between 15 and 16 152=225 162=256
13 x 15 = 195 . . . too small15 x 17 = 255 . . . too bigThere are none.The best we can do for you is: (14 x 16) = 224 .===================================For the advanced student:If you work it from the generic form of two consecutive odd numbers,then you would say(2x - 1) (2x + 1) = 225Whence . . . (4x2 - 1) = 225 . . . 4x2=226 . . . x2=56.5 . . . x=sqrt(56.5)The two numbers are 14.0333 and 16.0333 . (rounded)A truly revolting development. Although their product is nice and close to 225,there's no way you could pass them off as odd integers.
8. A way to work this out is figure out how many possible digits can go in each place value position (units, tens, hundreds) and multiply these together. Since there are 2 possible digits that can go in each position and there are three positions, you would go: 2 x 2 x 2 = 8 Want a list? 222 225 252 255 555 552 525 522
-225 - 6 = -225 + (-6) = -231
√225 = 15
112 and 113
224, 225
between 15 and 16 152=225 162=256
Let the first one be x, the second be x + 1, and the third one is x + 2. So we have,x + x+1 + x +2 = 2283x + 3 = 228 subtract 3 to both sides3x = 225 divide by 3 to both sidesx = 75x + 1 = 75 + 1 = 76x + 2 = 75 + 2 = 77Thus, the three consecutive integers are 75, 76, and 77.
If they can be same 15*15 =225 if they have to be distinct 13*17 =221
No. Integers are whole numbers
13 x 15 = 195 . . . too small15 x 17 = 255 . . . too bigThere are none.The best we can do for you is: (14 x 16) = 224 .===================================For the advanced student:If you work it from the generic form of two consecutive odd numbers,then you would say(2x - 1) (2x + 1) = 225Whence . . . (4x2 - 1) = 225 . . . 4x2=226 . . . x2=56.5 . . . x=sqrt(56.5)The two numbers are 14.0333 and 16.0333 . (rounded)A truly revolting development. Although their product is nice and close to 225,there's no way you could pass them off as odd integers.
It is: 225
600/8x3= 225 a=225
225
225 is the square of 15...
The least common factor of any set of integers is 1.