The common factors are: 1, 2
Oh, dude, math time! So, like, if you divide 120 by 14, you get 8 with a remainder of 8. So, technically, 14 goes into 120 about 8 times, but there's a little leftover, like when you eat too much pizza and have that last slice sitting there, tempting you.
The numbers are: 20 and -6
Not evenly because there will be a remainder of 8
The numbers are 6 and -20
Simultaneous equations A-B=14 (1) AB=120 (2) transposing(1) A=14+B substituting into(2) B(14+B)=120 B^2+14B =120 B^2+14B-120=0 solve quadratic (B+20)(B-6)=0 B=-20,6 one of the whole numbers is 6 and the other is 6+14=20
1, 2, 3 and 6 will go into both numbers
Oh, dude, you're hitting me with some math vibes! So, the two numbers we're looking for are -12 and -10. They multiply to 120 and add up to -14. It's like a math puzzle, but with numbers instead of jigsaw pieces.
14% of 120= 14% * 120= 0.14 * 120= 16.8
The HCF of both numbers is 14
The numbers are: 1 and 2
9 and 5
15