bSuppose a and b are two positive integers and d is an integ

b)Suppose a and b are two positive integers and d is an integer that divides 3a + 2b and 5a + 3b. Prove that d must divide both a and b.

c)Could any selection of the following numbers add up to 121? 9, 15, 18, 21, 36, 69, 72, 81, 87, 93

d is an integer that divides 3a + 2b and 5a + 3b.

(3a +2b)/d = k ;

3a +2b = dk1 ----(1)

multily by 3 : 9a +6b = 3dk1

(5a +3b)/d = k2

5a +3b = dk2 ----(2)

10a + 6b = 2dk2

subtract 1 from 2 : :a = 2d(k2 -k1)

we can see a is divisble by d

Now 3(2d(k2 -k1) + 2b = dk1

2b = dk1 - 6dk2 + 6dk1

b = d(7k1 -6k2)/2

So, b is also divisble by d

Hence proved