Question 205581: How many integers between 1 and 1000 (inclusive) do not contain the digits 8 or 9?
(A) 200 (B)488 (C)512 (D)521 (E)800
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! looking at a typical hundreds chart,
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in each set of 0 to 10,,(1-10,,11-20,,21-30,,etc) we have one 8, and one nine, for a total of two.
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this is times 8, for sets of 10 up to 80.
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then the 80's and 90's, give 10 each.
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for typical 100 set, 8*2 +10 +10 = 36, 8"s or 9"s
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this happens for 8,,,,100 sets,,(0-100,100-200, ....700-800),,,or 8*36 = 288
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all,,800's,,have an 8,,,or +100
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and all 900's have a 9,,,or +100
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This brings total to 288 +100+100=488,,,8"s or 9"s,,,up to 1000
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total without is 1000 - 488 = 512,,,or answer (c)
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