Find the numerical value of the letters given below-

(a) (b)

(a) Given, product of DE and 3 = FDE

Since DE is multiple of 3 we need to find the number such that numbers D, E exists in the product value. By trial and error method we get

D = 5 and E = 0

And the product of 50 × 3 = 150

∴ F = 1 and D, E numbers exists in the value 150

Hence, D = 5, E = 0 and F = 1

(b) Given, Product of GH and 6 = CGH

Since GH is multiple of 6 we need to find the number such that numbers G, H exists in the product value. By trial and error method we get

G = 2 and H = 0

And the product of 20 × 6 = 120

∴ C = 1 and G, H numbers exists in the value CGH

Hence, C = 1, G = 2 and H = 0