If n in belongs to 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?
Answers
Given : n have the property that the sum of the digits of n and the sum of digits of n + 1 are odd numbers
To Find : How many numbers
Solution:
Case 1 : Sum of digit of n + sum of digits of n + 1 is odd
if sum of digit of n is even then sum of digit of n + 1 is even + 1
Hence we get 2(even) + 1 = odd
if sum of digit of n is odd then sum of digit of n+1 is odd + 1
Hence we get 2(odd) + 1 = odd
But there are case where n + 1 digit does not have sum 1 extra
when ever it changes like 9 to 10 , 19 to 20 and so on
so numbers which does not satisfy
9 , 19 , 29 , 39 , 49 , 59 , 69 , 79 , 89 ( 99 is exceptional)
109 , 119 , 129 , 139 , 149 , 159 , 169 , 179 , 189 , 199
upto 1999
Numbers from 9 to 1999
1999 = 9 + (k - 1)10 => k = 200
subtract 1 for 99 hence 199
Out of 1999 numbers 199 does not satisfy
Hence 1999 - 199 = 1800 numbers have the property that the sum of the digits of n and the sum of digits of n + 1 are odd numbers?
Case 2 : Sum of digit of n is odd
also sum of digits of n + 1 is odd
n = 9 , 29 , 49 , 69 , 89 , 119 , 139 , 159 , 179 , 209 , 229 , 249 , 269 , 289 , 319 , 339 , 359 , 379 , 409 , 429 , 449 , 469 , 489 , 519 , 539 , 559 , 579 , 609 , 629 ,
So its like 9 numbers for every 200 numbers
Hence 90 such numbers.
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Answer:
For any number odd or even but have the sum of digits odd, the next number will have the sum of digits even.
However there are possibilities when the number ends with 9
Example 375 sum of digits = 3 + 7 + 5 = 15 ( odd) The next number 376 sum of digits = 3 + 7 + 6 = 16 (even) Thus it doesn't satisfy the condition.
Example 49 sum of digits = 4 + 9 = 13 (odd) The next number 50 sum of digits = 5 + 0 = 5 (odd) So we have a number 49 satisfying the conditions.
From 0 to 2000 the numbers ending with 9 and having sum of digits odd are 100 numbers. How?
From 0 - 99 9,29,49,69 89 [ five numbers ] From 100 - 199 119,139,159,179,199 [ five numbers]
Thus for every 100 numbers we have five numbers ending with 9 and the sum of digits odd. Therefore for 0 - 2000 we have 20 x 5 = 100 such numbers.
There are some exceptions, like 199, even though it ends with 9 and sum of digits odd, the next number 200 has the sum of digits even.
The expected exceptions are the numbers 199,399,599,799,1099.1299,1499,1699,1899. There are 9 exceptions. The number 999 is expected to be an exception in the above series but it is not as the next number 1000 is also gives odd sum.
Thus the numbers which satisfies the condition is 100 - 9 = 91 numbers.
Step-by-step explanation: