数字 1 在 1 到 500 之间出现了多少次？
数字系统是一种数学符号，用于计数和计算对象，以及执行算术计算。它是一种表示数字的书写系统。它给出了每个数字的特殊描述，并构成了数字的算术和代数形式。它允许我们进行算术运算，例如加法、减法、乘法和除法。
等式是用“=”符号连接两个具有相同值的代数表达式的声明。例如：在等式 8x + 4 = 7 中，8x + 4 是左侧表达式，7 是与“=”符号连接的右侧表达式。
什么是数字？
指定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。数字是由数字混合生成的值。这些数字用于表示代数数。数字是一组 10 个数字的指示，范围从 0、1、2、3、4、5、6、7、8 和 9。整数的任意组合表示一个数字。 Number 的大小取决于用于其增长的位数。例如：136、198、0.245、-16、98、96 等。
数字类型
数字有多种类型，具体取决于用于其发展的数字模式。数字中还放入了各种字符和规则，将它们分类为不同的类型，
整数
整数是一组整数加上自然数的负值。整数不包括分数，即它们不能写成 a/b 形式。 Integers 的范围是从负端的无穷大和正端的无穷大，包括零。整数用符号 Z 表示。整数是小数部分为 0 的数字，如 -5、-4、1、0、20、200。
自然数
自然数是范围从 1 到无穷大的数字。这些数字也被描述为正数或计数数。我们也可以用符号 N 表示自然数。所有大于 0 的整数都是自然数，可数的数如 5、6、7、8、9、10。
整数
整数类似于自然数，但它们也包括“零”。整数也可以用符号 W 表示。整数包括所有自然数和 0（零）。
素数和合数
所有只有两个确定成分的数字，即数字本身和 1，称为素数。除 0 外，所有非质数的数都称为合数。零既不是质数也不是合数。一些素数是 3、5、7、57、51、67 和 391。所有大于 1 的数都是合数。一些合数是 7、5、3、17、15 和 200。
分数
分数是对应于 a/b 形状的整数，其中 a 表示整数，b 表示自然数，即 b 永远不能为 0。分数的上半部分，即 a 被描述为分子，而下半部分即b部分称为分母。示例：-1/5、0.25、2/5、18/4、...
有理数
有理数是可以以分数形式显示的数字，即 a/b。这里，a和b都是数字，b不等于0。所有的分数都是有理数，但不是所有的有理数都是分数。示例：-2/5、0.54、1/5、13/4、...
无理数
无理数是不能以分数形式显示的数字，即它们不能写成 a/b。示例：√2、√3、√.434343、π…
实数和虚数
实数是可以以十进制形式显示的数字。这些数字涉及整数，整数，分数等。所有数字都属于实数，但所有实数都不属于整数。虚数是所有不是实数的数字。这些数字平方后将显示负数。 √-1 表示为 i。这些数字也称为复数。示例：√-2、√-5、…
数字 1 在 1 到 500 之间出现了多少次？
解决方案：
First we will take the digit in range of 10:
From the range of 0 to 10, the number 1 appears 2 time.
Now we calculate the integers in range of 100
From the range of digits 0 to 99, the number 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)
Check out that in digit 11, 1 seem two times.
Now from numbers 100 to 199 the digit 1 seem 120 times.
In this case the numbers at hundreds place is 1. Therefore there are 120 1’s from 100 to 199.
Explanation- we have to calculate 1’s from 100, 101, 102……….. 199.
At tens place 1 will come 10 times. At hundreds place of these 3 digit numbers 1 will come 100 times.
That is 110, 111, 112, 113, 114…… 119.
At units place 1 will appear 10 times(101, 111, 121, 131……. 191)
So add these:
100 + 10 + 10 = 120 so 1 comes 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 number 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
By adding all of these we get:
Total number of times = (20).(4) + 120
= 80 + 120
= 200
Therefore when we calculate digits from 1 to 500 the digit 1 appear 200 times.
Note: Don’t forget to check the number 1 twice in digits like 11, and also do not forget to check the number 1 in hundredth’s place in the digits from 100 to 199.
类似问题
问题 1：数字 2 在 1 到 500 之间会出现多少次？
解决方案：
First we will take the digit in range of 10:
From the range of 0 to 10, the number 2 appears 1 time.
Now we calculate the integers in range of 100
From the range of digits 0 to 99, the number 2 appears 20 times.
(They are: 2, 12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29 32, 42, 52, 62, 72, 82, 92)
Check out that in digit 22, 2 seem two times.
Now from 100 to 199 we have 102, 112, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 132, 142, 152, 162, 172,
182, 192, the number 2 appear 20 times.
Now from numbers 200 to 299 the digit 2 seem 120 times.
In this case the numbers at hundreds place is 2. Therefore there are 120 2’s from 200 to 299.
Explanation- we have to calculate 2’s from 200, 201, 202……….. 299.
At tens place 2 will come 10 times. At hundreds place of these 3 digit numbers 2 will come 100 times.
That is 220, 221, 222, 223, 224…… 229.
At units place 2 will appear 10 times(202, 212, 222, 232……. 292)
So add these:
100 + 10 + 10 = 120 so 2 comes 120 times.
Similarly,
From 300 to 399 we have 302,…312,….392. Again 20 times
From 400 to 499 we have 402,…412,…492. Again 20 times
By adding all of these we get:
Total number of times = (20).(4) + 120
= 80 + 120
= 200
Therefore when we calculate digits from 1 to 500 the digit 2 appear 200 times.
Note: Don’t forget to check the number 2 twice in digits like 22, and also do not forget to check the
number 2 in hundredth’s place in the digits from 200 to 299.
问题 2：数字 3 在 1 到 500 之间会出现多少次？
解决方案：
First we will take the digit in range of 10:
From the range of 0 to 10, the number 3 appears 1 time.
Now we calculate the integers in range of 100
From the range of digits 0 to 99, the number 3 appears 20 times.
(They are: 3, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 43, 53, 63, 73, 83, 93)
Check out that in digit 33, 3 seem two times.
From 100 to 199 we have 103, 113, 123, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 143, 153, 163, 173, 183,
193 Again 20 times
From 200 to 299 we have 203, 213, 223, 330, 231, 232, 233, 234, 235, 236, 237, 238, 239, 243, 253, 263, 273, 283,
293 Again 20 times
Now from numbers 300 to 399 the digit 3 seem 120 times.
In this case the numbers at hundreds place is 3. Therefore there are 120 3’s from 300 to 399.
Explanation- we have to calculate 3’s from 300, 301, 302……….. 399.
At tens place 3 will come 10 times. At hundreds place of these 3 digit numbers 3 will come 100 times.
That is 330, 331, 332, 333, 334…… 339.
At units place 3 will appear 10 times(302, 312, 322, 332……. 393)
So add these:
100 + 10 + 10 = 120 so 3 comes 120 times.
Similarly,
From 400 to 499 we have 403,…413,…493. Again 20 times
By adding all of these we get:
Total number of times = (20).(4) + 120
= 80 + 120
= 200
Therefore when we calculate digits from 1 to 500 the digit 3 appear 200 times.
Note: Don’t forget to check the number 3 twice in digits like 33, and also do not forget to check the
number 3 in hundredth’s place in the digits from 300 to 399.