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Sagot :
Answer:
Step-by-step explanation:
Numbers begin with ‘8’:
Integers from 800 to 899; that is 100 numbers.
Numbers end with ‘8’ from 600 to 900:
Options for hundred's place = 3 (digits ‘6’, ‘7’ and ‘8’).
Options for ten's place = 10 (all ten digits)
Options for unit's place = 1 (only ‘8’)
So, (3 * 10 * 1) = 30 numbers.
Now, among these (100 + 30) = 130 numbers; there are some which start and end with ‘8’; we need to eliminate them to avoid ‘double counting’ of these numbers. How many are they?
Numbers start and end with ‘8’:
Option for hundred's place = 1 (Only ‘8’)
Options for ten's place = 10 (all ten digits)
Option for unit's place = 1 (only ‘8’)
So, (1 * 10 * 1) = 10 numbers.
Therefore, there are (130 - 10) = 120 numbers from 600 to 900 which either start with ‘8’ or end with ‘8’.
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