Question

what number has 9 ten thousands, 2 fewer thousands than ten thousands, 7 fewer hundreds than ten thousands, no tens, and 6 fewer ones than thousands?

Explanation:

Step1: Determine ten - thousands place

The number of ten - thousands is 9, so the digit in the ten - thousands place is 9.

Step2: Determine thousands place

We know that the number of thousands is 2 fewer than ten - thousands. Since ten - thousands digit is 9, the number of thousands is $9 - 2=7$. So the digit in the thousands place is 7.

Step3: Determine hundreds place

The number of hundreds is 7 fewer than ten - thousands. Since ten - thousands digit is 9, the number of hundreds is $9 - 7 = 2$. So the digit in the hundreds place is 2.

Step4: Determine tens place

There are no tens, so the digit in the tens place is 0.

Step5: Determine ones place

The number of ones is 6 fewer than thousands. Since the thousands digit is 7, the number of ones is $7-6 = 1$. So the digit in the ones place is 1.

Now, we combine the digits from ten - thousands to ones place: 9 (ten - thousands), 7 (thousands), 2 (hundreds), 0 (tens), 1 (ones). So the number is 97201.

Answer:

97201