How many of the three digit natural numbers are
divisible by 5.
I will mark U as the Brainliest..
Answers
Answered by
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How many three-digit natural numbers are divisible by 5?
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23 ANSWERS

Doug Dillon, PhD Mathematics, Queen's University, Professional Bassist
Answered Feb 10, 2018 · Author has 5.2kanswers and 1.7m answer views
Originally Answered: How many three digit natural numbers are divisible by 5?
There are 900900 3-digit numbers starting at 100100and ending at 999999 and so 1515 or 180180 are divisible by 55.
1.6k Views · View 1 Upvoter
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Aman Konar, former Student
Answered Mar 9, 2018
Originally Answered: How many three digit numbers are divisible by 5?
A.P=100,105,110,………..,995
Therefore,a=100
d=5
a+(n-1)d=995
100+(n-1)(5)=995
(n-1)(5)=895
(n-1)=179
n=180
Therefore ,there are 180 three digit numbers which are divisible by 5
PLEASE MARK IT AS BRAINLIEST
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23 ANSWERS

Doug Dillon, PhD Mathematics, Queen's University, Professional Bassist
Answered Feb 10, 2018 · Author has 5.2kanswers and 1.7m answer views
Originally Answered: How many three digit natural numbers are divisible by 5?
There are 900900 3-digit numbers starting at 100100and ending at 999999 and so 1515 or 180180 are divisible by 55.
1.6k Views · View 1 Upvoter
RELATED QUESTIONS (MORE ANSWERS BELOW)
How many three-digit numbers are divisible by 13?
2,195 Views
How many two digit numbers are divisible by 7?
2,093 Views
How many three digit numbers are there which are after dividing by 9, sum of digits decrease by 9?
1,330 Views
How many three digit numbers are divisible by 7?
5,791 Views
How many three digit numbers less than 600 are divisible by the first 3 prime numbers?
1,019 Views
OTHER ANSWERS

Aman Konar, former Student
Answered Mar 9, 2018
Originally Answered: How many three digit numbers are divisible by 5?
A.P=100,105,110,………..,995
Therefore,a=100
d=5
a+(n-1)d=995
100+(n-1)(5)=995
(n-1)(5)=895
(n-1)=179
n=180
Therefore ,there are 180 three digit numbers which are divisible by 5
PLEASE MARK IT AS BRAINLIEST
Answered by
8
SOLUTION..........
Three Digits Numbers :-
100, 101, 102, ......... 999.
----- Three digits Numbers which are divisible by 5 :-
100, 105, 110, 115, 120,...............995.
The First term [a] = 100
and
The common difference [d] = 5
And
the last term [an] = 995.
Now,
an = a + [n -1 ] d
{ where; n is the no of digits which divisible by 5.}
995 = 100 + [n -1] × 5
995-100 = [n -1] × 5
[n-1] × 5 = 885
☛ ⛧⛧By, Ⓜr. Thakur ⛧⛧
Three Digits Numbers :-
100, 101, 102, ......... 999.
----- Three digits Numbers which are divisible by 5 :-
100, 105, 110, 115, 120,...............995.
The First term [a] = 100
and
The common difference [d] = 5
And
the last term [an] = 995.
Now,
an = a + [n -1 ] d
{ where; n is the no of digits which divisible by 5.}
995 = 100 + [n -1] × 5
995-100 = [n -1] × 5
[n-1] × 5 = 885
☛ ⛧⛧By, Ⓜr. Thakur ⛧⛧
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