9,000,000
9,999,999 is the last 7 digit number
1,000,000 is the first 7 digit number
when you subtract the first from the last and add one you get the above answer.
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
How many two digit numbers are there in which the tens digit is greater than the one’s digit ?
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
There are 84 such numbers.
There are 151 3-digit numbers that are divisible by 6.
Seven of them.
252
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
16, 25, 34, 43, 52, 61, 70 Seven of them.
1 million 1,000,000
That makes:* 8 options for the first digit * 8 options for second digit * 10 options for the third digit * ... etc. Just multiply all the numbers together.
It would be from 1111111 to 9999999, and because 9999999 - 1111111 = 8888888, the answer is 8888888.
How many two digit numbers are there in which the tens digit is greater than the one’s digit ?
Any pair of digits (not including 0), can be used to generate 14 four-digit numbers. If one of the digits is 0, only seven will start with a non-zero digit.
None. 3 digit numbers are not divisible by 19 digit numbers.
With a total of 90,000 five digit numbers, we can conclude that there is 45,000 EVEN five digit numbers.
There are five such numbers.