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Question 1197175: I am thinking of more than one number. If you add 1 to my numbers and
then multiply them by 5 you will get multiples of 5 between 40 and 65. What are the numbers that I am thinking of?
Found 3 solutions by greenestamps, math_tutor2020, ikleyn: Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!
Your number: n
1 added to your number: n+1
That number multiplied by 5: 5(n+1)
The resulting number is a multiple of 5 between 40 and 65:
40 < 5(n+1) < 65
8 < n+1 < 13
7 < n < 12
Since the number you ended up with was a multiple of 5, the original number n had to be an integer. So the numbers you are thinking of are the integers between 7 and 12.
ANSWER: Your numbers are 8, 9, 10, and 11.
NOTE: "between 40 and 65" in the statement of the problem is ambiguous; we don't know whether or not 40 and 65 are included. My interpretation is that they are not....
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
A supplement to the answer @greenestamps provided:
If you include each endpoint, then solves to which means your potential number is either 7, 8, 9, 10, 11, or 12. I agree that the term "between" is ambiguous.
Example:
Let's say your number was 8
Add 1 to get 8+1 = 9
Multiply by 5 to get 9*5 = 45
This fits the criteria since it's between 40 and 65
Answer by ikleyn(52792) (Show Source):
You can put this solution on YOUR website! .
Usually, even young students of the 4-th grade level can solve such problems using a backward method
and making inverse operations in the inverse order.
Solving this way, you make two steps:
(1) You divide 40 and 65 by 5, and you conclude that your numbers are integer numbers between 8 and 13.
(2) Then the next and the last step is to subtract 1 from the numbers, obtained in step (1).
Doing it, you conclude that your original numbers are integer numbers between 8-1 = 7 and 13-1 = 12.
ANSWER. Thus you found out that the original numbers are integer numbers between 7 and 12.
Solved for the 4-th grade level without using equations, with full explanations.
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