Thread: Finding how many numbers satisfy given restrictions on their digits

In a three-digit number, if the hundreds digit is twice the tens digit and tens digit is twice the units digit, then how many such numbers can be formed?

Since the digits have to be 9 or less, you're limited in your options. Also, since the tens and hundreds digits are even, you have even fewer choices.

So think about the ones digit. If it were 5, then the tens digit would be 10, which won't work. What is the largest value that the ones digit can be?

If the ones digit were 0, then we don't really have a "number". So what is the smallest value that the ones digit can be?

Do these same considerations with also the hundreds digit in mind (that is, repeat the above, but multiply again by 2 to find out if the hundreds digit will "work").

Since the tens and hundreds digits are completely determined by the ones digit, then you only really need to find the number of possibilities for the ones digit.

Given two numbers ‘x’ and ‘y’. If tens place of ‘x’ is 7 and tens place of ‘y’ is 6, then find the value in the tens place of the product?

Is this a different question...? (I'm not seeing how it relates to the topic of this thread, is why I ask.)

What have you tried so far? Where are you stuck?

Please be complete. Thank you!