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The sum of several consecutive natural numbers is 2012. How many numbers were in that sum​

User JoeyP
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1 Answer

8 votes

Answer:

8

Explanation:

first number = a, there are n numbers

so last number is a + n - 1

sum = [a + (a + n - 1)} * n/2 = (2a + n - 1) / 2 = 2012

n * (2a + n - 1) = 2012 * 2 = 4024

4024 is even, so n must be even, divisible by 2, 4, 8

4024 = 2 * 2012 = 4 * 1006 = 8 * 503

If n = 2, 2a + n - 1 = 2012, 2a + 2 - 1 = 2012, 2a = 2011, impossible.

If n = 4, 2a + n - 1 = 1006, 2a + 4 - 1 = 2a + 3 = 1006, 2a = 1003, impossible

If n = 8, 2a + n - 1 = 503, 2a + 8 - 1 = 2a + 7 = 503, a = 248

so n = 8, a = 248

User DGDD
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