by one by one
you multiply 3 digit numbers by killing yourself and giving away body parts.
How many two digit numbers are there in which the tens digit is greater than the oneβs digit ?
99*99=9801
There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
by one by one
You multiply the one digit number on the bottom to every number on the top starting at the right and so on with every other number on the bottom.
To multiply two digit numbers, multiply each place value of a factor by each place value digit and add the results.
Any multiple of 36 will do. Multiply 36 by different numbers, until you get a 5-digit number.Any multiple of 36 will do. Multiply 36 by different numbers, until you get a 5-digit number.Any multiple of 36 will do. Multiply 36 by different numbers, until you get a 5-digit number.Any multiple of 36 will do. Multiply 36 by different numbers, until you get a 5-digit number.
9*4*2 = 72
15
There are no such numbers.
Practice!
you multiply 3 digit numbers by killing yourself and giving away body parts.
first digit time second digit and second digit times first digit then repeat
You
Multiply the three-digit number by the one's digit, or last digit, of the two-digit number. That is your first part. Now multiply by the second-to-last digit, or ten's digit, and multiply the result by 10. That is your second part. Add the two parts and that is your answer.