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.
121.
2 or 62 no because 5 is not divisible in 2 or 62.
A number leaves reminder 6 when divided by 10. What is the remainder when the number is divided by 5? Justify your reasoning.
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
23457 divided by 23456 leaves a remainder of 1, and it is divisible by 7. So that is one of infinitely many possible answers.
Zero is even. An integer is even if, when divided by 2, it leaves no remainder. 0/2 = 0, with no remainder.
41
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.