The product of all prime numbers between 1 and 89 is divided by 10 what is the remainder?
To find the remainder when the product of all prime numbers between 1 and 89 is divided by 10, we need to consider the last digit of the product.
First, we start by listing all the prime numbers between 1 and 89: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, and 89.
To find the last digit of the product, we only need to consider the last digit of each prime number. We can see that all the prime numbers have either a last digit of 1, 3, 7, or 9.
If we multiply any odd number (such as 1, 3, 7, or 9) by itself, we always get an odd number. Similarly, if we multiply any even number (such as 2) by an odd number, we always get an even number.
We can use this fact to find the last digit of the product. The only even prime number is 2, and it appears once in the list. Therefore, the product of all prime numbers between 1 and 89 is even, and the last digit must be 0.
Thus, the remainder when the product is divided by 10 is 0.
Suppose that an and b are positive integers with a greater than b, and with lcm[a,b]=462 and gcd(a,b)=33. Find the largest prime factor of a tgat is not a prime factor of b
Recall that for any positive integers a and b, we have gcd(a,b) x lcm(a,b) = ab. Therefore, we have
33 x 462 = ab,
which implies that
a = 33m
and
b = 14n
for some positive integers m and n (since gcd(33,14) = 1). Note that 462 = 2 x 3 x 7 x 11, so the prime factors of a and b are 2, 3, 7, 11, and any other prime factors that may exist for a (but not for b).
Since a > b, we have m > 2n. Moreover, we want to maximize the largest prime factor of a that is not a prime factor of b. Note that if p is a prime factor of a that is not a factor of b, then p must divide m, and so we want to choose m as large as possible subject to the conditions above.
We observe that m cannot be divisible by 2, since otherwise b would have a larger power of 2 as a factor than a. Similarly, m cannot be divisible by 7 or 11. Therefore, m must be of the form
m = 3 x 13 x k
for some integer k > n. We choose k to be the smallest possible integer greater than n that makes m satisfy the condition m > 2n. Therefore, we choose k = n+1, and so
m = 3 x 13 x (n+1).
We can now compute a:
a = 33m = 33 x 3 x 13 x (n+1) = 1287(n+1).
The largest prime factor of a that is not a prime factor of b is then 13.
To verify that 13 is indeed a prime factor of a, note that 1287 is divisible by 13 (since the sum of its digits is divisible by 13), but not by any larger prime that is not a prime factor of b. Therefore, the answer is 13.
What is the greatest three digit multiple of 33 that can be written using three different digits
For a number to be divisible by 33, it must be divisible by both 3 and 11. Therefore, the sum of its digits must be divisible by 3, and the number formed by its last two digits minus the remaining digit must be a multiple of 11.
Since we want the greatest multiple of 33 that can be written using three different digits, we start by considering the digits 9, 8, and 7, which are the three largest digits. We can see that the sum of these digits is 24, which is divisible by 3, so we can permute them to form three-digit numbers that are divisible by 3.
We then check if any of these numbers satisfy the divisibility by 11 condition. We start with 987 and get 87 - 9 = 78, which is not a multiple of 11. We then try 978 and get 78 - 9 = 69, which is also not a multiple of 11. Finally, we try 879 and get 79 - 8 = 71, which is not a multiple of 11 either.
Since none of the numbers formed by the digits 9, 8, and 7 are divisible by 33, we move on to the next highest digit, which is 6. We can see that the only way to form a three-digit number using 6 and two other digits is to use 9 and 8. The sum of these digits is 23, which is not divisible by 3, so we cannot form a multiple of 33 using 6 as one of the digits.
We repeat this process for the digits 5, 4, 3, 2, and 1. We can see that only for the digits 3 and 2 is it possible to form a three-digit number using three different digits that is divisible by 33.
Using the digits 3, 2, and 1, we can form the numbers 321 and 312, both of which have a sum of digits that is divisible by 3. We then check if either of these numbers satisfy the divisibility by 11 condition. For 321, we get 21 - 3 = 18, which is not a multiple of 11. For 312, we get 12 - 3 = 9, which is also not a multiple of 11.
Therefore, the greatest three-digit multiple of 33 that can be written using three different digits is $\boxed{990}$.
Please write normally
The greatest three-digit multiple of 33 that can be written using three different digits is required to be found.
For a number to be divisible by 33, it must be divisible by both 3 and 11. Therefore, the sum of its digits must be divisible by 3, and the number formed by its last two digits minus the remaining digit must be a multiple of 11.
We observe that the digits 9, 8, and 7 are the three largest digits. By considering all permutations of these digits, we can form three-digit numbers that are divisible by 3. However, we realize that none of them satisfy the divisibility by 11 condition.
Moving on to the next highest digit, 6, we can see that the only way to form a three-digit number using 6 and two other digits is by using 9 and 8. The sum of these digits, 23, is not divisible by 3, so we cannot obtain a multiple of 33 using 6 as one of the digits.
We repeat this process for the digits 5, 4, 3, 2, and 1. We observe that only for the digits 3 and 2 is it possible to form a three-digit number using three different digits that is divisible by 33.
Using the digits 3, 2, and 1, we can form the numbers 321 and 312, both of which have a sum of digits that is divisible by 3. We check if either of these numbers satisfy the divisibility by 11 condition. For 321, we get 21 - 3 = 18, which is not a multiple of 11. For 312, we get 12 - 3 = 9, which is also not a multiple of 11.
Therefore, the greatest three-digit multiple of 33 that can be written using three different digits is 990.
990 has same digits THREE DIFFERENT DIGITS
DID U NOT UNDERSTAND
I apologize for the confusion caused earlier. You are correct that 990 does not have three different digits.
To find the greatest three-digit multiple of 33 with three different digits, we can start by noting that the first digit cannot be 9, since any multiple of 33 that starts with 9 would be a four-digit number. Therefore, we start with the digit 8.
We can see that the remaining digits must be either 1 and 4 or 2 and 7, since these are the only pairs of digits that add up to a multiple of 3 and whose difference is a multiple of 11. Given that we want a greater number, we choose the pair 2 and 7, which uses the larger digits.
We can now form the numbers 872 and 827 which are divisible by 33. The greater number among them is $\boxed{ 891}$, which is a multiple of 33 and has three different digits.