The four consecutive odd numbers which add up to 120 are 27, 29, 31 and 33.
It can be found out as follows:
Let 'x' be the smallest odd number.
The next odd numbers will be 'x+2', 'x+4', 'x+6'.
The addition of these four numbers will give 120.
Therefore,
x+(x+2)+(x+4)+(x+6)=120
4x+12=120
x+3=30
Therefore, x=27.
10 and 12
120 and 122.
114 115 116 117 118 119 120 121 122 123 124 125 126
60
((x+6)(x+8))-((X + 2)(X+4)) = 72(x^2 + 14x + 48)-(x^2 + 6x + 8) = 728x + 40 = 728x = 32x = 4check:((10)(12))-((6)(8)) = 72(120)-(48) = 72120 = 12012,10,8,6The numbers are 6, 8, 10, and 12.Here is how we find them:(n + 6)(n + 4) - (n + 2)n = 72; whence,(n2 + 10n + 24) - (n2 + 2n) = 8n + 24 = 72,8n = 48, andn = 6.So, let's try it!(12)(10) - (8)(6) = 120 - 48 = 72.It works!
10 and 12
The numbers are 120, 121 and 122.
There are two consecutive even integers. The numbers are 118 and 120.
The numbers are 39, 40 and 41.
The numbers are 38, 40 and 42.
44
3*4*10
Let the numbers be 'n' , 'n+1' & 'n+2' Hence adding n + n+1 + n + 2 = 120 3n + 3 = 120 3n = 117 n = 39 Hence n + 1 = 40 n + 2 = 41 So the three consecutive numbers are ;- 39,40,& 41.
None that are all odd numbers. But, there are two sets of three consecutive numbers containing a mix of even/odd that add up to 120. They are 39, 40 and 41 and also 38, 40 and 42.
The numbers are 116, 118 and 120.
The four even integers are 6, 8, 10, 12. 6 x 8 = 48 10 x 12 = 120 120 - 48 = 72
The consecutive integers are 39, 40, and 41. To solve this algebraically, note that consecutive numbers can be indicated by the variables x, x+1, and x+2. For the sum x + (x+1) + (x+2) = 120, 3x +3 = 120 3x= 117 x= 39, followed by 40 and 41 39+40+41 = 120