Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.
(A) 1050
(B) 10050
(C) 5050
(D) 5000
(E) 50000
Difficult Math Problem #73 - Arithmetic, Number Theory
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Those are multiples of 5 and odd numbers...
The multiples of 5 between 1 and 1000 are 200 numbers, but you must take only the odd numbers...
5x(1+3+5+...+197+199)=50000.
Regards,
PR
The multiples of 5 between 1 and 1000 are 200 numbers, but you must take only the odd numbers...
5x(1+3+5+...+197+199)=50000.
Regards,
PR
800guy wrote:Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.
(A) 1050
(B) 10050
(C) 5050
(D) 5000
(E) 50000
Where there's a will, there's a way...
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OA:
Consider 5 15 25 ... 995
l = a + (n-1)*d
l = 995 = last term
a = 5 = first term
d = 10 = difference
995 = 5 + (n-1)*10
thus n = 100 = # of terms
consider 5 10 15 20.... 995
995 = 5 + (n-1)*5
=> n = 199
Another approach...
Just add up 995 + 985 + 975 + 965 + 955 + 945 = 5820, so it has to be greater than 5050, and the only possible choices left are 1) & 4)
Also, series is 5 15 25.... 985 995
# of terms = 100
sum = (100/2)*(2*5 + (100-1)*10) = 50*1000 = 50000
Consider 5 15 25 ... 995
l = a + (n-1)*d
l = 995 = last term
a = 5 = first term
d = 10 = difference
995 = 5 + (n-1)*10
thus n = 100 = # of terms
consider 5 10 15 20.... 995
995 = 5 + (n-1)*5
=> n = 199
Another approach...
Just add up 995 + 985 + 975 + 965 + 955 + 945 = 5820, so it has to be greater than 5050, and the only possible choices left are 1) & 4)
Also, series is 5 15 25.... 985 995
# of terms = 100
sum = (100/2)*(2*5 + (100-1)*10) = 50*1000 = 50000
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- Junior | Next Rank: 30 Posts
- Posts: 11
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Consider 5 15 25 ... 995
l = a + (n-1)*d
l = 995 = last term
a = 5 = first term
d = 10 = difference
995 = 5 + (n-1)*10
thus n = 100 = # of terms
consider 5 10 15 20.... 995
995 = 5 + (n-1)*5
=> n = 199
Another approach...
Just add up 995 + 985 + 975 + 965 + 955 + 945 = 5820, so it has to be greater than 5050, and the only possible choices left are 1) & 4)
Also, series is 5 15 25.... 985 995
# of terms = 100
sum = (100/2)*(2*5 + (100-1)*10) = 50*1000 = 50000
l = a + (n-1)*d
l = 995 = last term
a = 5 = first term
d = 10 = difference
995 = 5 + (n-1)*10
thus n = 100 = # of terms
consider 5 10 15 20.... 995
995 = 5 + (n-1)*5
=> n = 199
Another approach...
Just add up 995 + 985 + 975 + 965 + 955 + 945 = 5820, so it has to be greater than 5050, and the only possible choices left are 1) & 4)
Also, series is 5 15 25.... 985 995
# of terms = 100
sum = (100/2)*(2*5 + (100-1)*10) = 50*1000 = 50000