Question

How many different numbers can be formed to satisfy all the conditions given below:

(a) The number is less than 2 x 10⁸ .

(b) The number is formed from the digits 0, 1 and 2.

(c) The number is divisible by 3.

Answer 1

Now 2 x 10⁸ = 200000000

Numbers less than 2 x 10⁸ are of the form a₁ a₂ a₃ a₄a₅a₆a₇a₈a₉ .

For all the nine-digit numbers a₁ =1 and a₂,a₃, ...or a₈ or (at present a₉ is deliberately not considered) can be one of 0, 1 or 2.

For all the 8-digit numbers,a₁ = 0,a₂ ≠ 0; for all the 7-digit numbers a₁ = 0,a₂ = 0, a₃ ≠ 0.

Thus, all numbers of 9-digits and less than 9-digits are included when a₁ is chosen in one of two ways (0 or 1); and a₂ , a_3, ... , a₈ are each chosen in one of 3 ways (0, 1 or 2)

Therefore, the choice of a₁,a₂, _....a₈ may be done 2 x 3⁷ ways. Suppose one such choice is a₁,a₂,...a₈ is already divisible by 3,3,a₉ (which was not considered before).

If a₁ + a₂ + ....+ a₈ is of the form 3p + 1, leaving a remainder 1 on division by 3, then a₉ = 2 .

If a₁ + a₂ + ....+ a₈ is of the form 3p + 2, leaving a remainder 2 on division by 3, then a₉ =1. Thus, in any case, there is only one way of choice for a₉ . The number of numbers, is therefore, 2 x 3⁷ x 1 and this includes 000000000, that is, zero. The required number of numbers is 2 x 3⁷ - 1 = 4373