Divisibility by 5: Finding 3-Digit Numbers

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Homework Help Overview

The discussion revolves around determining how many three-digit numbers are divisible by 5. Participants are exploring the criteria for divisibility by 5 and the range of three-digit numbers.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • One participant provides an initial answer of 136 and seeks verification. Others express confusion regarding this answer and question the reasoning behind it. A different participant estimates the count based on a rough calculation, suggesting around 180 three-digit numbers divisible by 5. Another participant outlines a method using arithmetic progression to arrive at a count of 180.

Discussion Status

The discussion is active, with multiple interpretations being explored. Some participants are questioning the initial answer of 136, while others provide alternative calculations and reasoning that lead to a different conclusion of 180. There is no explicit consensus on the correct number yet.

Contextual Notes

Participants are working within the constraints of the problem, focusing on three-digit numbers and the specific condition of divisibility by 5. There is an emphasis on verifying calculations and understanding the underlying reasoning.

Amith2006
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Homework Statement


How many 3 digit numbers are divisibIe by 5?


Homework Equations





The Attempt at a Solution


I get the answer as 136. could Someone please work it out and check my answer?
 
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Attempt to a solution?
 
I don't get 136. How did you come up with that answer?
 
I don't get 136 either. How did you get that?
 
There are 'roughly' 20 numbers in every 100 that are divisible by 5. There are 'roughly' 900 three digit numbers. So there are 'roughly' 180 three digit numbers divisible by 5. That's enough off from 136 to make me agree with Defennder and Tedjin. You must be wrong.
 
5, 10, 15, 20, 25, 30

Its all the numbers that end in 5 or zero, and then ignore the last one (1000) because it had the temerity to posses 4 digits.

As you start building the sequence, you see the rule easy enough.
 
an=nth term of an AP (Arithmetic progression)
a=1st term
d=common difference

Since the 1st 3 digit no. divisible by 5 is 100
a=100
the last 3 digit no. divisible by 5 is 995
an=995
common difference (d)=5
n=no. of 3 digit nos. divisible by 5

an=a+(n-1)d
995=100+(n-1)5
995-100=(n-1)5
895=(n-1)5
895/5=n-1
179=n-1
179+1=n
n=180

Therefore,
exactly (not roughly) 180 3 digit numbers are divisibe by 5

Absolutely correct Dick!

@Amith2006
How did you get 136?!
:confused:
 
Last edited:
Year that passed since the thread was started was enough to count these numbers using fingers.
 
Aha that made me laugh, insane bump
 

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