Firstly, I first encountered the question and it said “When the first number is a multiple of 2, the second is a multiple of 3 and the third is a multiple of 4?” As soon as it said “multiple of 2”, “multiple of 3” and “Multiple of 4”. I thought of finding the LCM of all of the numbers so I found the LCM of 2, 3, 4.
LCM of 2,3 and 4 is 12 using the LCM method.
After finding the LCM, I tried to do 12, 13, 14 but it did not work because 13/3 = decimal and 14/4 = decimal. Next, I tried to find it by guessing and after this, I got 14, 15, 16 and I know the LCM is 12 so I tried to do 12 + 2, which is 14.
Finally, I realised that multiples of 12 is the base number and the first number has to be a multiple of 2 so 12 + 2 = 14. Next, I did 12 + 3 = 15 and then 12 + 4 = 16 so then I got a list 14,15,16. I tried to add 12 separately to the number in the list:
14 + 12, 15 + 12, 16 + 12( 26,27, 28). I figured out it also works. Now, I know that the way you have to do is find the LCM and find what the multiple it needs to be for each number and add them each separately by the LCM.
First of all, to solve a problem like this, we can say that the first set of Numbers that satisfy the question is of course 1, 2, 3, 4… 10. However, if we want to find the next sequence of Numbers that satisfy the question, we must find the LCM, or the number that is divisible by all of the numbers, of 1, 2, 3, 4…10.
To find the LCM of all of the numbers from 1 to 10, we can use the LCM method to discover that the LCM is equal to 35 x 72, which is 2520.
Therefore, once we know that the LCM is equal to 2520, we can add one to get the number that is divisible by 1, which is 2521, which we can tell that it is divisible by 1, because all numbers are divisible by 1.
Next, we can add 2 to the LCM to get the number that is divisible by 2 we then get 2522 which is divisible by 2 because it is even.Using the same method we can add 3 to the number to get 2523, which is divisible by 3 because the digit sum is divisible by 3.
If we tried to check with every number, it would take too long to check, so let us check by adding 7 to 2520, which is 2527, which is divisible by 7 using short division/bustop.
We can give an educated guess by saying that this would work, because we haven’t checked all of the numbers, however we can be quite sure that it works.
To find the next set of numbers that are divisible by all of the numbers from 1-10, we can simply double or multiple 2520 by two to get 5040.
To check that this method will work, we can do the same method of checking that it will work, however it will be much shorter in explanation because we have already done it one time.
First number: 5040 + 1 = 5041/1 ✅by 1
Second number: 5040 +2 =5,042/2 ✅by 2
Third number: 5040 + 3 = 5043/3 ✅by 3
Fourth Number: 5040 + 4 + 5044/4 ✅ by 4
Fifth number: 5040 + 5 = 5045/5 ✅by 5
Sixth number: 5040 + 6 = 5046/6 ✅by 6
Seventh number: 5040 + 7 = 5047/7 ✅by 7
Eighth number: 5040 + 8 = 5048/8 ✅by 8
Nineth number: 5040 + 9 = 5049/9 ✅by 9
Last number: 5040 + 10 = 5050/10 ✅by 10
Or [5041,5042,5043,5044,5045,5046,5047,5048,5049,5050]
We can find more sets of numbers that are divisible by all of the numbers from 1 - 10, by going through the multiples of 2520, as checked above using the same method of addition.