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aliky
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I guess the easier way to do this would be.

400 numbers between - 200 & 600
Out of which half would be even, half odd.
Number of odd = 200. Of this.. the one with even tenth digit would be half again. hence 100.
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This has got nothing to do with the question, however according to the Sum formula aren't there 399 numbers between 200 and 600, i.e. 599-201 = 398 + 1 = 399????
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This has got nothing to do with the question, however according to the Sum formula aren't there 399 numbers between 200 and 600, i.e. 599-201 = 398 + 1 = 399????

It depends whether we include 200 and 600.

There are 401 integers from 200 to 600, both inclusive;
There are 400 integers from 200 to 600, including 600 or 200.
There are 399 integers from 200 to 600, both exclusive.

Hope it's clear.
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Yes thank you that is what I thought also, but then Asax is wrong? Isn't he or she?!
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KevinBrink
Yes thank you that is what I thought also, but then Asax is wrong? Isn't he or she?!

I guess asax meant that there are 400 integers from 201 to 600, both inclusive.
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Ok thanks Bunuel for the clear explanation!
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Yes thank you that is what I thought also, but then Asax is wrong? Isn't he or she?!

even when you exclude 200 and 600 there will be 200 odd numbers and 100 out of which will have tenths digit as even
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Total brain fart. Ever get those instances where your brain just stares blankly at the page and nothing is being processed? 6 minutes.

Pattern recognition is key here.

Tens digit is even means 0,2,4,6,8
3-digit integer must also be odd

201 202 203 204 205 206 207 208 209 <--- Only 5 of these satisfy the condition.
221 222 223... <--- Only 5 of these satisfy

For every X00 series where X is 2,3,4,5 there are a total of 25 digits that satisfy the condition

4 x 25 = 100

Answer is C.
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