How many numbers are there between 99 and 1000 such that the digit 8 occupies the units plac?
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7
This question asks for number of occurrences of digit 88 in numbers between 9999 and 10001000.
As the limits(9999 and 10001000) doesn’t contain the digit 8,8, both the upper limit and lower limit can be excluded and the range converges to all three digit numbers
Following are the various combination with single occurrence of 8:
8__8__ : 88 in Hundred’s place
For this case, Ten’s and Unit’s place can be filled up with digits 0–90–9 except 8.8.This gives us 9∗9=819∗9=81 numbers with 88 in hundered’s place.
_8__8_ : 8 in Ten’s place
Hundred’s place can be filled up with any digit from 1–9,1–9,barring 88, i.e. 8 such digits. Units place can be filled up with any digit from 00–9,–9, barring 88 i.e. 9 such digits.
This gives us 99∗8=72∗8=72numbers, with 88 in tens place.
__8__8 : 8 in Unit’s place
This gives us 8∗9=728∗9=72 numbers, with 88 in units place.
As the limits(9999 and 10001000) doesn’t contain the digit 8,8, both the upper limit and lower limit can be excluded and the range converges to all three digit numbers
Following are the various combination with single occurrence of 8:
8__8__ : 88 in Hundred’s place
For this case, Ten’s and Unit’s place can be filled up with digits 0–90–9 except 8.8.This gives us 9∗9=819∗9=81 numbers with 88 in hundered’s place.
_8__8_ : 8 in Ten’s place
Hundred’s place can be filled up with any digit from 1–9,1–9,barring 88, i.e. 8 such digits. Units place can be filled up with any digit from 00–9,–9, barring 88 i.e. 9 such digits.
This gives us 99∗8=72∗8=72numbers, with 88 in tens place.
__8__8 : 8 in Unit’s place
This gives us 8∗9=728∗9=72 numbers, with 88 in units place.
Answered by
23
from 99 -200 9 numbers are there so from 99-1000 9×10=90eight unit numbers are there
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