Find Numbers Below 43,125: Odd Tens of Thousands Digit Challenge

Q: How do I find the tens of thousands digit in any number?

Count five positions from the right! In 32,100: ones(0), tens(0), hundreds(1), thousands(2), tens of thousands(3). The fifth digit is always the tens of thousands place.

Q: What if the number has fewer than 5 digits?

Numbers with fewer than 5 digits have 0 in the tens of thousands place. For example, 2,450 effectively has 02,450, so the tens of thousands digit is 0 (which is even).

Q: How can I remember which digits are odd and even?

Odd digits: 1, 3, 5, 7, 9 (they end in these numbers when you count). Even digits: 0, 2, 4, 6, 8. Remember: 0 is even!

Q: Why isn't 54,123 correct even though 5 is odd?

Even though 54,123 has an odd tens of thousands digit (5), it fails the first condition: it must be less than 43,125. Since 54,123 > 43,125, it's automatically eliminated.

Place Value Identification with Digit Constraints

Which number is less than 43,125 and has an odd tens of thousands digit?

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Which number is less than 43,125 and has an odd tens of thousands digit?

Step-by-step solution

To solve this problem, let's follow these steps to determine the correct choice:

Step 1: Identify if the number is less than 43,125.
Step 2: Check if the tens of thousands digit is odd.

Step 1: Let's evaluate each given choice based on the condition that the number must be less than 43,125:

Option 1: $26,999$ is less than 43,125.
Option 2: $32,100$ is less than 43,125.
Option 3: $54,123$ is greater than 43,125.
Option 4: $2,450$ is less than 43,125.

Step 2: Now let's check the tens of thousands digit for oddness:

Option 1: $26,999$ has a tens of thousands digit of 2 (even).
Option 2: $32,100$ has a tens of thousands digit of 3 (odd).
Option 3: $54,123$ has a tens of thousands digit of 5 (odd) but doesn't satisfy condition 1.
Option 4: $2,450$ has a tens of thousands digit of 0 (even).

From these evaluations, only $32,100$ is less than 43,125 and has an odd tens of thousands digit of 3.

Therefore, the solution to the problem is $32,100$ .

Final Answer

$32,100$

Key Points to Remember

Essential concepts to master this topic

Place Value Rule: Tens of thousands digit is the fifth digit from right
Technique: Check 32,100: tens of thousands digit is 3 (odd)
Verification: Confirm both conditions: 32,100 < 43,125 and digit 3 is odd ✓

Common Mistakes

Avoid these frequent errors

Confusing place value positions when counting digits
Don't count from the left or mix up thousands with tens of thousands = wrong digit identification! Students often grab the wrong digit position, leading to incorrect odd/even analysis. Always count digit positions from right to left starting with ones place.

Practice Quiz

Test your knowledge with interactive questions

What number do the units shown below represent?

Write the units in the place value chart and convert into a number.

FAQ

Everything you need to know about this question

How do I find the tens of thousands digit in any number?

Count five positions from the right! In 32,100: ones(0), tens(0), hundreds(1), thousands(2), tens of thousands(3). The fifth digit is always the tens of thousands place.

What if the number has fewer than 5 digits?

Numbers with fewer than 5 digits have 0 in the tens of thousands place. For example, 2,450 effectively has 02,450, so the tens of thousands digit is 0 (which is even).

How can I remember which digits are odd and even?

Odd digits: 1, 3, 5, 7, 9 (they end in these numbers when you count). Even digits: 0, 2, 4, 6, 8. Remember: 0 is even!

Why isn't 54,123 correct even though 5 is odd?

Even though 54,123 has an odd tens of thousands digit (5), it fails the first condition: it must be less than 43,125. Since 54,123 > 43,125, it's automatically eliminated.

Can a number satisfy one condition but not the other?

Yes! For example, 26,999 is less than 43,125 but has an even tens of thousands digit (2). Both conditions must be met simultaneously for the correct answer.

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