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Sum of number < 500 that have a reminder of 1 when div by 14

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Sum of number < 500 that have a reminder of 1 when div by 14 [#permalink] New post 05 Nov 2009, 07:15
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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 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.

499/14 = 35*14 + 9
..
..
492/14 = 35*14 + 2
491/14 = 35*14 + 1
490/14 = 35*14 + 0
489/14 = 34*14 + 13
..
..
1/14 = 0*14 + 1

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] New post 05 Nov 2009, 07:40
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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

average (or median) is 17.5 * 14 = 245 + 1 = 246

multiply by 36 and you get 8856
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Re: Sum of number < 500 that have a reminder of 1 when div by 14 [#permalink] New post 05 Nov 2009, 07:55
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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.

499/14 = 35*14 + 9
..
..
492/14 = 35*14 + 2
491/14 = 35*14 + 1
490/14 = 35*14 + 0
489/14 = 34*14 + 13
..
..
1/14 = 0*14 + 1

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.


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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Re: Sum of number < 500 that have a reminder of 1 when div by 14 [#permalink] New post 05 Nov 2009, 08:13
Thanks guys. :-D
I had forgotten about the arithmetic sum formula.

Sn=n*(2a1+d(n-1))/2

where a1 = first term, n = number of terms, d = difference between terms.
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Re: Sum of number < 500 that have a reminder of 1 when div by 14 [#permalink] New post 05 Nov 2009, 08:22
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drukpaGuy wrote:
Thanks guys. :-D
I had forgotten about the arithmetic sum formula.

Sn=n*(2a1+d(n-1))/2

where a1 = first term, n = number of terms, d = difference between terms.


We could use another formula for it as well: Sn=n*(a1+an)/2 (an is the last term 491 in our case).
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Sum of number < 500 that have a reminder of 1 when div by 14   [#permalink] 05 Nov 2009, 08:22
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