What is the smallest integer greater than 5 that leaves a remainder of 5 when divided by any of the integers 6,8, and 10?
To find the smallest integer greater than 5 that leaves a remainder of 5 when divided by 6, 8, and 10, we can start by considering the common multiples of these three numbers.
First, let's find the common multiples of 6, 8, and 10:
- The multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...
- The multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, ...
- The multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, ...
Looking at the lists, we can see that the multiples of 6, 8, and 10 coincide when we reach the number 24. However, we are looking for a number greater than 5 that leaves a remainder of 5, so 24 doesn't satisfy this condition.
To find the smallest integer greater than 5 that leaves a remainder of 5 when divided by any of the integers 6, 8, and 10, we need to find the next common multiple. We can find it by finding the least common multiple (LCM) of 6, 8, and 10.
To find the LCM, you can use different methods like prime factorization or the ladder method. Let's use the ladder method:
Start with the largest number, 10:
10, 20, 30, 40, 50, 60, ...
Since 8 is not a multiple of 10, move to the next multiple of 8:
80, ...
Since 6 is not a multiple of 80, move to the next multiple of 6:
90, ...
So, the LCM of 6, 8, and 10 is 90.
Now, to find the smallest integer greater than 5 that leaves a remainder of 5 when divided by 6, 8, and 10, we can start with the first multiple of 90 that is greater than 5 and add 5 to it:
90 + 5 = 95.
Therefore, the smallest integer greater than 5 that leaves a remainder of 5 when divided by any of the integers 6, 8, and 10 is 95.