You can write that as 1 teaspoon, or 1.0 teaspoon.
93 and 7 would be an example.
6 times
There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.
There are 9 single digits (1, 2, 3, 4, 5, 6, 7, 8, 9) that are repeated throughout the numbers from 1 to 1000000. Each single digit appears 100,000 times (except for 0, which appears 100,000 times as well). Therefore, there are a total of 900,000 single digits in the range from 1 to 1000000.
You can write that as 1 teaspoon, or 1.0 teaspoon.
189 pennies.
That looks like it should be the same as 10 times (the number of times the digit 1 occursin the units place in the numbers from 10 to 100), which would work out to 90 .
You would use a zero 11 times in writing the numbers 1 to 100 once - so doing it five times gives you 55 zeros.
It would help to know which digit. 0 appears in 9 numbers and each of the others in 18 numbers.
£1.92 (or $1.92, of you mean American penny)
Without restrictions, it was would numbers 000-000-0000 through 999-999-9999. So that would be 9,999,999,999 + 1 = 10 billion different 10-digit phone numbers. Ex: If there existed single digit phone numbers, there would be 10, because the digits are 0 through 9. If there existed only double digit phone numbers, then it would be 00 through 99 which would be 100 total two-digit numbers. Therefore the total possible combinations for an X digit phone number would be: 10^X
Technically, that is not possible. A digit is another word for a number and it would be impossible for a single number to contain ten numbers.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
That would be the numbers in the form "32x" (where "x" can be any digit). In other words, ten numbers.
10,000 of them, from 0000 to 9999. However, most people would not count numbers with leading 0s, so that there would be 9000 4-digit numbers.
93 and 7 would be an example.