The six multiples are: 57 x 1 = 57 57 x 2 = 114 57 x 3 = 171 57 x 4 = 228 57 x 5 = 285 57 x 6 = 342
0.57x6%
1 x 114, 2 x 57, 3 x 38, 6 x 19, 19 x 6, 38 x 3, 57 x 2, 114 x 1
-9.5 x -6
57
1 x 228, 2 x 114, 3 x 76, 4 x 57, 6 x 38, 12 x 19 (19 x 12, 38 x 6, 57 x 4, 76 x 3, 114 x 2, 228 x 1)
1 x 228, 2 x 114, 3 x 76, 4 x 57, 6 x 38, 12 x 19.
The six multiples are: 57 x 1 = 57 57 x 2 = 114 57 x 3 = 171 57 x 4 = 228 57 x 5 = 285 57 x 6 = 342
0.57x6%
1, 2, 3, 6, 19, 38, 57, 114Prime Factors of 114=2x3x19.1,2,3,6,19,38,57,1141 x 114, 2 x 57, 3 x 38, 6 x 19.
Let the lowest odd number equal x, thus: x + x + 2 + x + 4 = 57 3x + 6 = 57 3x = 57 - 6 3x = 51 x = 51/3 x = 17 The three numbers are therefore 17, 19 and 21.
1 x 114, 2 x 57, 3 x 38, 6 x 19, 19 x 6, 38 x 3, 57 x 2, 114 x 1
-9.5 x -6
Yes, but not in rational terms. The factorisation is: {x - [3-sqrt(57)]/6} * {x - [3+sqrt(57)]/6}
57
6 with 5 remaining 347 - 5 = 342 = 57 x 6
Nine and a half. 9 x 6 = 54, and 10 x 6 = 60, so 6 will go into 57 nine times in total with half left over; the answer would be expressed 9.5.