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117 is a composite. I know because suming the digits is 9. A number is a multiple of 117 if it's a multiple of 9 and 13 at the same time
To keep it simple: Write a main loop that goes through all the numbers, starting with 2, and incrementing one at a time. Determine whether each number is a prime number. If it is, increment a counter. To determine whether each number is a prime number, either use an inner loop, or a separate function. Test divisibility of the number "n" by every number from 2 to n-1. If you find a factor, then it is not a prime number. Note that you can test divisibility by using the "%" operator. For example: if (number % factor == 0) // number is divisible by factor else // it isn't
Yes, 104 is a multiple of four. You can tell by using the divisibility rules : if the last two digits of a number are divisible by four, then the whole number is divisible by four. Now lets go back, in case you don't know, divisibility is simply if a smaller number, like four, can go into a larger number, like 104, perfectly;without having to use decimals.
all digits in the number must total to a number divisible by 3 when added up e.g 156=1+5+6=12 which is divisible by 3 you can proove this using other numbers eg 39,24,1107 e.t.c
Yes, you can. Using the divisibility rules, you can quickly tell that 87 is divisible by 3 (the digits add up to 15, which is divisible by 3). 3 x 29 = 87
In a calculater or by hand do 295/8 OR by using the divisibility rule of 8 if the last 3 digits are divisible by 8, so is the entire number. So you would still do 295/8 but now you know why.
117 is a composite. I know because suming the digits is 9. A number is a multiple of 117 if it's a multiple of 9 and 13 at the same time
You can determine if a number is divisible by 7 by using the following method: E.g. 161 - Double the last digit (2 x 1 = 2) Subtract 2 from the number formed by the the remaining digits in this case, 16 -2 = 14 If the result is divisible by 7 (14 /7 =2); then the original number is divisible by 7. 161 / 7 = 23
To keep it simple: Write a main loop that goes through all the numbers, starting with 2, and incrementing one at a time. Determine whether each number is a prime number. If it is, increment a counter. To determine whether each number is a prime number, either use an inner loop, or a separate function. Test divisibility of the number "n" by every number from 2 to n-1. If you find a factor, then it is not a prime number. Note that you can test divisibility by using the "%" operator. For example: if (number % factor == 0) // number is divisible by factor else // it isn't
If you have a few different numbers that you are using, divide them each by 8 and if you get a whole number, that number is divisible. If you are trying to figure out what is divisible by 8, you can use a divisibility test.A number is divisible by 8 if:the number formed by the last three digits is divisible by 8.So, an example of this would be:7, 120.This number is divisible by 8 because 120 (the last 3 digits) is divisible by 8!
Using the divisibility rules, we can tell that 6111 is divisible by 3 (The digits when added together equal 9, which is divisible by 3, so 6111 is divisible by 3). Since 3 is a factor, it is a composite number (a number with more factors than 1 and itself).
Yes, 104 is a multiple of four. You can tell by using the divisibility rules : if the last two digits of a number are divisible by four, then the whole number is divisible by four. Now lets go back, in case you don't know, divisibility is simply if a smaller number, like four, can go into a larger number, like 104, perfectly;without having to use decimals.
all digits in the number must total to a number divisible by 3 when added up e.g 156=1+5+6=12 which is divisible by 3 you can proove this using other numbers eg 39,24,1107 e.t.c
Yes, you can. Using the divisibility rules, you can quickly tell that 87 is divisible by 3 (the digits add up to 15, which is divisible by 3). 3 x 29 = 87
No. 189 is only evenly divisible by 3 and 9 (from the set provided). Using the following rules of divisibility on the number 189: Divisible by 2? No - the number is not even Divisible by 3? Yes - the sum of the digits (1 + 8 + 9 = 18) is divisible by 3 Divisible by 4? No - the last two digits are not evenly divisible by 4 Divisible by 5? No - the last digit is not a 0 or a 5 Divisible by 6? No - the number is not even Divisible by 9? Yes - the sum of the digits is divisible by 9 Divisible by 10? No - the number is not divisible by 2 or 5
You can always check on the divisibility of a number by dividing it into another number. But if you know the divisibility rules, you can get that information easier and faster.
You can elimate the even numbers 72 and 42 right away because the only even prime number is 2. Using the divisibility rules, you can eliminate 87; 8 + 7 = 15, which is divisible by 3, which means 87 is divisible by 3. 59 is the prime number.