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):
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