89, 91, and 93
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
To find out how many times 15 can go into 273, you would divide 273 by 15. When you perform the division, 273 ÷ 15 equals 18.2, which means 15 can go into 273 a total of 18 full times.
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.