That is false. This type of statement is only true for prime numbers, not for compound numbers such as 6. Counterexample: 2 x 3 = 6
The statement cannot be proven because it is FALSE. If one of x and y is odd and the other is even then x2 + y2 MUST be odd. Also if x and y are even then x2 + y2 MUST be divisible by 4. The statement is only true if x and y are odd integers. Whether or not they are positive makes little difference.
The integers are 99, 100 and 101. There is also a set of consecutive even integers whose sum is 300. That set is 98, 100 and 102.
positive integers
Oh, dude, numbers divisible by both 3 and 9 are multiples of the least common multiple of 3 and 9, which is 9. So, any number that is a multiple of 9 is also divisible by 3 and 9. Like, it's just basic math, nothing to lose sleep over.
To determine if 4908 is divisible by 3, we need to sum its digits. 4 + 9 + 0 + 8 = 21. Since 21 is divisible by 3 (21 ÷ 3 = 7), 4908 is also divisible by 3. Therefore, 4908 is divisible by 3.
It is divisible by their factors. It is also divisible by their product.
8 of them.
The statement cannot be proven because it is FALSE. If one of x and y is odd and the other is even then x2 + y2 MUST be odd. Also if x and y are even then x2 + y2 MUST be divisible by 4. The statement is only true if x and y are odd integers. Whether or not they are positive makes little difference.
Integers divisible by 8 have their last 3 digits divisible by 8. So: 2934829387957008 is divisible by 8 because the last three digits are divisible by 8 (008 / 8 = 1). 2348012934801298304400 is also (400 / 8 = 50). 123006 is not (006 / 8 = 0.75).
Talking in Integers only, then no. Example, 3 is not divisible by 6. Talking all natural numbers... then yes. 3 / 6 = 0.5
Yes. All numbers rationally divisible by 4 are also rationally divisible by 2.All numbers that when divided by 4 result in answers that are integers will also give answers that are integers when divided by 2, and so forth.For example: Let x/4=p,x/2=2.(x/4)=>x/2=2p.from the above illustration,it is clear that all the multiples of 4 are the multiples of 2.hence every number which is divisible by 4 is divisible by 2.
Oh, dude, let me break it down for you. So, first, you find how many numbers are divisible by 6, which is 166. Then you find how many are divisible by 21, which is 47. But wait, don't forget to subtract the ones divisible by both 6 and 21, which is 7. So, the total is 166 + 47 - 7 = 206 integers that fit the bill. Easy peasy!
18
False. The question consists of two parts: - a number is divisible by 6 if it is divisible by 3? False. It must also be divisible by 2. - a number is divisible by 6 only if it is divisible by 3? This is true but the false part makes the whole statement false.
1. 0 because it has to be a divisible by 5 also it has to be divisible by 2.
13 and 14. Also negative 13 and negative 14.
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