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A year is called an Additive year when the sum of the number formed by the first two digits and the number formed by the last two digits add up to the number formed by the middle two digits. For example, considering the year 1978 we observe that when 19 is added to 78, the result is 97 which is the number formed by the middle two digits.

(i) Determine the total number of Additive years from years 1000 to 9999 inclusively.

(ii) What are the respective first and last Additive years from years 2000 to 2999 inclusively?

(iii) What are the respective first and last Additive years in the period covered under (i)?

 No Solution Yet Submitted by K Sengupta Rating: 5.0000 (1 votes)

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 Solution | Comment 1 of 6
To solve this problem we let the number be abcd.  Where a,b,c and d are integers between 0 and 9.

Therefore a number is additive if 10a + b + 10c + d = 10b + c.
Therefore,
10a + d = 9(b-c)

This implies that (10a + d) must be a multiple of 9.

For a = 1,
10 + d is a multiple of 9 if d = 8, and (b-c) = 2.
There are 8 combinations of b and c such that b-c = 2.

For a = 2,
10*2 + d is a multiple of 9 if d = 7 and (b-c) = 3.
There are 7 combinations of b and c such that b-c = 3.

This pattern continues...
...
...

For a = 8,
10*8 + d is a multiple of 9  if d = 1 and (b-c) = 9.
There is 1 combination of b and c such that b-c = 9.

For a = 9,
10*9 + d is a multiple of 9 if (d=0, or d = 9).
If d = 0, then (b-c)=10.  There are 0 such combinations.
If d =9, then (b-c) = 11.  There are 0 such combinations.

Therefore answer = 8+7+6+5+4+3+2+1 = 36

 Posted by Chris, PhD on 2012-08-08 11:22:57

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