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989.

If there is a remainder of 2 when divided by 3, the number is one less than a multiple of 3.

If there is a remainder of 4 when divided by 5, the number is one less than a multiple of 5.

Thus the number required is one less than a multiple of the lowest common multiple of 3 and 5 (that is 15).

So what is needed is an even multiple of 15 less than or equal to 1000:

1000 ÷ 15 = 662/3

Thus the highest even multiple of 15 not greater than 1000 is 66 x 15 = 990, and the required number is 989.

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โˆ™ 2011-08-04 08:47:19
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Q: What is the largest number less than 1000 which is odd leaves a remainder of 2 when divided by 3 and a remainder of 4 when divided by 5?
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