It is the integer that you get by dividing the three digit number by 10. Alternatively, the number of the multiples of ten is that you get when you truncate the rightmost number.
for example assume a three digit number 517
Divide 517 by ten, you get 51.7
The integer number is 51
Alternatively, if you trnucate the rightmost number (7) you get 51 also
then the answer the three digit number 517 has 51 multiples of 10.
In base 10, all numbers whose last digit is a 5 or a 0
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
There are 90 two-digit numbers... starting with 10 and ending with 99.
There are 9*10*9 = 810 such numbers.
720 (10*9*8)
In base 10, all numbers whose last digit is a 5 or a 0
I believe there are 81. There are 9 after every multiple of 10, including 90.
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
the right answer is 10 cause there is o,1,2,3,4,5,6,7,8,9 so there 10 numbers in total which have 1 digit
There are 90 two-digit numbers... starting with 10 and ending with 99.
The only positive two-digit multiple of 12 and 10 is 60.
ten factorial = 10! = 3,628,800
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
72. (with the range of two digit numbers being from 10 to 99).
3
There are 9*10*9 = 810 such numbers.
There are 900 three digit numbers. (99 - 1000) (# of possible numbers in the first position = 9) (# of possible numbers in the second position = 10) (# of possible numbers in the third position = 10) 9 *10 *10 = 900