The number of all 3-digit numbers in each of which the sum of the digi

1.	The number of all 3-digit numbers in each of which the sum of the digits is even is (a) 450 (b) 375 (c) 365 (d) 250
Answer» (a) 450 Out of the 10 digits 0, 1, 2, 3, ..., 9, five digits, i.e., 0, 2, 4, 6, 8 are even and five digits, i.e., 1, 3, 5, 7 and 9 are odd. The sum of the three digits D₁, D₂, D₃ of the number D₁D₂D₃ will be even if : (i) All the three digits are even: ∴ Number of such numbers = 4 × 5 × 5 = 100 (Here the position D₁ cannot be occupied by 0) (ii) One of the digits is even and the rest two are odd : (a) D₁ is even, D₂ is odd, D₃ is odd ∴ Number of such numbers = 4 × 5 × 5 = 100 (Again since D₁ ≠ 0, so only 4 choices for D₁) (b) D₁ is odd, D₂ is even, D₃ is odd ∴ Number of such numbers = 5 × 5 × 5 = 125 (c) D₁ is odd, D₂ is odd, D₃ is even ∴ Number of such numbers = 5 × 5 × 5 = 125 ∴ Total number of required numbers = 100 + 100 + 125 + 125 = 450.

The number of all 3-digit numbers in each of which the sum of the digits is even is (a) 450 (b) 375 (c) 365 (d) 250

Answer»

(a) 450

Out of the 10 digits 0, 1, 2, 3, ..., 9, five digits, i.e., 0, 2, 4, 6, 8 are even and five digits, i.e., 1, 3, 5, 7 and 9 are odd.

The sum of the three digits D₁, D₂, D₃ of the number D₁D₂D₃ will be even if :

(i) All the three digits are even:

∴ Number of such numbers = 4 × 5 × 5 = 100

(Here the position D₁ cannot be occupied by 0)

(ii) One of the digits is even and the rest two are odd :

(a) D₁ is even, D₂ is odd, D₃ is odd

∴ Number of such numbers = 4 × 5 × 5 = 100

(Again since D₁ ≠ 0, so only 4 choices for D₁)

(b) D₁ is odd, D₂ is even, D₃ is odd

∴ Number of such numbers = 5 × 5 × 5 = 125

∴ Total number of required numbers

= 100 + 100 + 125 + 125 = 450.