Suppose the smallest number is x. Then the six numbers are x, x+1, x+2, x+3, x+4 and x+5. Their sum is 6x+15 which is 273. So 6x = 273-15 = 258 So x = 258/6 = 43. The six numbers are 43, 44, 45, 46, 47 and 48.
3 and 91 multiplied together make 273
Yes. 273 is evenly divisible by three.
273, 93, 3, 3The prime numbers are 33
273 and 3/4
The numbers are 89, 91 and 93 and 90, 91 and 92
They are: 90+91+92 = 273
Suppose the first number is n. Then the next two are n+1 and n+2 All of them add up to n + (n+1) + (n+2) = 3n+3 3n+3 = 273 Now you finish it...
Divide the sum of the three consecutive odd integers by 3: -273 /3 = -91. The smallest of these integers will be two less than -91 and the largest will be two more than -91, so the three consecutive odd integers will be -89, -91, and -93.
91+31+151=273
The numbers one and three go into 273 and six evenly.
Suppose the smallest number is x. Then the six numbers are x, x+1, x+2, x+3, x+4 and x+5. Their sum is 6x+15 which is 273. So 6x = 273-15 = 258 So x = 258/6 = 43. The six numbers are 43, 44, 45, 46, 47 and 48.
7 x 39 = 273 13 x 21 = 273
3 + 131 + 139 = 273 73 + 97 + 103 = 273 37 + 43 + 193 = 273
3 and 91 multiplied together make 273
1, 3, 7, 13, 21, 39, 91, 273.
Yes. 273 is evenly divisible by three.