If you had tried the first such set you would have seen that the answer is no. 1+2+3+4 = 10 which is not divisible by 4.
No. 26 for instance the sum of the digits is 8 but not divisible by 4. 32 the sum of the digits is 5 but divisible by 4 The rules for some other numbers are 2 all even numbers are divisible by 2 3 The sum of the digits is divisible by 3 4 The last 2 numbers are divisible by 4 5 The number ends in a 0 or 5 6 The sum of the digits is divisible by 3 and is even 7 no easy method 8 The last 3 numbers are divisible by 8 9 The sum of the digits is divisible by 9
The sum of three consecutive odd numbers must be divisible by 3. As 59 is not wholly divisible by 3 the question is invalid. PROOF : Let the numbers be n - 2, n and n + 2. Then the sum is 3n which is divisible by 3. If the question refers to three consecutive numbers then a similar proof shows that the sum of these three numbers is also divisible by 3. Again, the question would be invalid.
There are eight pairs of numbers divisible by three that sum to 150, and 75 is also divisible by three, so: sum = 17 * 75 = 750 + 525 = 1275
A number is divisible by 3 if the sum of its digits is divisible by 3.
If you had tried the first such set you would have seen that the answer is no. 1+2+3+4 = 10 which is not divisible by 4.
No. 26 for instance the sum of the digits is 8 but not divisible by 4. 32 the sum of the digits is 5 but divisible by 4 The rules for some other numbers are 2 all even numbers are divisible by 2 3 The sum of the digits is divisible by 3 4 The last 2 numbers are divisible by 4 5 The number ends in a 0 or 5 6 The sum of the digits is divisible by 3 and is even 7 no easy method 8 The last 3 numbers are divisible by 8 9 The sum of the digits is divisible by 9
The sum of three consecutive odd numbers must be divisible by 3. As 59 is not wholly divisible by 3 the question is invalid. PROOF : Let the numbers be n - 2, n and n + 2. Then the sum is 3n which is divisible by 3. If the question refers to three consecutive numbers then a similar proof shows that the sum of these three numbers is also divisible by 3. Again, the question would be invalid.
There are eight pairs of numbers divisible by three that sum to 150, and 75 is also divisible by three, so: sum = 17 * 75 = 750 + 525 = 1275
All multiples of 6 Numbers are evenly divisible by 6 if they are evenly divisible by both 2 AND 3. Even numbers are always evenly divisible by 2. Numbers are evenly divisible by 3 if the sum of all the individual digits is evenly divisible by 3. For example, the sum of the digits for the number 3627 is 18, which is evenly divisible by 3 but 3627 is an odd number so the number 3627 is not evenly divisible by 6
All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.
No since only numbers in which its sum can be divisible by 3 is divisible by 3. 2 + 4 + 4 is 10, and 10 is not divisible by 3.
If the sum of its' digits are divisible by 3 for all numbers greater than 9.
A number is divisible by 3 if the sum of its digits is divisible by 3.
38
yes 1428/3 = 476 if you add numbers up and you can divide sum by 3 then it is divisible by 3 1+4+2+8 = 15 which is divisible by 3
All multiples of 6 Numbers are evenly divisible by 6 if they are evenly divisible by both 2 AND 3. Even numbers are always evenly divisible by 2. Numbers are evenly divisible by 3 if the sum of all the individual digits is evenly divisible by 3. For example, the sum of the digits for the number 3627 is 18, which is evenly divisible by 3 but 3627 is an odd number so the number 3627 is not evenly divisible by 6