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Find all positive integers less than 5000 which leave remainders 2, 4, and 8 upon division by 9, 10, and 11 respectively.
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10 should be easy...
i am trying to figure out a way for 9 and 11.
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Count off all the numbers that meet the first condition until you find one that also meets the second.
2, 11, 20, 29, 38, 47, 56, 65, 74
That means that any number that fits the first two conditions must give a remainder of 74 when divided by 90. (Look up the Chinese Remainder Theorem if you're interested in why)
You would now make another list to find a number that fits all three conditions, but luckily 74 does that. So, by the Chinese Remainder Theorem again, a number that fits all three conditions will give a remainder of 74 when divided by 990.
Now simply list the numbers less than 5000 for which this is true (there will be five of them), and add them up to get your answer.
Why did the vector cross the road?
It wanted to be normal.
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