yes!
1, 3, 263, 789.
No.
no
No.
To determine the least number to be subtracted from 789 to make it divisible by 56, first, calculate the remainder when 789 is divided by 56. Performing the division, 789 ÷ 56 gives a quotient of 14 and a remainder of 5. To make 789 divisible by 56, subtract this remainder from 56, which is 56 - 5 = 51. Therefore, the least number to be subtracted from 789 is 51.
1, 3, 263, 789.
Nope, divisible by 3 (263 times)
2,367 is evenly divisible by 1, 3, 9, 263, 789 and 2367.
No.
no
No.
264
To determine the least number to be subtracted from 789 to make it divisible by 56, first, calculate the remainder when 789 is divided by 56. Performing the division, 789 ÷ 56 gives a quotient of 14 and a remainder of 5. To make 789 divisible by 56, subtract this remainder from 56, which is 56 - 5 = 51. Therefore, the least number to be subtracted from 789 is 51.
5 789 is divisible by 7 so there is no remainder. The answer is 827.
0.0038
No. It isn't. If it doesn't have an even number in the last one of a number, it isn't
try 3. In looking for primes, a good place to start (after seeing if it even or not,) is to add the digits in the number together. In this case it would be 7+8+9 equals 24. If the sum is divisible by 3, then the original number is divisible by three.