SOLUTION: Find all the natural numbers, which when divided by 7, leave the same number for the quotient and the remainder.

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Question 1170387: Find all the natural numbers, which when divided by 7, leave the same number for the quotient and the remainder.
Found 2 solutions by ikleyn, Solver92311:
Answer by ikleyn(52800)   (Show Source): You can put this solution on YOUR website!
.

Let  "n"  be a natural number the problem asks for.


Then according to the condition, we can write


    n = 7*m + m


where an integer number  "m"  is the quotient and the remainder, at the same time.


As a remainder, the number  "m"  is under  inequalities  0 <= m <= 6;  so "m" may have these and only these values

     m = 0, 1, 2, 3, 4, 5, 6.


Accordingly, the number "n" may have these and only these values


    n = 7*0 + 0 = 0

    n = 7*1 + 1 = 8

    n = 7*2 + 2 = 16

    n = 7*3 + 3 = 24

    n = 7*4 + 4 = 32

    n = 7*5 + 5 = 40

    n = 7*6 + 6 = 48


ANSWER.  The possible values are  0, 8, 16, 24, 32, 40, 48.

Solved, answered and explained. And completed.



Answer by Solver92311(821)   (Show Source): You can put this solution on YOUR website!


I'm assuming that you mean the quotient and remainder are equal otherwise your question doesn't make sense. The question is very poorly worded, however. You should also define "Natural Number" since it could mean either the positive integers or the non-negative integers. I will assume the former.

If a natural number divided by another natural number has a quotient and remainder , then the following relation must hold:



But if and are equal, then



For your problem,

So


John

My calculator said it, I believe it, that settles it

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