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Sum of number < 500 that have a reminder of 1 when div by 14
[#permalink]
05 Nov 2009, 08:15
1
Kudos
s3017789 wrote:
Can someone explain the division test for 14. If there is one.
The test seems to be if the number is divisible by 2 and 7 since 2*7=14. ex: 56 is divisible by 14 because 56/2=28 and 56/7=8
[quote] I had a test that wanted to know the sum of numbers 500, so we are interested in values under 36. Try 35*14=490. So we are interested in values 35 to 1. Now any number divided by 14 can have remainders 0,1,2,3,4...13.
So the thing is we are interested in values like 491, i.e (multiple of 14) + 1 The sum would look something like this; (35*14) + 1 + (34*14) + 1 + (34*14) + 1 + ..... + (14*1) + 1 + (14*0) + 1 = 36 + 14(35 + 34 + 33 + ... +1) = 36 + 14( (35+1)(35)/2 ) = 36 + (14*18*35) = 36 + 8820 = 8856
I get the 36 by adding up all the 1's And for the sum of 35+34..+1, I'm just using the formula n(n+1)/2. I think it's part of HH's notes. However I haven't verified the answer, and I might be miscounting some where along the line, so this answer is half baked :wink:
UPDATE: Ok I plugged it into the computer and that's the answer I'm getting. I'd hate to get this on an exam though.
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Re: Sum of number < 500 that have a reminder of 1 when div by 14
[#permalink]
05 Nov 2009, 08:40
1
Kudos
I don't know about a rule of 14 but here is my take
Well with these problems it always helps to find the first two values (to get a formula and confirm what I did is correct) and the last value:
First value: 1 because 1/14 has remainder 1 2nd value: 15
formula: 14x + 1 < 500 starting at x = 0 (0 being the 1st value that gives remainder 1 when plugged into the formula)
To get the last value solve 14x + 1 < 500 for x and the answer is 35 (the 35th number); the 35th number is the value 491
the number of integers between 0 and 35 inclusive is 36
the median is the number (35 + 0)/2 = 17.5; that means 17.5th number which must be multiplied by 14x and then add 1 (the formula from above) to get the actual number
when numbers are evenly spaced the mean = median and when multiplied by the number of integers you get the sum
Re: Sum of number < 500 that have a reminder of 1 when div by 14
[#permalink]
05 Nov 2009, 08:55
2
Kudos
Expert Reply
drukpaGuy wrote:
Quote:
I had a test that wanted to know the sum of numbers <500 that had a remainder of 1 when divided by 14.
For this one the only solution I could come up with is as follows:
36*14=504 which is >500, so we are interested in values under 36. Try 35*14=490. So we are interested in values 35 to 1. Now any number divided by 14 can have remainders 0,1,2,3,4...13.
So the thing is we are interested in values like 491, i.e (multiple of 14) + 1 The sum would look something like this; (35*14) + 1 + (34*14) + 1 + (34*14) + 1 + ..... + (14*1) + 1 + (14*0) + 1 = 36 + 14(35 + 34 + 33 + ... +1) = 36 + 14( (35+1)(35)/2 ) = 36 + (14*18*35) = 36 + 8820 = 8856
I get the 36 by adding up all the 1's And for the sum of 35+34..+1, I'm just using the formula n(n+1)/2. I think it's part of HH's notes. However I haven't verified the answer, and I might be miscounting some where along the line, so this answer is half baked
UPDATE: Ok I plugged it into the computer and that's the answer I'm getting. I'd hate to get this on an exam though.
OK first let's determine how many such numbers are <500, that lives a remainder of 1 when divided by 14.
Formula for these numbers is x=14p+1 (where p is an integer >=0). 14p+1<500 --> p<35.6 as p can be zero too, thus we have total of 36 such numbers 1, 15, 29....491 (36th number)
Now since the question is not given there can be two cases: A. We are asked to determine the sum of these 36 numbers 1,15,29,..491 In this case: as we have the arithmetic progression with firs term 1 and common difference of 14, then sum Sn=n*(2a1+d(n-1))/2 (a1 first term, 1 in our case; n number of terms 36 in our case; and d common difference 14 in our case). S=36*(2+14*35)/2=8856.
B. If we are asked to determine just the sum of theses 36 meaning 1+2+3+...+36, then S=n*(1+n)/2=36*37/2=666. _________________
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