Easiest way to answer this is list the multiples of 8 and then spot the one with a 2 in it.
Multiples of eight:
8, 16, 24, 32, 40, 48, 56...etc
Therefore you could say the answer is:
24
or
32
Divisibility Rules 2 if it is even 3 if the sum of digits is a multiple of 3 4 if the last two digits is a multiple of 4 6 if it is divisible by 2 and 3 8 if the last three digits is multiple of 8
The unit digits are 2/4/6/8/0
64
Start with the smallest multiple of 13 and continue with the next smallest until finding one that fits the specifications. 13, sum of digits is 1 + 3 = 4 which is not prime 26, sum of digits is 2 + 6 = 8 which is not prime 39, sum of digits is 3 + 9 = 12 which is not prime 52, sum of digits is 5 + 2 = 7 which is prime So, 52 is the smallest positive multiple of 13 for which the sum of its digits is prime.
Here are two solutions: 1234567890 and 9876543210. There are many more.To be a multiple of 6, the number must be both a multiple of 3 and a multiple of 2. If the sum of the digits is a multiple of 3, then the number is a multiple of 3. Adding up all digits (0 through 9) gives a sum of 45, which is a multiple of 3. Since addition is commutative, rearranging the digits in any order will still have a sum of 45, so it's still a multiple of 3. For it to be a multiple of 2, the last digit (the one's place) must be {0,2,4,6, or 8}
Divisibility Rules 2 if it is even 3 if the sum of digits is a multiple of 3 4 if the last two digits is a multiple of 4 6 if it is divisible by 2 and 3 8 if the last three digits is multiple of 8
72
The unit digits are 2/4/6/8/0
8 and 1
0 or 2 or 4 or 6 or 8 .
64
no, 2+8+4+2 (the digits) is 16 which is not a multiple of three. So it's not a multiple of 6 also
64
7800 is divisible by:2 - because it is 2 if it is even3 - the sum of its digits (7+8+0+0 = 15) is a multiple of 3 :4 - the last two digits is a multiple of 45 - it ends or 5 with 06 - it is divisible by 2 and 38 - the last three digits (800) is multiple of 810 - it ends with 0
9 + 1 + 8 + 2 + 7 + 3 + 6 + 4 = 40
By 'after 8000' it looks like 'greater than 8000' is what you want.So to get multiple of 4, last 2 digits must be a multiple of 4. {This is because 100 and every multiple of 100 are multiples of 4, so you only have to look at the last two digits}.For multiple of 9, digits must sum up to 9 (or multiple of 9). So 8001 sums up to 9, but it's not multiple of 4. 8010 doesn't work either. So let's look at digits sum to 18. We already have 8, so the last 2 digits sum to 10 would work, and the last 2 digits multiple of 4:19 : 1+9 = 10, but not multiple of 4.28 : 2+8 = 10, and yes it's a multiple of 4.8028 is the answer.
Start with the smallest multiple of 13 and continue with the next smallest until finding one that fits the specifications. 13, sum of digits is 1 + 3 = 4 which is not prime 26, sum of digits is 2 + 6 = 8 which is not prime 39, sum of digits is 3 + 9 = 12 which is not prime 52, sum of digits is 5 + 2 = 7 which is prime So, 52 is the smallest positive multiple of 13 for which the sum of its digits is prime.