Numbers & Place Value

Problem

List two-digit numbers from digit conditions

I need to build two-digit numbers. The tens digit has to be 2 or more but less than 4, and the ones digit has to be more than 6 but no greater than 8. I want to count how many different numbers fit both rules.
Base-ten numbers
Your answer
How to solve
Strategy Make a Systematic List — First I figure out exactly which digits are allowed in each place, then I list every two-digit number by pairing each allowed tens digit with each allowed ones digit. Listing them in order makes sure I miss none and repeat none. Once I see the list is built by pairing 2 choices with 2 choices, the pattern shows the count is 2 times 2.
1STEP 1

Find the allowed tens digits

"At least 2" means 2 is allowed, and "less than 4" means 4 is not allowed. So the tens digit can be 2 or 3.

2 ≤ tens < 4 → tens ∈ {2, 3}
2STEP 2

Find the allowed ones digits

"Greater than 6" means 6 is not allowed, and "at most 8" means 8 is allowed. So the ones digit can be 7 or 8.

6 < ones ≤ 8 → ones ∈ {7, 8}
3STEP 3

List every number by pairing the digits

Pairing each tens digit with each ones digit lists 27, 28, 37 and 38.

{27, 28, 37, 38}
4STEP 4

Count the list and notice the pattern

There are 4 numbers in the list. The pattern is that 2 tens choices each give 2 ones choices, so the total is 2 groups of 2.

2 × 2 = 4
Answer
4 numbers
The answer counts whole numbers, so a small whole number like 4 makes sense. Each of the 4 numbers (27, 28, 37, 38) really does have a tens digit of 2 or 3 and a ones digit of 7 or 8, and no other two-digit number fits, so 4 is complete.
Takeaway

Find which digits are allowed in each place, then pair them up and count: 2 tens choices times 2 ones choices gives 4 numbers.

  • Find the allowed tens digits
  • Find the allowed ones digits
  • List every number by pairing the digits
  • Count the list and notice the pattern
Where next?
Another one like thissuggested

▶ Practice — 12 problems