数字 1 在数字 1 到 1,000 中出现了多少次？
Number Systems 是用于计数和测量对象以及执行算术计算的值。它是一种表达数字的书写方法。它为每个数字提供唯一的表示，并呈现数字的算术和代数形式。它使我们能够进行算术运算，例如加法、减法、乘法和除法。
等式是用“=”符号链接两个相等值的代数表达式的声明。例如：在方程 8x + 4 = 7 中，8x + 4 是左侧表达式，7 是与“=”符号链接的右侧表达式。
什么是数字？
表示数量的单词或符号称为数字。数字 4、6、8 等是偶数，而 3、5、7 等是奇数。数字是由数字组合形成的值。这些数字用于表示代数量。整数是一组 10 个字符的指示，范围从 0、1、2、3、4、5、6、7、8 和 9。整数的任何合并表示一个数字。数字的大小取决于用于其开发的位数。例如：126、128、0.356、-12、78、94 等。
整数
整数与自然数相同，但也包括“零”。我们可以通过符号 W 捐赠整数。整数包括所有自然数和 0（零）。
我们知道数字 1、2、3、4、5 是自然数。数字 0、1、2、3、4、5、6 .. 等都是整数，因为它们也包括 0。整数表示为 W，自然数表示为 N。因此，可以这样说，
W= {N} + 0
在数轴上，0 右侧的所有内容（包括 0）都在整数之下。
数字 1 在数字 1 到 1,000 中出现了多少次？
解决方案：
Hint: The easy method is to write the numbers in the range of 10, 100 and then count the number of 1’s in them
Steps to find out the solution:
First we will take the numbers in range of 10:
From the range of numbers 0 to 10, the digit 1 appears 2 time.
Now we add the numbers in range of 100
From the range of numbers 0 to 99, the digit 1 appears 20 times.
(They are: 1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91)
Observe that in number 11, 1 appears two times.
Now from 100 to 199 the digit 1 appears 120 times.
In this case the digit at hundreds place is 1. Therefore there are 120 1’s from 100 to 199.
Explanation- we have to find 1’s from 100, 101, 102……….. 199. At hundreds place of these 3 digit numbers 1 will come 100 times. At tens place 1 will come 10 times
That is in cases of 110, 111, 112, 113, 114…… 119. At units place 1 will come 10 times(101, 111, 121, 131……. 191)
So add these:
100 + 10 + 10 = 120 so 1 appears 120 times.
Now from 200 to 299 we have 201, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 221, 231, 241, 251, 261, 271, 281, 291 the digit 1 appear 20 times.
Similarly,
From 300 to 399 we have 301,…311,….391. Again 20 times
From 400 to 499 we have 401,…411,…491. Again 20 times
From 500 to 599 we have 501,…511,…591. Again 20 times.
From 600 to 699 we have 601,…611,…691. Again 20 times
From 700 to 799 we have 701,…711,…791. Again 20 times
From 800 to 899 we have 801,…811,…891. Again 20 times
From 900 to 999 we have 901,…911,…991. Again 20 times
and then we have 1000 in which digit 1 appears 1 time
By adding all of the above, we get:
Total number of times = (20).(9) + 120 + 1
= 180 + 120 + 1
= 301
Therefore when we list numbers from 1 to 1000 the digit 1 is written 301 times.
Note: Don’t forget to count the digit 1 twice in numbers like 11, and also do not forget to count the digit 1 in hundredth’s place in the numbers from 100 to 199 and also in the thousand place number like 1000.
类似问题
问题 1：列出 1 到 1000 的数字时，数字 5 出现了多少次。
回答：
Steps to find the solution:
First, we will take the numbers in the range of 10:
From the range of numbers 0 to 10, the digit 5 appears 1 time.
Now we add the numbers in the range of 100
From the range of numbers 0 to 99, the digit 5 appears 20 times.
(They are: 5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 65, 75, 85, 95)
Observe that in number 55, 5 appears two times.
Now from 100 to 199 we have: 105, 115, 125, 135, 145, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 165, 175, 185, 195. The digit 5 appears 20 times.
Similarly,
From 200 to 299 we have 205,215…….245, 250,…255,…..,295. Again 20 times.
From 300 to 399 we have 305,…355,…,395. Again 20 times
From 400 to 499 we have 405,…455,…,495. Again 20 times
From 600 to 699 we have 605,…655,…,695. Again 20 times
From 700 to 799 we have 705,…755,…,795. Again 20 times
From 800 to 899 we have 805,…855,…,895. Again 20 times
From 900 to 999 we have 905,…955,…,995. Again 20 times
Now the remaining range is 500 to 599:
Now from 500 to 599, the digit 5 appears 120 times.
In this case, the digit at hundred’s places is 5. Therefore there are 120 5’s from 500 to 599.
Explanation: We have to find 5’s from 500, 501, 502……….. 599. At hundred’s place of these 3 digit numbers, 5 will come 100 times. At tens place, 5 will come 10 times
That is in cases of 550, 551, 552, 553, 554…… 559. At units place 5 will come 10 times(505, 555, 525, 535……. 595)
So add these:
100 + 10 + 10 = 120 so 5 appear 120 time’s.
By adding all above, we get:
Total number of times = (20).(9) + 120
= 180 + 120
= 300
Therefore when we list numbers from 1 to 1000 the digit 5 is written 300 times.
Note: Don’t forget to count the digit 5 twice in numbers like 55, and also do not forget to count the digit 5 in hundredth’s place in the numbers from 500 to 599.
问题 2：数字 2 在 1 到 100 的数字中出现了多少次？
回答：
Complete step-by-step solution:
We have to find the number of times the digit 2 appears in numbers from 1 to 100.
From 1 to 10, the digit 2 appears only once for 2.
From 11 to 20, the digit 2 appears two times for 12 to 20.
From 21 to 30, the digit 2 appears in 21, 22, 23, 24, 25, 26, 27, 28, 29 and in 22 it
appears twice. So, it appears ten times.
From 31 to 40, the digit 2 appears only once for 32.
From 41 to 50, the digit 2 appears only once for 42.
From 51 to 60, the digit 2 appears only once for 52.
From 61 to 70, the digit 2 appears only once for 62.
From 71 to 80, the digit 2 appears only once for 72.
From 81 to 90, the digit 2 appears only once for 82.
From 91 to 100, the digit 2 appears only once for 92.
So, the total number of times of repetition of 2 is 1+2+10+1+1+1+1+1+1+1 = 20.
Hence, the digit 2 appears in numbers from 1 to 100 for 20 times.