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How many numbers from 10 to 200 have exactly two identical digits. How is the math done?
Wiki User
∙ 8y ago
There are 63 of them.
Between 10 and 99 all with repeated 2 digits are solutions, there are 1..9 repeated, ie 9 of them.
Between 100 and 199 the repeated digits can be the first and second (11x), the first and third (1x1), or the second and third digit (1yy). For the first two cases, the missing digit (x) cannot be the same as the first digit(1), which means there are 9 of them in each case since there are 10 digits {0, 1, ..., 9} less the one digit that is the first digit; for the last case, the repeated digits (yy) cannot be the same as the first digit (1), which means there are 9 of them. This makes 9 + 9 + 27 = 27 of them in total
Between 200 and 299 the same argument shows there are 27 of them.
Thus there are 9 + 27 + 27 = 63 in total.
The 63 numbers are:
11, 22, 33, 44, 55, 66, 77, 88, 99
110, 112, 113, 114, 115, 116, 117, 118, 119
101, 121, 131, 141, 151, 161, 171, 181, 191
100, 122, 133, 144, 155, 166, 177, 188, 199
220, 221, 223, 224, 225, 226, 227, 228, 229
202, 212, 232, 242, 252, 262, 272, 282, 292
200, 211, 233, 244, 255, 266, 277, 288, 299
Wiki User
∙ 8y ago
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