120 times.
To determine the number of 3-digit numbers that are multiples of 5, we need to find the first and last 3-digit multiples of 5. The first 3-digit multiple of 5 is 100, and the last 3-digit multiple of 5 is 995. To find the total number of such multiples, we can use the formula (Last - First) / 5 + 1 = (995 - 100) / 5 + 1 = 180. Therefore, there are 180 3-digit numbers that are multiples of 5.
The number of combinations possible for taking a specified sub-set of numbers, r, from a set, n, isC(n,r) = n!/[r!(n-r)!]In this case, n = 4 and r = 1, soC(n,r) = 4!/1!3!C(n,r) = 24/6C(n,r) = 4Therefore, there are four possible combinations of the numbers 3, 7, 9 and 4.==================================This contributor strongly disagrees, but is leaving the original answer hereso that others can shop and compare.With their commas, parentheses, and factorials, the formulas are certainly impressive.Only the conclusion is wrong.The question specified "4-digit" combinations, so 'n' and 'r' are both 4.Now, to come down off the pedestal and make it understandable as well asformally rigorous ..."Combination" really means different groups of 4 digits that you can select outof the four you gave us. There's only one of those groups.If you actually want to know how many different 4-digit numbers you can makefrom them, then what you want is called "permutations" of the four digits, andyou can think of it this way, without using 'n', 'r', or parentheses or factorials:The first digit can be any one of 4. For each of those . . .The second digit can be any one of the remaining 3. For each of those . . .The third digit can be either one of the remaining 2.So the total number of different ways to line them up is (4 x 3 x 2) = 24 different 4-digit numbers.
You look at the digit in the one millions position, which is a 4. That causes the number to be rounded down to 530000000.
Let us assume you can repeat numbers. That is to say, we allow 101 and are not bothered by the repeat 1, or 111 for that matter. Let's do it first so that the order matters, that is to say 123 is not the same as 321. The possible numbers for the first digit are 0,1,2,...9 so there are 10 of them, Same for the second. Now for the digit "in front", for example in 123 the digit 1, we must decide if we can allow a 0. To make things easy let's say yes for now. Then there are 10 choices for each digits and 10^3 total choices=1000 Now, lets narrow things down just in case we have some restrictions. If exclude 0 from the digit in front, then there are only 9 choices for that place and we have 900 possibilities. Now you said combinations and that usually means that order does not matter. A permutation is where order matters. So if order does not matter 123 is the same as 321 or 213. This means we over counted and now must divided. Each group of 3 numbers such as 123 can be arranged in 6 different ways, which is 3! for example 123 132 213 232 312 321 We we divided by 3! Assuming the leading 0 was allowed we have 1000/3! combinations, if not we have 900/3! combinations.
they are written upside down
The number 6
a palindrome
9116
basicly if you write five numbers together, two that are the same over but only have five numbers on your calculator and tip it upside down then you will be happy :) 9696 or 6969 tip it updi down then it is the same :) Thanks Pengbeargirls!
Convert to Arabic number (IX = 9) and turn the digit upside down (6).
6
Because their process changes
The answer is 888.Also:101111181808818
11 and 88
'8' becomes '8' And '6' becomes '9'.
The closest you can get is with the numbers: 316. When you hold the calculator upside down, you will see something similar to "Pie."