496
4 with 2 remaining 62 - 2 = 60 = 15 x 4
4
Suppose the first number is x. Then the second number is x-4 and the third is 4*x, or 4x So, x+(x-4)+4x = 62 ie 6x-4= 62 or 6x = 66 or x = 11. Then x-4 = 7 and 4x = 44 So the 3 numbers are 11, 7 and 44.
31 x 10 = 310, 62 x 5 = 310, 1 x 310 = 310 there will be others this is just a selection
1, 2, 4, 31, 62, 124: (1 x 124, 2 x 62, 4 x 31, 31 x 4, 62 x 2, 124 x 1)
The first ten positive integer multiples of 62 are: 1 x 62 = 62 2 x 62 = 124 3 x 62 = 186 4 x 62 = 248 5 x 62 = 310 6 x 62 = 372 7 x 62 = 434 8 x 62 = 496 9 x 62 = 558 10 x 62 = 620
The first four positive integer multiples of 62 are: 62 x 1 = 62 62 x 2 = 124 62 x 3 = 186 62 x 4 = 248
496
248/4 = 62 so 62 x 4 = 248
4 with 2 remaining 62 - 2 = 60 = 15 x 4
4
if x= -62 then -x= 62
8 x 4/5 = 32/5 or 62/5
1 x 124, 2 x 62, 4 x 31.
Suppose the first number is x. Then the second number is x-4 and the third is 4*x, or 4x So, x+(x-4)+4x = 62 ie 6x-4= 62 or 6x = 66 or x = 11. Then x-4 = 7 and 4x = 44 So the 3 numbers are 11, 7 and 44.
The equation of a circle centered in the origin, with radius 6, is x2 + y2 = 62. To move the coordinates of the center from (0,0) to (4,7), you change each x by (x-4), and each y by (y-7); thus, you obtain: (x-4)2 + (y-7)2 = 62.