A number was divided successively in order by 4 , 5 and 6 leaving out the…

Answer: 214

Let the original number be p. From the successive divisions we have:

• When p is divided by 4 the remainder is 2, so p = 4q + 2.

• When q is divided by 5 the remainder is 3, so q = 5r + 3.

• When r is divided by 6 the remainder is 4, so r = 6s + 4.

Substitute successively to express p in terms of s:

• r = 6s + 4

• q = 5r + 3 = 5(6s + 4) + 3 = 30s + 23

• p = 4q + 2 = 4(30s + 23) + 2 = 120s + 94

Thus p = 120s + 94 for some integer s ≥ 0. The smallest positive solution occurs at s = 0 giving p = 94, but 94 is not among the choices. For s = 1 we get p = 214, which is one of the options.

Verify p = 214: 214 ÷ 4 = 53 remainder 2; 53 ÷ 5 = 10 remainder 3; 10 ÷ 6 = 1 remainder 4. All required remainders match, so 214 is the correct choice from the given options.