Question

If n €N and n < 2000, how many numbers n have the property that the sum of the digits of n and the sum of digits of n +1 are odd numbers?​

Answer 1

Answer:

As the hint suggests that the last digit must be 9, let abcd¯¯¯¯¯¯¯¯¯¯ be 4 digits number where a,b,c,d can be 0. We have d=9 and abc¯¯¯¯¯¯¯ must have its digit sum even.

Thus it is equivalent to find the number of triples (a,b,c) satisfying 0≤a≤1,0≤b,c≤9 that a+b+c=0,2,4,…18, because a+b+c≤1+9+9=19. Using inclusion-exclusion principle and the number of solutions (a,b,c) with a,b,c≥0 in a+b+c=n equals (n+22), we obtain the number of numbers that we want to be

∑0≤n≤9{(2n−72)−(2n−92)−2(2n−172)+2(2n−192)}=55.

Step-by-step explanation:

idk if this is the answer...i tryd