That's not possible... * * * * * How about -5 and 25?
One such set is {-4, 5}.
54 x 5 = 270 54 + 5 = 59 The two factors in question are therefore 54 and 5.
It is 10 whose factors are: 1, 2, 5 and 10
Two, whose digital sum is 5.
Factors for 20 are: 1, 2, 4, 5, 10, 20 Prime Factors: 2, 5 sum of prime factors = 2 + 5 = 7
Factors of 20 are 1, 2, 4, 5, 10 and 20. And their sum is 1 + 2 + 4 + 5 + 10 + 20 = 32.
95
The prime factors are: 2 and 5 (2+5 = 7)
The factors of 10 are 1, 2, 5, and 10. Among these, the pairs whose sum is 7 are 1 and 6, and 2 and 5. These pairs are factors of 10 as they divide evenly into 10 without leaving a remainder.
That's not possible... * * * * * How about -5 and 25?
One such set is {-4, 5}.
1 + 2 + 4 + 5 + 10 + 20 = 42
How about: 5-4 = 1
54 x 5 = 270 54 + 5 = 59 The two factors in question are therefore 54 and 5.
20 and 1 10, 5, 4 and 2
The sum of -5 and 20 is 15. The distance between -5 and 20 is 25. The sum of the numbers between -5 and 20 is 180. The sum of the numbers from -5 to 20 is 195.