Question

Find all pairs of natural numbers which can be the solution to the equation a+b=42.

Answer 1

Answer:

(1,41), (2,40), (3,39), (4,38), (5,37), (6,36), (7,35), (8,34), (9,33), (10,32), (11,31), (12,30), (13,29), (14,28), (15,27), (16,26), (17,25), (18,24), (19,23), (20,22), (21,21), (22,20), (23,19), (24,18), (25,17), (26,16), (27,15), (28,14), (29,13), (30,12), (31,11), (32,10), (33,9), (34,8), (35,7), (36,6), (37,5), (38,4), (39,3), (40,2), (41,1).

Step-by-step explanation:

To find all pairs of natural numbers which are solution to a+b=42, we need to first choose a value to one of them (let's choose 'a'), and that value will be the lowest possible, so we begin with a=1, and then we increase to 2, and 3, and so on.

For each value of a, we calculate the value of b using the equation.

So, starting with a=1, we have that b=41

Then, with a=2, we have that b=40, and so on.

Doing this until a=41 (so b=1 will be the lowest value possible for b), we have the pairs of natural number we want:

(1,41), (2,40), (3,39), (4,38), (5,37), (6,36), (7,35), (8,34), (9,33), (10,32), (11,31), (12,30), (13,29), (14,28), (15,27), (16,26), (17,25), (18,24), (19,23), (20,22), (21,21), (22,20), (23,19), (24,18), (25,17), (26,16), (27,15), (28,14), (29,13), (30,12), (31,11), (32,10), (33,9), (34,8), (35,7), (36,6), (37,5), (38,4), (39,3), (40,2), (41,1).