Mathcounts State Perparations: Counting problems

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Problems (solutions below) 
 #1:How many times does the digit 7 appear in the list of all the integers from 1 to 1000 inclusive?

#2. How many integers from 1 to 1000 inclusive do not contain the digit 7?

#3. How many integers from 1 to 1000 inclusive contain the digit 7?

#4.How many four-digit positive integers have at least one digit that is a 1, 3 or 5?

#5. How many three-digit numbers exist of which the middle digit is the average of the first and the last digits?

 

Solutions:

#1. Every 100 will give you 20 “7”s – 10 for the unite digit (07, 17, 27…97) and 10 for the tenth

digit (70, 71, 72…79) so from 1 to 1000 there will be 20 times 10 = 200 times the number 7 appear on the tenth and unit digits.

However, there are 100 times the number “7” appear on the hundredth digit. (from 700 to 799), so the answer is 300.

#2. There are 8 single digit numbers that do not contain the digit 7.

There are 8 x 9 = 72 two digit numbers that do not contain the digit 7.

There are 8 x 9 x 9 = 648 three digit numbers that do not contain the digit 7.

Add them together and + 1 (the number 1000 itself) and you get 729.

#3. Do complementary counting. Use the total minus no "7" appeared.
1000 – 729 (from the previous answer) = 271

#4. Do complementary counting. There are 9 x 10 x 10 x 10 = 9000 four-digit positive integers.

Of those, there are 6 (exclude 1, 3, or 5) x 7 (again exclude 1, 3, or 5) x 7 x 7= 2058  integers without 2 3 or 5.

9000 – 6 x 7 x 7 x 7 = 6942

5. For the middle digit to be an integer, the first and last digits must be both odd or both even.

(Why?) There are 5 x 5 = 25 numbers that the hundredth and unit digit are odd and there are 4 (no zero at the hundredth place) x 5 = 20 numbers that the hundredth and unit digit are both even.

The total is 45.