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Difficult Math Problem #73 - Arithmetic, Number Theory

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Junior | Next Rank: 30 Posts
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Joined: Sun Dec 03, 2006 2:45 pm

Re: Difficult Math Problem #73 - Arithmetic, Number Theory

by projase » Fri Dec 08, 2006 6:38 pm
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
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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Master | Next Rank: 500 Posts
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OA

by 800guy » Mon Dec 11, 2006 5:52 pm
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
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Junior | Next Rank: 30 Posts
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Re: Difficult Math Problem #73 - Arithmetic, Number Theory

by jasminekaur280 » Wed Jun 03, 2020 12:29 am
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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