The total number of seven-digit numbers whose | vedclass

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(B) Let the seven-digit number be represented by $x_1x_2x_3x_4x_5x_6x_7$.The first digit $x_1$ can take values from $\{1, 2, 3, ..., 9\}$ ($9$ choices

Vedclass · Accepted Answer

$4500000$

Vedclass · Answer

(B) Let the seven-digit number be represented by $x_1x_2x_3x_4x_5x_6x_7$.The first digit $x_1$ can take values from $\{1, 2, 3, ..., 9\}$ ($9$ choices).The digits $x_2, x_3, x_4, x_5, x_6$ can each take values from $\{0, 1, 2, ..., 9\}$ ($10$ choices each).For any fixed values of $x_1, x_2, x_3, x_4, x_5, x_6$,the sum $S = x_1 + x_2 + x_3 + x_4 + x_5 + x_6$ is either even or odd.We need the total sum $S + x_7$ to be even.If $S$ is even,$x_7$ must be even (i.e.,$x_7 \in \{0, 2, 4, 6, 8\}$),which gives $5$ choices.If $S$ is odd,$x_7$ must be odd (i.e.,$x_7 \in \{1, 3, 5, 7, 9\}$),which also gives $5$ choices.Thus,for any combination of the first six digits,there are exactly $5$ choices for $x_7$ to make the total sum even.The total number of such seven-digit numbers is $9 	imes 10 	imes 10 	imes 10 	imes 10 	imes 10 	imes 5 = 4500000$.

The total number of seven-digit numbers whose sum of digits is even is

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