Q: The sum of four consecutive odd integers is 248. What is the smallest integer?

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If the largest integer is subtracted from four times the smallest, the result is 4 more than twice the middle integer. Let the smallest integer be x, then the others are x + 2 and x+ 4. Therefore 4x - (x + 4) = 2 (x + 2 ) + 4 Expanding, we get 4x -x -4 = 2x + 4 + 4 Gathering terms: x = 12 Thus the three integers are 12, 14 and 16.

Let the first consecutive integer be x. So that:the second integer is x + 1,the third integer is x + 2, andthe fourth integer is x + 3.We have:(x + 1) + (x + 3) = 1322x + 4 = 1322x = 128x = 64 the first integerThus, the four consecutive integers are 64, 65, 66, and 67.

The simplest way is to solve 4n + 6 = 138 for the smallest of them.

14, 15, 16 and 17.

There is no set of four consecutive odd integers for 204. The only set is even: 48, 50, 52 and 54.

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The integers are 47, 49, 51, and 53. 47 * 53 = 2491

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If n is the smallest of the four integers, their sum is 4n+6.

The sum is four.

The integers would begin with 10.

9,11,13

17 , 19 , 21 and 23 are the odd integers whose sum is 80 and the least integer is 17

If the largest integer is subtracted from four times the smallest, the result is 4 more than twice the middle integer. Let the smallest integer be x, then the others are x + 2 and x+ 4. Therefore 4x - (x + 4) = 2 (x + 2 ) + 4 Expanding, we get 4x -x -4 = 2x + 4 + 4 Gathering terms: x = 12 Thus the three integers are 12, 14 and 16.

Let the first consecutive integer be x. So that:the second integer is x + 1,the third integer is x + 2, andthe fourth integer is x + 3.We have:(x + 1) + (x + 3) = 1322x + 4 = 1322x = 128x = 64 the first integerThus, the four consecutive integers are 64, 65, 66, and 67.

Suppose the middle integer is 2a. Then the smallest is 2a-2 and the biggest is 2a+2. 4 times the smallest is 8a-8 So largest subtracted from the smallest is (8a-8) - (2a+2) = 6a-10 So, 6a-10 = 2*2a = 4a so that 2a = 10 So the integers are 8, 10 and 12.