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 Nil Carry Nuance (Posted on 2014-12-04)
Consider the set S ={1000, 1001, ...., 2013, 2014}.

For how many pairs of consecutive integers in S is no carrying required when the two integers are added?

 No Solution Yet Submitted by K Sengupta No Rating

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A pair of consecutive integers in S will have no carrying if the second of the two integers meets the following conditions:

The ones digit is 5 or less, and

The tens digit is 4 or less, plus the cases when it first switches from 4 to 5 (e.g. 1049 + 1050), and

The hundreds digit is 4 or less, plus the cases when it first switches from 4 to 5 (e.g. 1499 + 1500).

Just considering the 1000 integers from 1001 to 2000:

600 of them have 5 or less in the ones place.

Of these, half have 4 or less in the tens place.  An additional 10 must be included for the 10 cases that switch from 4 to 5 (1049+1050, 1149+1150, etc.).  So 310 in total meet the first two conditions.

Of these, half have 4 or less in the hundreds place.  We must add 1 for the case 1499 + 1500, so 156 in total meet all three conditions.

Considering the remaining integers from 2001 to 2014, ten of them meet the first condition, and they all meet all of the other conditions as well.

Thus the grand total is 166.

 Posted by tomarken on 2014-12-04 07:56:29

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