You can do this by finding all of the prime factors of every number less than or equal to 12, then multiplying them by the largest number of times that they occur in any given number, and then multiplying them all together
1 = 1
2 = 2
3 = 3
4 = 2 * 2
5 = 5
6 = 2 * 3
7 = 7
8 = 2 * 2 * 2
9 = 3 * 3
10 = 2 * 5
11 = 11
12 = 3 * 2 * 2
So we need to multiply out:
1 * 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 = 27720
but you could go down to points.
2520
13
100015
There can be no such number. Suppose n is a number that is divisible by k numbers. Suppose p is any integer greater than 1. Then n*p is a number that is divisible by at least k+2 numbers.In this way it is always possible to find another number that is divisible by more numbers.
Oh, what a delightful little puzzle we have here! To find a four-digit number that is divisible by both 8 and 9, we can look for a number that is divisible by their least common multiple, which is 72. So, let's find a multiple of 72 that is a four-digit number, like 1008. Such a harmonious number, don't you think?
2520
13
100015
To find the least 4-digit number divisible by 2, 3, 4, 5, 6, and 7, we need to find the least common multiple (LCM) of these numbers. The LCM of 2, 3, 4, 5, 6, and 7 is 420. The smallest 4-digit number divisible by 420 is 1050. Therefore, 1050 is the least 4-digit number divisible by 2, 3, 4, 5, 6, and 7.
There can be no such number. Suppose n is a number that is divisible by k numbers. Suppose p is any integer greater than 1. Then n*p is a number that is divisible by at least k+2 numbers.In this way it is always possible to find another number that is divisible by more numbers.
Oh, what a delightful little puzzle we have here! To find a four-digit number that is divisible by both 8 and 9, we can look for a number that is divisible by their least common multiple, which is 72. So, let's find a multiple of 72 that is a four-digit number, like 1008. Such a harmonious number, don't you think?
If the last two digits of a number are divisible by 4, the whole number is divisible by 4.
A 4-digit number divisible by both 5 and 9 must be divisible by their least common multiple, which is 45. To find a 4-digit number divisible by 45, we need to find a number that ends in 0 and is divisible by 45. The smallest 4-digit number that fits these criteria is 1005 (45 x 22 = 990, and adding 15 gives us 1005).
To find a number that is divisible by both 2458 and 10, we need to find the least common multiple (LCM) of the two numbers. The LCM of 2458 and 10 is the smallest number that is a multiple of both 2458 and 10, which is 24580. Therefore, 24580 is the number that is divisible by both 2458 and 10.
To find the least number that must be added to 37969 to make it exactly divisible by 65, first, we calculate the remainder when 37969 is divided by 65. The remainder is 44 (since 37969 ÷ 65 = 584 with a remainder of 44). To make it divisible by 65, we need to add (65 - 44 = 21). Thus, the least number that must be added is 21.
To find if a number is divisible by 5, the digit in the ones place has to be 0 or 5.
nothing it is divisible by nothing well at least i couldn't find anything i could divided it by to give me a whole number so if anyone differs please do tell =)