An interesting question. The answer is 80%. A number must end in 0 or 5 to be dividable by 5. From 100 to 109 (10 numbers in total), there will be 2 numbers that are dividable by 5. If I then consider the entire set from 100 to 999 is 900 numbers, you have 90 sets of 10 numbers, each set with 2 dividable number, or 180/900 = 2/10 =20%. If 20% are dividable, 80% aren't. Suppose I extended the question, and say my set is all whole positive numbers and zero with n digits or less, where I pick n randomly from 1 to a million. If from this set, a number is picked randomly, what is the chance that it is dividable by 5. Answer: 20%.
Chat with our AI personalities
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
It is possible to create a 3-digit number, without repeated digits so the probability is 1.
The question is ambiguous. It is in complete.
9/10
A three digit number can be chosen in 9*10*10 = 900ways No. of ways to choose a three digit number without 7 is 8*9*9 = 648 ways probability of picking a three digit number that includes atleast one digit that is 7 is 1- (648/900) = 252/900 = 7/25
49/9000
There is 100% chance.
10,000,000 to one.
1 in 4,782,969
None of the digits can be 10, so the probability is 0.
The answer is 9*9!/9*109 = 0.0003629 approx.
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
It is possible to create a 3-digit number, without repeated digits so the probability is 1.
Yes, yes, and no. 3- sum of digits must be multiple of 3. 6- sum of digits must be multiple of 3 and number must be even (multiple of 2). 9- sum of digits must be multiple of 9. (The sum of the digits here is 21.)
2: The number ends in 0, 2, 4, 6 or 8 4 : The last 2 digits are a multiple of 4 The 10s digit is even and the last digit is 0, 4 or 8 The 10s digit is odd and the last digit is 2 or 6 8: The last 3 digits are a multiple of 8 The 100s digit is even and the last 2 digits are a multiple of 8 The 100s digit is odd and the last 2 digits are 4 times an odd number 16: The last 4 digits are a multiple of 16 The 1,000s digit is even and the last 3 digits are a multiple of 16 The 1,000s digit is odd and the last 3 digits are 8 times an odd number 32: The last 5 digits are a multiple of 32 The 10,000s digit is even and the last 4 digits are a multiple of 32 The 10,000s digit is odd and the last 4 digits are 16 times an odd number 64: The last 6 digits are a multiple of 64 The 100,000s digit is even and the last 5 digits are a multiple of 64 The 100,000s digit is odd and the last 5 digits are 32 times an odd number 128: The last 7 digits are a multiple of 128 The 1,000,000s digit is even and the last 6 digits are a multiple of 128 The 1,000,000s digit is odd and the last 6 digits are 64 times an odd number 256: The last 8 digits are a multiple of 256 The 10,000,000s digit is even and the last 7 digits are a multiple of 256 The 10,000,000s digit is odd and the last 7 digits are 128 times an odd number 512: The last 9 digits are a multiple of 512 The 100,000,000s digit is even and the last 8 digits are a multiple of 512 The 100,000,000s digit is odd and the last 8 digits are 256 times an odd number 1,024: The last 10 digits are a multiple of 1,024 The 1,000,000,000s digit is even and the last 9 digits are a multiple of 1,024 The 1,000,000,000s digit is odd and the last 9 digits are 512 times an odd number
you forgot the last 2 digits of your user ID for a games website. You know that both digits are odd. Find the probability that you type the correct last digits by randomly typing 2 odd numbers
22