ) is the right step because the inequality must hold for all ( x ), so the right-hand side must be less than or equal to the minimum of the left-hand side.4. Factoring ( a^2 - 3a - 4 ) into ( (a - 4)(a + 1) ) is correct.5. Testing the intervals and critical points seems thorough, and the conclusion that ( a ) is between -1 and 4 is correct.So, I think my solution is solid. The range of ( a ) is ( [-1, 4] ), which corresponds to option A.**Final Answer**boxed{A}

answer：Alright, so I've got this problem here where Chinese characters represent digits from 1 to 9, and each character stands for a different digit. I know that "北" is 1 and "京" is 9. The equation given is:frac{text { 北 }}{text { 京 }} = frac{text { 奥运会 }}{text { 心想事成 }}First, I need to understand what this equation is saying. It looks like a proportion where the fraction of "北" over "京" is equal to the fraction of "奥运会" over "心想事成". Since "北" is 1 and "京" is 9, the left side of the equation simplifies to:frac{1}{9}So, the equation becomes:frac{1}{9} = frac{text { 奥运会 }}{text { 心想事成 }}This means that "奥运会" divided by "心想事成" should equal 1/9. In other words, "心想事成" should be 9 times "奥运会". Now, "奥运会" is a three-digit number, and "心想事成" is a four-digit number because when you multiply a three-digit number by 9, you can get a four-digit number. For example, 100 * 9 = 900, which is still three digits, but 112 * 9 = 1008, which is four digits. So, "奥运会" should be a three-digit number such that when multiplied by 9, it results in a four-digit number "心想事成".Given that all characters represent different digits from 1 to 9, and "北" is 1 and "京" is 9, we need to ensure that none of the digits in "奥运会" and "心想事成" repeat these digits or each other.Let me think about the possible three-digit numbers for "奥运会". Since "奥运会" multiplied by 9 gives "心想事成", and "心想事成" is a four-digit number, "奥运会" must be between 112 and 111 (wait, that doesn't make sense). Let me correct that.Actually, the smallest three-digit number is 100, and 100 * 9 = 900, which is still three digits. To get a four-digit number, "奥运会" must be at least 112 because 112 * 9 = 1008. So, "奥运会" should be between 112 and 1111/9, which is approximately 123.44. Wait, that can't be right either.Wait, no. If "奥运会" is a three-digit number, then "心想事成" is a four-digit number. So, "奥运会" must be such that when multiplied by 9, it becomes a four-digit number. Therefore, "奥运会" must be greater than or equal to 1000 / 9, which is approximately 111.11. So, "奥运会" must be at least 112.But since "奥运会" is a three-digit number, it can be from 112 up to 999. However, considering that "心想事成" is a four-digit number, "奥运会" must be such that 9 * "奥运会" is a four-digit number. So, "奥运会" must be between 112 and 1111 / 9, which is approximately 123.44. Wait, that still doesn't make sense because 123 * 9 = 1107, which is four digits. So, "奥运会" can be from 112 up to 1111 / 9, which is approximately 123.44. Therefore, "奥运会" must be between 112 and 123.Wait, that seems too restrictive. Let me think again. If "奥运会" is a three-digit number, and "心想事成" is a four-digit number, then "奥运会" must be such that 9 * "奥运会" is a four-digit number. So, "奥运会" must be greater than or equal to 1000 / 9, which is approximately 111.11. Therefore, "奥运会" must be at least 112.But 112 * 9 = 1008, which is four digits. So, "奥运会" can be from 112 up to 999, but since "心想事成" is a four-digit number, "奥运会" must be such that 9 * "奥运会" is less than 10000. So, "奥运会" must be less than 10000 / 9, which is approximately 1111.11. But since "奥运会" is a three-digit number, it can be up to 999.Wait, I'm getting confused. Let's clarify:- "奥运会" is a three-digit number: 100 ≤ "奥运会" ≤ 999- "心想事成" is a four-digit number: 1000 ≤ "心想事成" ≤ 9999- Given that "奥运会" * 9 = "心想事成"So, "奥运会" must satisfy:1000 ≤ 9 * "奥运会" ≤ 9999Therefore,1000 / 9 ≤ "奥运会" ≤ 9999 / 9Which is approximately:111.11 ≤ "奥运会" ≤ 1111But since "奥运会" is a three-digit number, it must be:112 ≤ "奥运会" ≤ 999But also, "奥运会" must be such that when multiplied by 9, it results in a four-digit number. So, "奥运会" must be at least 112, as 112 * 9 = 1008.Now, considering that "北" is 1 and "京" is 9, and all characters represent different digits, we need to ensure that the digits in "奥运会" and "心想事成" do not include 1 or 9, except for "北" and "京" which are already assigned.Wait, no. The problem states that "北" represents 1 and "京" represents 9, but it doesn't say that these are the only digits they represent. So, "奥运会" and "心想事成" can include other digits, but all digits from 1 to 9 must be used without repetition.Wait, actually, the problem says: "the Chinese characters represent the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9. Different characters represent different digits." So, each character is a unique digit from 1 to 9. So, "北" is 1, "京" is 9, and the other characters in "奥运会" and "心想事成" must be the remaining digits: 2,3,4,5,6,7,8.But "奥运会" is a three-digit number, so it uses three digits, and "心想事成" is a four-digit number, so it uses four digits. Together, they use 3 + 4 = 7 digits. Since we have 9 digits in total (1-9), and "北" and "京" are already assigned, that leaves 7 digits to be assigned to "奥运会" and "心想事成".So, the digits available for "奥运会" and "心想事成" are 2,3,4,5,6,7,8. Each must be used exactly once in "奥运会" and "心想事成".Therefore, "奥运会" is a three-digit number using three of these digits, and "心想事成" is a four-digit number using the remaining four digits.Given that "奥运会" * 9 = "心想事成", and all digits are unique and from 2,3,4,5,6,7,8.So, we need to find a three-digit number "奥运会" composed of digits from 2,3,4,5,6,7,8, with no repetition, such that when multiplied by 9, it results in a four-digit number "心想事成", which also uses the remaining four digits from 2,3,4,5,6,7,8, with no repetition.Moreover, "奥运会" must be such that 9 * "奥运会" is a four-digit number, so "奥运会" must be between 112 and 1111 / 9 ≈ 123.44. Wait, that can't be right because 112 * 9 = 1008, which is four digits, and 123 * 9 = 1107, which is still four digits. But 124 * 9 = 1116, which is also four digits. Wait, actually, 112 * 9 = 1008, and 111 * 9 = 999, which is three digits. So, "奥运会" must be at least 112 to get a four-digit "心想事成".But considering that "奥运会" is a three-digit number, and "心想事成" is a four-digit number, and all digits are unique and from 2,3,4,5,6,7,8, we need to find "奥运会" such that:- "奥运会" is a three-digit number using three unique digits from 2,3,4,5,6,7,8- 9 * "奥运会" = "心想事成", which is a four-digit number using the remaining four digits from 2,3,4,5,6,7,8- All digits in "奥运会" and "心想事成" are unique and do not include 1 or 9So, let's list possible "奥运会" numbers:Start with the smallest possible "奥运会" which is 112, but 112 uses digit 1, which is already used by "北". So, we need to find "奥运会" numbers starting from 234 upwards, using digits 2,3,4,5,6,7,8 without repetition.Let's try 234:234 * 9 = 2106But 2106 includes digit 1, which is already used by "北", so invalid.Next, 235:235 * 9 = 2115 → includes 1, invalid.236:236 * 9 = 2124 → includes 1, invalid.237:237 * 9 = 2133 → includes 1 and repeated 3, invalid.238:238 * 9 = 2142 → includes 1 and repeated 2, invalid.243:243 * 9 = 2187 → includes 1, invalid.245:245 * 9 = 2205 → includes 1 and repeated 2, invalid.246:246 * 9 = 2214 → includes 1 and repeated 2, invalid.247:247 * 9 = 2223 → repeated 2 and 3, invalid.248:248 * 9 = 2232 → repeated 2 and 3, invalid.253:253 * 9 = 2277 → repeated 2 and 7, invalid.254:254 * 9 = 2286 → repeated 2, invalid.256:256 * 9 = 2304 → includes 0, which is not allowed, invalid.257:257 * 9 = 2313 → includes 1 and repeated 3, invalid.258:258 * 9 = 2322 → repeated 2 and 3, invalid.263:263 * 9 = 2367 → digits: 2,3,6,7. Check if all are unique and not including 1 or 9."奥运会" = 263, digits: 2,6,3"心想事成" = 2367, digits: 2,3,6,7But "奥运会" uses 2,6,3, and "心想事成" uses 2,3,6,7. So, digits 2,3,6 are repeated, which is invalid.Next, 264:264 * 9 = 2376 → digits: 2,3,7,6"奥运会" = 264, digits: 2,6,4"心想事成" = 2376, digits: 2,3,7,6Again, digits 2,6 are repeated, invalid.265:265 * 9 = 2385 → includes 8 and 5"奥运会" = 265, digits: 2,6,5"心想事成" = 2385, digits: 2,3,8,5Digits 2 is repeated, invalid.267:267 * 9 = 2403 → includes 0, invalid.268:268 * 9 = 2412 → includes 1 and repeated 2, invalid.273:273 * 9 = 2457 → digits: 2,4,5,7"奥运会" = 273, digits: 2,7,3"心想事成" = 2457, digits: 2,4,5,7Digits 2 is repeated, invalid.274:274 * 9 = 2466 → repeated 6, invalid.275:275 * 9 = 2475 → digits: 2,4,7,5"奥运会" = 275, digits: 2,7,5"心想事成" = 2475, digits: 2,4,7,5Digits 2,7,5 are repeated, invalid.276:276 * 9 = 2484 → repeated 4, invalid.278:278 * 9 = 2502 → includes 0, invalid.283:283 * 9 = 2547 → digits: 2,5,4,7"奥运会" = 283, digits: 2,8,3"心想事成" = 2547, digits: 2,5,4,7Digits 2 is repeated, invalid.284:284 * 9 = 2556 → repeated 5, invalid.285:285 * 9 = 2565 → repeated 5 and 6, invalid.286:286 * 9 = 2574 → digits: 2,5,7,4"奥运会" = 286, digits: 2,8,6"心想事成" = 2574, digits: 2,5,7,4Digits 2 is repeated, invalid.287:287 * 9 = 2583 → digits: 2,5,8,3"奥运会" = 287, digits: 2,8,7"心想事成" = 2583, digits: 2,5,8,3Digits 2,8 are repeated, invalid.293:293 * 9 = 2637 → includes 6,3,7"奥运会" = 293, but 9 is already used by "京", invalid.Wait, 293 includes 9, which is already assigned to "京", so invalid.Similarly, any number with 9 is invalid.So, moving on to numbers starting with 3:324:324 * 9 = 2916 → includes 9 and 1, invalid.325:325 * 9 = 2925 → includes 9 and 1, invalid.326:326 * 9 = 2934 → includes 9 and 1, invalid.327:327 * 9 = 2943 → includes 9 and 1, invalid.328:328 * 9 = 2952 → includes 9 and 1, invalid.342:342 * 9 = 3078 → includes 0, invalid.345:345 * 9 = 3105 → includes 0 and 1, invalid.346:346 * 9 = 3114 → includes 1, invalid.347:347 * 9 = 3123 → includes 1, invalid.348:348 * 9 = 3132 → includes 1, invalid.352:352 * 9 = 3168 → includes 1, invalid.354:354 * 9 = 3186 → includes 1, invalid.356:356 * 9 = 3204 → includes 0, invalid.357:357 * 9 = 3213 → includes 1, invalid.358:358 * 9 = 3222 → repeated 2, invalid.362:362 * 9 = 3258 → digits: 3,2,5,8"奥运会" = 362, digits: 3,6,2"心想事成" = 3258, digits: 3,2,5,8Digits 3 and 2 are repeated, invalid.364:364 * 9 = 3276 → digits: 3,2,7,6"奥运会" = 364, digits: 3,6,4"心想事成" = 3276, digits: 3,2,7,6Digits 3,6 are repeated, invalid.365:365 * 9 = 3285 → digits: 3,2,8,5"奥运会" = 365, digits: 3,6,5"心想事成" = 3285, digits: 3,2,8,5Digits 3 is repeated, invalid.367:367 * 9 = 3303 → includes 0 and repeated 3, invalid.368:368 * 9 = 3312 → includes 1 and repeated 3, invalid.372:372 * 9 = 3348 → repeated 3 and 4, invalid.374:374 * 9 = 3366 → repeated 3 and 6, invalid.375:375 * 9 = 3375 → repeated 3 and 7, invalid.376:376 * 9 = 3384 → repeated 3 and 8, invalid.378:378 * 9 = 3402 → includes 0, invalid.382:382 * 9 = 3438 → repeated 3 and 4, invalid.384:384 * 9 = 3456 → digits: 3,4,5,6"奥运会" = 384, digits: 3,8,4"心想事成" = 3456, digits: 3,4,5,6Digits 3 and 4 are repeated, invalid.385:385 * 9 = 3465 → digits: 3,4,6,5"奥运会" = 385, digits: 3,8,5"心想事成" = 3465, digits: 3,4,6,5Digits 3 is repeated, invalid.386:386 * 9 = 3474 → repeated 4, invalid.387:387 * 9 = 3483 → repeated 3 and 8, invalid.392:392 * 9 = 3528 → includes 9, invalid.Wait, 392 includes 9, which is already used by "京", so invalid.Moving on to numbers starting with 4:423:423 * 9 = 3807 → includes 0, invalid.425:425 * 9 = 3825 → digits: 3,8,2,5"奥运会" = 425, digits: 4,2,5"心想事成" = 3825, digits: 3,8,2,5Digits 2 and 5 are repeated, invalid.426:426 * 9 = 3834 → repeated 3 and 8, invalid.427:427 * 9 = 3843 → repeated 3 and 8, invalid.428:428 * 9 = 3852 → digits: 3,8,5,2"奥运会" = 428, digits: 4,2,8"心想事成" = 3852, digits: 3,8,5,2Digits 2 and 8 are repeated, invalid.432:432 * 9 = 3888 → repeated 8, invalid.435:435 * 9 = 3915 → includes 9 and 1, invalid.436:436 * 9 = 3924 → includes 9, invalid.437:437 * 9 = 3933 → includes 9 and repeated 3, invalid.438:438 * 9 = 3942 → includes 9, invalid.452:452 * 9 = 4068 → includes 0, invalid.453:453 * 9 = 4077 → includes 0 and repeated 7, invalid.456:456 * 9 = 4104 → includes 0 and 1, invalid.457:457 * 9 = 4113 → includes 1, invalid.458:458 * 9 = 4122 → includes 1 and repeated 2, invalid.462:462 * 9 = 4158 → includes 1, invalid.463:463 * 9 = 4167 → includes 1, invalid.465:465 * 9 = 4185 → includes 1, invalid.467:467 * 9 = 4203 → includes 0, invalid.468:468 * 9 = 4212 → includes 1 and repeated 2, invalid.472:472 * 9 = 4248 → repeated 4, invalid.473:473 * 9 = 4257 → digits: 4,2,5,7"奥运会" = 473, digits: 4,7,3"心想事成" = 4257, digits: 4,2,5,7Digits 4 is repeated, invalid.475:475 * 9 = 4275 → digits: 4,2,7,5"奥运会" = 475, digits: 4,7,5"心想事成" = 4275, digits: 4,2,7,5Digits 4,7,5 are repeated, invalid.476:476 * 9 = 4284 → repeated 4 and 8, invalid.478:478 * 9 = 4302 → includes 0, invalid.482:482 * 9 = 4338 → repeated 3 and 8, invalid.483:483 * 9 = 4347 → repeated 4 and 3, invalid.485:485 * 9 = 4365 → digits: 4,3,6,5"奥运会" = 485, digits: 4,8,5"心想事成" = 4365, digits: 4,3,6,5Digits 4 is repeated, invalid.486:486 * 9 = 4374 → repeated 4 and 7, invalid.487:487 * 9 = 4383 → repeated 4 and 8, invalid.492:492 * 9 = 4428 → repeated 4 and 2, invalid.493:493 * 9 = 4437 → repeated 4 and 3, invalid.495:495 * 9 = 4455 → repeated 4 and 5, invalid.496:496 * 9 = 4464 → repeated 4 and 6, invalid.497:497 * 9 = 4473 → repeated 4 and 7, invalid.498:498 * 9 = 4482 → repeated 4 and 8, invalid.Now, moving on to numbers starting with 5:523:523 * 9 = 4707 → includes 0, invalid.524:524 * 9 = 4716 → includes 1, invalid.526:526 * 9 = 4734 → repeated 4 and 7, invalid.527:527 * 9 = 4743 → repeated 4 and 7, invalid.528:528 * 9 = 4752 → digits: 4,7,5,2"奥运会" = 528, digits: 5,2,8"心想事成" = 4752, digits: 4,7,5,2Digits 5 and 2 are repeated, invalid.532:532 * 9 = 4788 → repeated 8, invalid.534:534 * 9 = 4806 → includes 0, invalid.536:536 * 9 = 4824 → repeated 4 and 8, invalid.537:537 * 9 = 4833 → repeated 3 and 8, invalid.538:538 * 9 = 4842 → repeated 4 and 8, invalid.542:542 * 9 = 4878 → repeated 8, invalid.543:543 * 9 = 4887 → repeated 8, invalid.546:546 * 9 = 4914 → includes 9 and 1, invalid.547:547 * 9 = 4923 → includes 9, invalid.548:548 * 9 = 4932 → includes 9, invalid.562:562 * 9 = 5058 → includes 0, invalid.563:563 * 9 = 5067 → includes 0, invalid.564:564 * 9 = 5076 → includes 0, invalid.567:567 * 9 = 5103 → includes 0 and 1, invalid.568:568 * 9 = 5112 → includes 1, invalid.572:572 * 9 = 5148 → includes 1, invalid.573:573 * 9 = 5157 → includes 1 and repeated 5, invalid.574:574 * 9 = 5166 → includes 1 and repeated 6, invalid.576:576 * 9 = 5184 → includes 1, invalid.578:578 * 9 = 5202 → includes 0, invalid.582:582 * 9 = 5238 → digits: 5,2,3,8"奥运会" = 582, digits: 5,8,2"心想事成" = 5238, digits: 5,2,3,8Digits 5 and 2 are repeated, invalid.583:583 * 9 = 5247 → digits: 5,2,4,7"奥运会" = 583, digits: 5,8,3"心想事成" = 5247, digits: 5,2,4,7Digits 5 is repeated, invalid.584:584 * 9 = 5256 → repeated 5 and 2, invalid.586:586 * 9 = 5274 → digits: 5,2,7,4"奥运会" = 586, digits: 5,8,6"心想事成" = 5274, digits: 5,2,7,4Digits 5 and 2 are repeated, invalid.587:587 * 9 = 5283 → digits: 5,2,8,3"奥运会" = 587, digits: 5,8,7"心想事成" = 5283, digits: 5,2,8,3Digits 5,8 are repeated, invalid.592:592 * 9 = 5328 → includes 9, invalid.593:593 * 9 = 5337 → includes 9 and repeated 3, invalid.594:594 * 9 = 5346 → includes 9, invalid.596:596 * 9 = 5364 → includes 9, invalid.597:597 * 9 = 5373 → includes 9 and repeated 3, invalid.598:598 * 9 = 5382 → includes 9, invalid.Moving on to numbers starting with 6:623:623 * 9 = 5607 → includes 0, invalid.624:624 * 9 = 5616 → includes 1 and repeated 6, invalid.625:625 * 9 = 5625 → repeated 5 and 6, invalid.627:627 * 9 = 5643 → digits: 5,6,4,3"奥运会" = 627, digits: 6,2,7"心想事成" = 5643, digits: 5,6,4,3Digits 6 is repeated, invalid.628:628 * 9 = 5652 → repeated 5 and 6, invalid.632:632 * 9 = 5688 → repeated 8, invalid.634:634 * 9 = 5706 → includes 0, invalid.635:635 * 9 = 5715 → includes 1 and repeated 5, invalid.637:637 * 9 = 5733 → repeated 3 and 5, invalid.638:638 * 9 = 5742 → digits: 5,7,4,2"奥运会" = 638, digits: 6,3,8"心想事成" = 5742, digits: 5,7,4,2Digits 5,7,4,2 are unique and not used in "奥运会", so this is a possible solution.Let me check:"奥运会" = 638, digits: 6,3,8"心想事成" = 5742, digits: 5,7,4,2All digits are unique and from 2,3,4,5,6,7,8. No repetition, and none include 1 or 9.So, this is a valid solution.Next, 639:639 * 9 = 5751 → includes 1 and 5, invalid.642:642 * 9 = 5778 → repeated 7, invalid.643:643 * 9 = 5787 → repeated 7 and 8, invalid.645:645 * 9 = 5805 → includes 0, invalid.647:647 * 9 = 5823 → digits: 5,8,2,3"奥运会" = 647, digits: 6,4,7"心想事成" = 5823, digits: 5,8,2,3Digits 5,8,2,3 are unique and not used in "奥运会", so this is another valid solution.Check:"奥运会" = 647, digits: 6,4,7"心想事成" = 5823, digits: 5,8,2,3All digits unique, no repetition, no 1 or 9.Next, 648:648 * 9 = 5832 → digits: 5,8,3,2"奥运会" = 648, digits: 6,4,8"心想事成" = 5832, digits: 5,8,3,2Digits 8 is repeated, invalid.652:652 * 9 = 5868 → repeated 8 and 6, invalid.653:653 * 9 = 5877 → repeated 7, invalid.654:654 * 9 = 5886 → repeated 8 and 6, invalid.657:657 * 9 = 5913 → includes 9 and 1, invalid.658:658 * 9 = 5922 → includes 9 and repeated 2, invalid.672:672 * 9 = 6048 → includes 0, invalid.673:673 * 9 = 6057 → includes 0, invalid.674:674 * 9 = 6066 → includes 0 and repeated 6, invalid.675:675 * 9 = 6075 → includes 0, invalid.678:678 * 9 = 6102 → includes 0 and 1, invalid.682:682 * 9 = 6138 → includes 1, invalid.683:683 * 9 = 6147 → includes 1, invalid.684:684 * 9 = 6156 → includes 1 and repeated 6, invalid.685:685 * 9 = 6165 → includes 1 and repeated 6, invalid.687:687 * 9 = 6183 → includes 1, invalid.689:689 * 9 = 6201 → includes 0 and 1, invalid.692:692 * 9 = 6228 → includes 9 and repeated 2, invalid.693:693 * 9 = 6237 → includes 9, invalid.694:694 * 9 = 6246 → includes 9 and repeated 6, invalid.695:695 * 9 = 6255 → includes 9 and repeated 5, invalid.697:697 * 9 = 6273 → includes 9, invalid.698:698 * 9 = 6282 → includes 9 and repeated 2, invalid.Moving on to numbers starting with 7:723:723 * 9 = 6507 → includes 0, invalid.724:724 * 9 = 6516 → includes 1 and repeated 6, invalid.725:725 * 9 = 6525 → repeated 5 and 6, invalid.726:726 * 9 = 6534 → digits: 6,5,3,4"奥运会" = 726, digits: 7,2,6"心想事成" = 6534, digits: 6,5,3,4Digits 6 is repeated, invalid.728:728 * 9 = 6552 → repeated 5 and 6, invalid.732:732 * 9 = 6588 → repeated 8, invalid.734:734 * 9 = 6606 → includes 0 and repeated 6, invalid.735:735 * 9 = 6615 → includes 1 and repeated 6, invalid.736:736 * 9 = 6624 → repeated 6 and 2, invalid.738:738 * 9 = 6642 → repeated 6 and 4, invalid.742:742 * 9 = 6678 → repeated 6 and 7, invalid.743:743 * 9 = 6687 → repeated 6 and 8, invalid.745:745 * 9 = 6705 → includes 0, invalid.746:746 * 9 = 6714 → includes 1, invalid.748:748 * 9 = 6732 → digits: 6,7,3,2"奥运会" = 748, digits: 7,4,8"心想事成" = 6732, digits: 6,7,3,2Digits 7 is repeated, invalid.752:752 * 9 = 6768 → repeated 6 and 7, invalid.753:753 * 9 = 6777 → repeated 7, invalid.754:754 * 9 = 6786 → repeated 6 and 7, invalid.756:756 * 9 = 6804 → includes 0, invalid.758:758 * 9 = 6822 → repeated 2, invalid.762:762 * 9 = 6858 → repeated 8 and 6, invalid.763:763 * 9 = 6867 → repeated 6 and 7, invalid.764:764 * 9 = 6876 → repeated 6 and 7, invalid.765:765 * 9 = 6885 → repeated 8 and 6, invalid.768:768 * 9 = 6912 → includes 9 and 1, invalid.782:782 * 9 = 7038 → includes 0, invalid.783:783 * 9 = 7047 → includes 0, invalid.784:784 * 9 = 7056 → includes 0, invalid.785:785 * 9 = 7065 → includes 0, invalid.786:786 * 9 = 7074 → includes 0, invalid.789:789 * 9 = 7101 → includes 0 and 1, invalid.792:792 * 9 = 7128 → includes 9, invalid.793:793 * 9 = 7137 → includes 9 and repeated 7, invalid.794:794 * 9 = 7146 → includes 9, invalid.795:795 * 9 = 7155 → includes 9 and repeated 5, invalid.796:796 * 9 = 7164 → includes 9, invalid.798:798 * 9 = 7182 → includes 9, invalid.Moving on to numbers starting with 8:823:823 * 9 = 7407 → includes 0, invalid.824:824 * 9 = 7416 → includes 1, invalid.825:825 * 9 = 7425 → digits: 7,4,2,5"奥运会" = 825, digits: 8,2,5"心想事成" = 7425, digits: 7,4,2,5Digits 2 and 5 are repeated, invalid.826:826 * 9 = 7434 → repeated 4 and 7, invalid.827:827 * 9 = 7443 → repeated 4 and 7, invalid.829:829 * 9 = 7461 → includes 1, invalid.832:832 * 9 = 7488 → repeated 8, invalid.834:834 * 9 = 7506 → includes 0, invalid.835:835 * 9 = 7515 → includes 1 and repeated 5, invalid.836:836 * 9 = 7524 → digits: 7,5,2,4"奥运会" = 836, digits: 8,3,6"心想事成" = 7524, digits: 7,5,2,4Digits 7,5,2,4 are unique and not used in "奥运会", so this is a valid solution.Check:"奥运会" = 836, digits: 8,3,6"心想事成" = 7524, digits: 7,5,2,4All digits unique, no repetition, no 1 or 9.Next, 837:837 * 9 = 7533 → repeated 3 and 7, invalid.839:839 * 9 = 7551 → includes 1 and repeated 5, invalid.842:842 * 9 = 7578 → repeated 7 and 8, invalid.843:843 * 9 = 7587 → repeated 7 and 8, invalid.845:845 * 9 = 7605 → includes 0, invalid.846:846 * 9 = 7614 → includes 1, invalid.847:847 * 9 = 7623 → digits: 7,6,2,3"奥运会" = 847, digits: 8,4,7"心想事成" = 7623, digits: 7,6,2,3Digits 7 is repeated, invalid.849:849 * 9 = 7641 → includes 1, invalid.852:852 * 9 = 7668 → repeated 6 and 7, invalid.853:853 * 9 = 7677 → repeated 7, invalid.854:854 * 9 = 7686 → repeated 6 and 7, invalid.856:856 * 9 = 7704 → includes 0, invalid.857:857 * 9 = 7713 → includes 1 and repeated 7, invalid.859:859 * 9 = 7731 → includes 1 and repeated 7, invalid.862:862 * 9 = 7758 → repeated 7 and 8, invalid.863:863 * 9 = 7767 → repeated 7 and 6, invalid.864:864 * 9 = 7776 → repeated 7 and 6, invalid.865:865 * 9 = 7785 → repeated 7 and 8, invalid.867:867 * 9 = 7803 → includes 0, invalid.869:869 * 9 = 7821 → includes 1, invalid.872:872 * 9 = 7848 → repeated 8 and 7, invalid.873:873 * 9 = 7857 → repeated 7 and 8, invalid.874:874 * 9 = 7866 → repeated 6 and 7, invalid.875:875 * 9 = 7875 → repeated 7 and 8, invalid.876:876 * 9 = 7884 → repeated 8 and 7, invalid.879:879 * 9 = 7911 → includes 1 and 9, invalid.892:892 * 9 = 8028 → includes 0 and repeated 8, invalid.893:893 * 9 = 8037 → includes 0, invalid.894:894 * 9 = 8046 → includes 0, invalid.895:895 * 9 = 8055 → includes 0 and repeated 5, invalid.896:896 * 9 = 8064 → includes 0, invalid.897:897 * 9 = 8073 → includes 0, invalid.898:898 * 9 = 8082 → includes 0 and repeated 8, invalid.899:899 * 9 = 8091 → includes 0 and 9, invalid.Finally, numbers starting with 9 are invalid because "京" is already 9.So, after checking all possible three-digit numbers, the valid "奥运会" numbers are:- 638, with "心想事成" = 5742- 647, with "心想事成" = 5823- 836, with "心想事成" = 7524These are the only three-digit numbers that satisfy the given conditions.

answer：First, I'll define variables to represent the number of each type of bill. Let ( x ) be the number of 20 bills.Since the theater received twice as many 10 bills as 20 bills, the number of 10 bills is ( 2x ).Additionally, there were 20 more 5 bills than 10 bills, so the number of 5 bills is ( 2x + 20 ).Next, I'll calculate the total amount of money collected from each type of bill:- From 20 bills: ( 20x ) dollars- From 10 bills: ( 10 times 2x = 20x ) dollars- From 5 bills: ( 5 times (2x + 20) = 10x + 100 ) dollarsThe total amount collected is the sum of these amounts:[20x + 20x + 10x + 100 = 50x + 100]Since the theater sold 300 tickets at 40 each, the total revenue is:[300 times 40 = 12,000 text{ dollars}]Setting up the equation:[50x + 100 = 12,000]Solving for ( x ):[50x = 12,000 - 100][50x = 11,900][x = frac{11,900}{50} = 238]Now, I'll find the number of each type of bill:- Number of 20 bills: ( x = 238 )- Number of 10 bills: ( 2x = 2 times 238 = 476 )- Number of 5 bills: ( 2x + 20 = 2 times 238 + 20 = 476 + 20 = 496 )Finally, the total number of bills is:[238 + 476 + 496 = 1,210]

answer：Okay, so I have this quadratic equation: ( x^2 + 2x - 5 = 0 ), and I'm told that ( m ) and ( n ) are the roots of this equation. I need to find the value of ( m^2 + mn + 3m + n ). Hmm, let's see how to approach this.First, I remember that for any quadratic equation of the form ( ax^2 + bx + c = 0 ), the sum of the roots ( m + n ) is equal to ( -b/a ), and the product of the roots ( mn ) is equal to ( c/a ). In this case, ( a = 1 ), ( b = 2 ), and ( c = -5 ). So, let me calculate the sum and product of the roots.Sum of roots: ( m + n = -b/a = -2/1 = -2 ).Product of roots: ( mn = c/a = -5/1 = -5 ).Alright, so I have ( m + n = -2 ) and ( mn = -5 ). Now, I need to find ( m^2 + mn + 3m + n ). Hmm, that expression looks a bit complicated, but maybe I can break it down using what I know about ( m ) and ( n ).Let me write down the expression again: ( m^2 + mn + 3m + n ). I notice that ( m^2 ) is part of the quadratic equation itself. Since ( m ) is a root of the equation ( x^2 + 2x - 5 = 0 ), substituting ( m ) into the equation gives ( m^2 + 2m - 5 = 0 ). So, I can solve for ( m^2 ) here.From ( m^2 + 2m - 5 = 0 ), I can rearrange it to get ( m^2 = -2m + 5 ). That's helpful because I can substitute this back into the original expression.So, substituting ( m^2 = -2m + 5 ) into ( m^2 + mn + 3m + n ), I get:( (-2m + 5) + mn + 3m + n ).Now, let's simplify this expression step by step. First, combine like terms:- The terms with ( m ): ( -2m + 3m = m ).- The constant term: ( 5 ).- The terms with ( mn ) and ( n ): ( mn + n ).So, the expression simplifies to:( m + 5 + mn + n ).Now, I can rearrange this as:( (m + n) + mn + 5 ).Wait, that's interesting because I already know ( m + n ) and ( mn ) from earlier. Let me substitute those values in.( (m + n) + mn + 5 = (-2) + (-5) + 5 ).Calculating that:( -2 - 5 + 5 = (-7) + 5 = -2 ).So, the value of ( m^2 + mn + 3m + n ) is ( -2 ).Let me double-check my steps to make sure I didn't make any mistakes. I used the sum and product of roots correctly, substituted ( m^2 ) from the quadratic equation, and then simplified the expression step by step. It all seems to add up correctly. I think my answer is right.**Final Answer**boxed{B}

answer：Okay, so I have this problem where a random variable X follows a normal distribution with mean 1 and variance σ², which is written as N(1, σ²). The problem says that the probability of |X - 1| being less than 1 is 2/3, and I need to find the probability that X is greater than or equal to 0. Hmm, let me try to break this down step by step.First, I know that when dealing with normal distributions, it's often helpful to standardize the variable to use the standard normal distribution table or Z-table. The standardization formula is Z = (X - μ)/σ, where μ is the mean and σ is the standard deviation. In this case, μ is 1, so Z = (X - 1)/σ.The problem gives me P(|X - 1| < 1) = 2/3. Let me rewrite that inequality without the absolute value to understand it better. So, |X - 1| < 1 means that X - 1 is between -1 and 1. Therefore, X is between 0 and 2. So, P(0 < X < 2) = 2/3.Now, since X is normally distributed with mean 1, the distribution is symmetric around 1. That means the probability that X is less than 1 is 0.5, and the probability that X is greater than 1 is also 0.5. So, the total probability from 0 to 2 is 2/3, which is more than half. That suggests that the tails on either side of this interval must each have a probability of (1 - 2/3)/2 = 1/6. So, P(X < 0) = 1/6 and P(X > 2) = 1/6.Wait, let me make sure I'm thinking about this correctly. The total probability outside the interval (0, 2) is 1 - 2/3 = 1/3. Since the distribution is symmetric around 1, the probabilities on both sides of the interval should be equal. So, each tail beyond 0 and beyond 2 should have a probability of 1/6. Therefore, P(X < 0) = 1/6 and P(X > 2) = 1/6.Now, the question asks for P(X ≥ 0). That would be the probability that X is greater than or equal to 0. Since the total probability is 1, P(X ≥ 0) = 1 - P(X < 0). We already found that P(X < 0) = 1/6, so P(X ≥ 0) = 1 - 1/6 = 5/6.Let me double-check this reasoning. If the probability between 0 and 2 is 2/3, and the distribution is symmetric around 1, then the areas to the left of 0 and to the right of 2 should each be (1 - 2/3)/2 = 1/6. So, yes, P(X < 0) = 1/6, and thus P(X ≥ 0) = 5/6.I think that makes sense. But just to be thorough, maybe I should also consider standardizing the variable and using Z-scores to confirm.So, let's standardize X. Let Z = (X - 1)/σ. Then, P(0 < X < 2) = P((0 - 1)/σ < Z < (2 - 1)/σ) = P(-1/σ < Z < 1/σ). We know this probability is 2/3.Looking up in the standard normal distribution table, the Z-scores that correspond to the middle 2/3 of the distribution. The total area in the tails is 1 - 2/3 = 1/3, so each tail has 1/6. The Z-score that leaves 1/6 in the upper tail is approximately 1.09 (since Φ(1.09) ≈ 0.86, which is 1 - 0.14, but wait, maybe I need to be more precise here).Wait, actually, if the area to the left of Z is 1 - 1/6 = 5/6, then the Z-score corresponding to 5/6 is approximately 0.84. Hmm, that doesn't seem right because 0.84 corresponds to about 0.7995, which is close to 0.8. Maybe I need to use a more accurate method.Alternatively, using the inverse of the standard normal distribution, if the area between -Z and Z is 2/3, then the area in each tail is 1/6. So, the Z-score such that Φ(Z) = 1 - 1/6 = 5/6. Looking up 5/6 in the Z-table, which is approximately 0.8416. So, Z ≈ 0.8416.Therefore, 1/σ = 0.8416, so σ = 1 / 0.8416 ≈ 1.188.But wait, do I actually need the value of σ to find P(X ≥ 0)? Because in my initial reasoning, I didn't use σ, but just used the symmetry of the distribution. Maybe I don't need σ after all.Let me think again. Since the distribution is symmetric around 1, and the interval (0, 2) is symmetric around 1, the probabilities outside this interval on both sides are equal. Therefore, P(X < 0) = P(X > 2) = 1/6. So, P(X ≥ 0) = 1 - P(X < 0) = 1 - 1/6 = 5/6.Yes, that seems consistent. So, even without calculating σ, I can use the symmetry of the normal distribution to find the required probability.Just to make sure, let's consider the standardized variable again. If Z = (X - 1)/σ, then P(X < 0) = P(Z < (0 - 1)/σ) = P(Z < -1/σ). Since we found that P(Z < -1/σ) = 1/6, and from the standard normal table, P(Z < -0.8416) ≈ 1/6. Therefore, -1/σ ≈ -0.8416, so σ ≈ 1.188.But again, do I need σ? It seems that I don't because the question only asks for P(X ≥ 0), which we can find using the symmetry and the given probability.So, to summarize, since P(0 < X < 2) = 2/3, and the distribution is symmetric around 1, the probabilities outside this interval on both sides are equal, each being 1/6. Therefore, P(X < 0) = 1/6, and thus P(X ≥ 0) = 1 - 1/6 = 5/6.I think that's solid. I don't see any mistakes in this reasoning. The key was recognizing the symmetry of the normal distribution around the mean and using the given probability to find the tail probabilities.

answer：Okay, so I have this problem where there are 10 plus signs and 15 minus signs on a board. I can erase any two signs and write a plus sign if they are the same, and a minus sign if they are different. I need to figure out what sign will remain after performing 24 such operations.Alright, let's break this down step by step. First, I need to understand the rules clearly. If I erase two plus signs, I write a plus sign. If I erase two minus signs, I also write a plus sign. But if I erase one plus and one minus, I write a minus sign. So, in essence, erasing two identical signs gives me a plus, and erasing two different signs gives me a minus.Now, I have 10 plus signs and 15 minus signs. That's a total of 25 signs. Each operation reduces the total number of signs by one because I erase two and write one. So, after 24 operations, there should be 25 - 24 = 1 sign left. So, the question is, will that last sign be a plus or a minus?Hmm, okay, so I need to figure out the parity or something related to the number of plus and minus signs. Let's think about the number of minus signs. Initially, there are 15 minus signs, which is an odd number. Now, let's see how the operations affect the number of minus signs.If I erase two minus signs, I write a plus sign. So, that reduces the number of minus signs by two. If I erase two plus signs, I write a plus sign, which doesn't affect the number of minus signs. If I erase one plus and one minus, I write a minus sign. So, that operation removes one minus and one plus, but then adds a minus, so the number of minus signs remains the same.Wait, so erasing two minus signs reduces the number of minus signs by two, which doesn't change the parity (since subtracting two from an odd number keeps it odd). Erasing two plus signs doesn't affect the minus signs. Erasing one plus and one minus doesn't change the number of minus signs either because you remove one and add one.So, the number of minus signs remains odd throughout all operations. Since we start with 15 minus signs, which is odd, and each operation preserves the parity, the number of minus signs will always stay odd. Therefore, when we're left with just one sign, it has to be a minus sign because the number of minus signs is odd.Let me double-check that. If I have an odd number of minus signs and each operation either keeps the number of minus signs the same or reduces it by two, which preserves the oddness, then yes, the last sign must be a minus.So, after 24 operations, there will be one minus sign left on the board.