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All the numbers between 1947 and 2008 are written, including 1947 and 2008. From the list, all the multiples of 3 and 5 are struck off. The sum of the remaining numbers is
The sum of all 62 numbers is $$\frac{62(1947 + 2008)}{2} = 122605$$. The multiples of 3 are $$1947, 1950, \ldots, 2007$$, 21 numbers with sum $$21 \times 1977 = 41517$$; the multiples of 5 are $$1950, 1955, \ldots, 2005$$, 12 numbers with sum 23730; the multiples of 15 are $$1950, 1965, 1980, 1995$$, sum 7890. The multiples of 3 or 5 sum to $$41517 + 23730 - 7890 = 57357$$, so the remaining sum is $$122605 - 57357 = 65248$$.
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