The number must be a multiple of 5. There are only two prime numbers which satisfy the last requirement: 23 & 29. 29 x 5 = 145, which is too big and 23 x 5 = 115.
Any number that ends with 5 is a multiple of 5.
Any number that ends with 0, 2, 4, 6, or 8 is a multiple of 2.
Any number that ends with 0, 2, 4, 6, or 8 is a multiple of 2!
Any number that ends with 0, 2, 4, 6, or 8 is a multiple of 2.
No... Every whole number that ends with a 5 includes 5 as a factor (is a multiple of 5).
Every other number that ends in 5 is a multiple of 5 this is why
Any number that ends with 0, 2, 4, 6, or 8 is a multiple of 2.
Any number that ends with 0, 2, 4, 6, or 8 is a multiple of 2.
Any number that ends with 5 is a multiple of 5.
Any number that ends with 0, 2, 4, 6 or 8 is a multiple of 2.
If a number ends in 0, 2, 4, 6 or 8 it is even. It is, therefore, the number 2 or a multiple of 2. 92 is not the number 2; so it is a multiple of 2 and therefore composite.
No. Go to www.ask.com and enter your question in the search box. First result there is this which gives you the answer. http://www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-factorization.php?number=176 If the number ends in 0, 2, 4, 6, or 8, it is divisible by two. If it ends in 0 or 5, it is divisible by five. If its digits total some multiple of 3, it is divisible by 3.