It is only divisible by itself and one so therefore 31 is a prime number
Yes. 1,305/3=435. In the future, you can check the divisibility of numbers as follows: 2=digit in one's place is even 3=sum of digits is a multiple of 3 4=last 2 digits are divisible by 4 5=digit in one's place is either 5 or 0 6=satisfies divisibility rule for both 2 and 3
A one digit number is a number with only one number. For example: 12 is a two digit number. It has two numbers. 3 is a one digit number because it only has one number.
Divide the two-digit number by the one-digit number. If the remainder is zero then the 2-digit number is a multiple and if not, it is not.
To find the greatest four-digit number divisible by more than one number, start with the largest four-digit number, which is 9999. Check its divisibility by the desired numbers (e.g., 2, 3, 5, etc.). If 9999 is not divisible by those numbers, decrement by 1 and check again until you find a number that meets the criteria. This process continues until you identify the largest four-digit number that is divisible by at least two specified numbers.
there is one for seven although its not very useful. take the last digit trunccate it and double it then subtract the rest. If the resulting number is divisible by seven, so was the origanil
There are two parts to the rule: Part 1: If the last but one digit is even, the last digit must be 0, 4 or 8 If the last but one digit is odd, then the last digit must be 2 or 6. Part 2: Find the sum of the first, third, fifth, etc digits. Find the sum of the second, fourth, sixth, etc digits. The difference between these two sums must be a multiple of 11 ie 0, 11, 22, 33 etc. Part 1 ensures divisibility by 4 Part 2 ensures divisibility by 11 Together, they ensure divisibility by 44.
It is only divisible by itself and one so therefore 31 is a prime number
It is only divisible by itself and one so therefore 41 is a prime number
The number must end in one of the following: 00, 20, 40, 60, 80.
14...Checking for divisibility by 14 would combine the rules for 2 and 7, since 2x7 = 14The rule for 2 is that the last digit of the number is even (0,2,4,6 or 8).There is no real rule for 7, but there is one way of testing divisibility by 7 that I have come up with, although rather complicated. First, write your number, then beneath it, under each digit from right to left, write out the numbers 1,3,2,6,4,5 in that order and keep repeating the sequence until each digit of your number has another digit beneath it. Use only the amount of digits needed. Then, multiply each digit of your number by the digit beneath it, and add all of these new numbers. If the number you get is a multiple of 7, the number you began with was also a multiple of 7.Example for 7:We are checking 3192847 for divisibility by 7...3 1 9 2 8 4 71 5 4 6 2 311x3 + 5x1 + 4x9 + 6x2 + 2x8 + 3x4 + 1x7= 3 + 5 + 36 + 12 + 16 + 12 + 791We can test 91 again to find divisibility...9 13 13x9 + 1x1 = 2828 is a multiple of 7 (it is 4 x 7), so 91 and 3192847 must also be multiples of 7. 3192847 is actually 456121 x 7.If both of these tests work, the number must be a multiple of 14.15...Divisibility by 15 can be tested by combining the rules for 3 and 5.The rule for 3 is that the sum of all the digits in the number is a multiple of 3.The rule for 5 is that the number ends in 5 or 0.If both of these tests work, the number must be a multiple of 15.18...Checking for divisibility by 18 combines the divisibility rules for 2 and 9.Once again, a number is divisible by 2 if its last digit is even.A number is divisible by 9 if the sum of its digits is a multiple of 9.If both of these tests work, the number must be divisible by 18.
Yes. 1,305/3=435. In the future, you can check the divisibility of numbers as follows: 2=digit in one's place is even 3=sum of digits is a multiple of 3 4=last 2 digits are divisible by 4 5=digit in one's place is either 5 or 0 6=satisfies divisibility rule for both 2 and 3
For a number to be divisible by 105 it must be divisible by 3, by 5 and by 7. So, divisibility by 3 requires all three of the following to be satisfied:Sum the digits together. Repeat if necessary. If the answer is 0, 3, 6 or 9 the original number is divisible by 3.If the final digit of the number is 0 or 5, the original number is divisible by 5.Take the number formed by all but the last digit. From it subtract double the last digit. Keep going until there is only one digit left. If it is 0 or 7 then the original number is divisible by 7.
There isn't really one that's any use. It's much quicker just to go ahead and do the division!
Oh honey, buckle up. The divisibility rule for 91 is a doozy. Basically, if a number is divisible by 91, it's also divisible by 13 and 7. So, you better grab a calculator and start crunching those numbers if you want to see if 91 can divide into your digits. Good luck, you're gonna need it!
A one digit number is a number with only one number. For example: 12 is a two digit number. It has two numbers. 3 is a one digit number because it only has one number.
Divide the two-digit number by the one-digit number. If the remainder is zero then the 2-digit number is a multiple and if not, it is not.