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Q: What are three consecutive integers with the sum of 96?

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31, 32, 33

31 32 33

31, 32, 33

If three consecutive integers have the sum of 96, then the problem can be expressed with the equation... N + (N+1) + (N+2) = 96 ...Simplify that and solve and you get... 3N + 3 = 96 3N = 93 N = 31 ... so the three integers are 31, 32, and 33.

31, 32 and 33

-31, -32, -33

The integers are 47 and 49.

Average is 96/3 ie 32 so integers are 30, 32 and 34.

They are odd consecutive integers: 21, 23, 25 and 27.

The integers are -98 and -96.

31, 32, 33

To figure this out, divide 96 by three, ignoring the remainder for now. This gives the number 30. The remainder is 6. Three consecutive numbers that add up to six are 1, 2 and 3. So the three consecutive integers are 31, 32, and 33.

30, 32 and 34.

Divide 96 by 3= 32 your three numbers are now: 32 -1 = 31 32 = 32 32 + 1=33 31 + 32 +33 = 96

31 + 32 + 33 = 96

96

96

im not 100 percent sure but i thing its: (x+1)+(x+2)+(x+3)=96

96

31+32+33 = 96

94+95+96 96

The integers are 96 and 97.

31, 32, 33

Three consecutive integers are [X], [X+1], and [X+2].x + (x+1) + (x+2) = 963x + 3 = 963x = 93x = 31The numbers are 31, 32, and 33.

There is no set of two consecutive even numbers that total -96. There is only a set of odds. The numbers are -49 and -47.