Sum of three consecutive prime numbersd is 121. what is the 50% of the highest number
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let the numbers be x, x+2,x+4
x+x+2+x+4 =121
3x=115
x=115\3
and 50% is 115\3*50\100
x+x+2+x+4 =121
3x=115
x=115\3
and 50% is 115\3*50\100
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