-3
9.
90 of them.
Divide the two-digit number by the one-digit number. If the remainder is zero then the 2-digit number is a multiple and if not, it is not.
find the diagonal method of two digit number and three digit number
99
-3
one
It is: 10
9.
They're every other counting number beginning with 11 and ending with 99 .
90 of them.
- There are an infinite number of "counting numbers" that are greater than 70. - There are also an infinite number of "counting number" that are multiples of 10. So.... since you used an "OR" statement, this mean how many number are true for both statements above. That would be: AN INFINITE NUMBER of counting numbers. In fact, if you had said "AND", it still would be an infinite number: 80, 90, 100, ... and so on FOREVER. * * * * * The above answer has interpreted the questions as "two-digit counting numbers greater than 70" OR "a multiple of ten". Apart from the fact that there are not an infinite number of two-digit counting numbers greater than 70, the answer would be correct. But the answer could be interpreted as "two digit counting numbers" that are "greater than 70" OR "a multiple of ten". In that case, the first set is {71, 72, ... 99} and the second is {10, 20, 30, ... 90} with an intersection set consisting of {80 and 90} So there are 29 + 9 - 2 = 36 such number.
11 and 88
-99 is the smallest two digit number.-99 is the smallest two digit number.-99 is the smallest two digit number.-99 is the smallest two digit number.
81
Divide the two-digit number by the one-digit number. If the remainder is zero then the 2-digit number is a multiple and if not, it is not.