301
41
Yes. If x is not divisible by 3 then x leaves a remainder of 1 or 2 when it is divided by 3. That is, x is of the form 3y+z where z = 1 or 2. Then x2 = (3y+z)2 = 9y2 + 6yz + z2 = 3(3y2 + 2yz) + z2 The first part of this expression is clearly a multiple of 3, but z2 is not. Whether z = 1 or 2, z2 leaves a remainder of 1 when divided by 3.
Yes, 15298 is divisible by 2 because it is an even number. In mathematical terms, a number is divisible by 2 if it leaves no remainder when divided by 2. In this case, 15298 divided by 2 equals 7649, which is a whole number without any remainder.
121.
2 or 62 no because 5 is not divisible in 2 or 62.
It is an integer which, when divided by 2, leaves a remainder of 1.
6172839
It leaves a remainder of 1 when divided by 2. Or, if the number is n, then n-1 or n+1 is even (divisible by 2).
When a number is divided by 36 and leaves no remainder.
The integer is 157. 157/3 = 52 remainder 1 157/5 = 31 remainder 2 157/7 = 22 remainder 3
85
234568
41
41
Zero is even. An integer is even if, when divided by 2, it leaves no remainder. 0/2 = 0, with no remainder.
23457 divided by 23456 leaves a remainder of 1, and it is divisible by 7. So that is one of infinitely many possible answers.
Yes. If x is not divisible by 3 then x leaves a remainder of 1 or 2 when it is divided by 3. That is, x is of the form 3y+z where z = 1 or 2. Then x2 = (3y+z)2 = 9y2 + 6yz + z2 = 3(3y2 + 2yz) + z2 The first part of this expression is clearly a multiple of 3, but z2 is not. Whether z = 1 or 2, z2 leaves a remainder of 1 when divided by 3.