What is the maximum number of 3-digit conscequetive integers that have atleast one odd digit?

2 Answers
Mar 8, 2016

997, 998 and 999.

Explanation:

If the numbers have at least one odd digit, in order to get the highest numbers let's chose 9 as the first digit. There is no restriction on the other digits, so the integers can be 997, 998 and 999.

Or you wanted to say at THE MOST one odd digit.

So let's chose 9 again. The other digits can not be odd. Since in three consecutive numbers, at least one must be odd, we cannot have three consecutive numbers in which 9 is the first digit.

So, we have to reduce the first digit to 8. If the second digit is 9, we can't have three consecutive numbers only with even numbers,unless the last of these numbers i 890, and the others are 889 and 888.

Mar 9, 2016

111

Explanation:

If I am interpreting the question correctly, it is asking for the length of the longest sequence of consecutive 3-digit integers such that each integer contains at least one odd digit.

Any such sequence would necessarily include either 100199, 300399, 500599, 700799, or 900999.

We can discard 100=199 as for any other sequence we gain additional values by subtracting from the lower end, whereas for 100 we would go into 2-digit integers, which are not allowed.

As adding 1 to any of 399,599,799,999 generates either an integer with no odd digits or with more than 3 digits, one of those will be the greatest integer in the sequence. As there is no benefit to choosing one over another, we can choose one at random, say, 399.

Counting down, as all of the 300s have the first digit as odd, we only need to pay attention when we enter the 200s. As we count down, all of the 290s have the second digit as odd, and 289 has the third digit as odd. Beyond that, we hit 288 which would break the sequence. Similarly, if we tried with any other starting point, we would find that the longest sequence we could generate would be one of

289399, 489599, 689799, or 889999.

each of which has a length of 111.