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Q: What is the least whole number that is divisible by 7 but leaves a remainder of 1 when divided by any integer 2 - 6?

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It is an integer which, when divided by 2, leaves a remainder of 1.

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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.

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It is an integer which, when divided by 2, leaves a remainder of 1.

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It is an integer which, when divided by 2, leaves a remainder of 1.

It is an integer which, when divided by 2, leaves a remainder of 1.

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

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234568

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23457 divided by 23456 leaves a remainder of 1, and it is divisible by 7. So that is one of infinitely many possible answers.

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