Q: What are three consecutive odd integers whose sum is 336?

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There are four consecutive odd integers: 81, 83, 85 and 87.

The integers are 81, 83, 85 and 87.

111 + 112 + 113 = 336

Any odd integer can be represented as 2n+1, where n is any integer. The integers, starting with one, always proceed as odd, even, odd, even, etc. Therefore, a consecutive odd would be two more than the first. So 2n+1, 2n+3, 2n+5, and 2n+7 summed to 336. Simple algebra gives n=40, and the four numbers as 81, 83, 85, and 87.

Standard notation is 336. Word notation is three hundred thirty-six.

Related questions

There are four consecutive odd integers: 81, 83, 85 and 87.

The integers are 81, 83, 85 and 87.

111 + 112 + 113 = 336

x + x + 2 + x + 4 +x + 6 = 336 4x + 12 = 336 4x = 324 x = 81 81,83,85, 87

111, 112 and 113.

Any odd integer can be represented as 2n+1, where n is any integer. The integers, starting with one, always proceed as odd, even, odd, even, etc. Therefore, a consecutive odd would be two more than the first. So 2n+1, 2n+3, 2n+5, and 2n+7 summed to 336. Simple algebra gives n=40, and the four numbers as 81, 83, 85, and 87.

1x2x3=6 2x3x4=24 3x4x5=60 4x5x6=120 5x6x7=210 6x7x8=336 7x8x9=504 8x9x10=720 9x10x11=990

The number 336 is "three hundred thirty-six."

As of the end of the 2008 NFL season Jeff Feagles the punter for the NY Giants has played in 336 consecutive games. That is 21 seasons of 16 games each.

Standard notation is 336. Word notation is three hundred thirty-six.

0.1786

The total number of integers that are multiples of both 12 and 28 is infinite. Here are the first few: 84, 168, 252, 336, 420, 504, 588, 672, 756, 840 . . .