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| If integer k is equal to the sum of all even multiples of 15 |
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Ron Purewal
MGMAT STAFF
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first thing to realize here is that 'even multiples of 15' is just a fancy schmancy way of saying 'multiples of 30'. (if you don't see this at first, just make a list of multiples of 15 and notice that the even ones are precisely the multiples of 30.)
so this problem is really asking you about the sum of the multiples of 30, from 300 to 600. there are at least 3 ways you could address this sum: (1) use the formula for arithmetic series, sum = (average) x (number of terms). this is the best way, both because it's the fastest and because it's already factored somewhat: the sum is 450 x 11 - no need to multiply out! (2) factor out 30 to give 30(10 + 11 + 12 + ... + 20), which is 30 x 165. even if you have to just add these numbers up, you can easily do so within the allotted time. (3) arguably the worst way, but you could still manage it in time if you get started right away: just add up 300 + 330 + ... + 600, giving 4950. no matter which way you do the sum, the factorization will give 2 x 3 x 3 x 5 x 5 x 11, so the greatest prime factor is 11. |
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| If integer k is equal to the sum of all even multiples of 15 |
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