Divided : 100 / 10 = 10 10 / 10 = 1 1 / 1 =1 You Take Away The Zeros/Add a decimal point. To divide in general, you see how many times one number can be put into another. (Eg. 50/5 = 10) 50/5 - To divide 50 by 5, you must see how many times 5 can be put into 50. You can count up to 50 in 5's: 5 (1), 10 (2), 15 (3), 20 (4), 25 (5), 30 (6), 35 (7), 40 (8), 45 (9), 50 (10). This means that the answer for 50/5 is 10. To divide a 2 didgit number by 1 didgit numbers, "you see how many times one number can be put into another.". If you mean to divide a 1 didgit number by a 2 didgit number, a possible way would be like this: ___ 60|5 (60/5) 0. As 60 does not go into 5, we write "0." as part of the answer. We also change the "5" into a "50" Now we can work out 60/5. 10x 50 < This now means we write 0.12 (12x) --------- 10 2x 10 We add the "--x" together, (10x + 2x), --------- "10 + 2". 0 This gives us "12" We now put that 12 after the decimal point. 0.12. This is the answer. This method is referred to as "chunking".
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.
By using the 3 digits of a number we can form 3 different two digit numbers. 3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3
If you want 4-digit numbers, there are 24 of them.
6! = 720
There are 200 positive four digit integers that have 1 as their first digit and 2 or 5 as their last digit. There are 9000 positive four digit numbers, 1000 through 9999. 1000 of them have 1 as the first digit, 1000 through 1999. 200 of them have 2 or 5 as their last digit, 1002, 1005, 1012, 1015, ... 1992, and 1995.
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There are 28706 such combinations. 5456 of these comprise three 2-digit numbers, 19008 comprise two 2-digit numbers and two 1-digit numbers, 4158 comprise one 2-digit number and four 1-digit numbers and 84 comprise six 1-digit 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.
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Highest 2-digit number = 99 Highest 1-digit number = 9 Highest possible sum from 2, 2-digit numbers = 198 Highest possible sum from 2, 1-digit numbers = 18
There is no such ratio that applies for all single-digit and double-digit integers.
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to do it the long way do it by long division to do it the short way put it in the calculator 2 digits numbers are just the numbers between10-99 and one digits numbers are just the numbers between 0-9.
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.
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Answer1 x 9 (number of 1 digit numbers) plus 2 x 89 (number of 2 digit numbers) plus 3(number of 3 digit numbers) x 347 = 1228 The correct answer is 1323. Here is how you do it.1 digit numbers = 92 digit numbers = 90 (99-9) number of 2 digit numbers - number of 1 digit numbers3 digit numbers = 378 (477-99) number of 3 digit numbers - number of 1 & 2 digit numbersSo, the solution is:1x9 = 92x90 = 1803x378 = 11349+180+378 = 1323