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x is a positive integer less than 500. When x is divided by

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x is a positive integer less than 500. When x is divided by [#permalink] New post 05 Nov 2006, 12:47
x is a positive integer less than 500. When x is divided by 7, the
remainder is 1; when x is divided by 3, the remainder is 2. How many
such numbers are possible?
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 [#permalink] New post 05 Nov 2006, 12:59
7m + 1 =3n +2

7m = 3n+1

if n= 2,9,16......

Total 71 numbers
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 [#permalink] New post 05 Nov 2006, 13:21
Here's a helpful thm for these kinds of problems:

Notation: A number n when divided by k has remainder r, we say that
n = r Mod k

E.g., 28 = 4 mod 6
4 = 1 mod 3
0 = n mod n

Thm: If n = r Mod k then n*t = r*t Mod k (t an integer obviously)

E.g., (for this problem)

7 = 1 mod 3 so 7*t = t Mod 3 so 7 = 1 Mod 3, 14 = 7*2 =2 Mod 3, 21 = 7*3 = 3 Mod 3 = 0 Mod 3, 28 = 7*4 = 4 Mod 3 = 1 mod 3, 35 = 2 Mod 3, ...

So the sequence is {1, 2, 0, 1, 2, 0, ...}

The question asks " How many + numbers n < 500 of the form 7*k + 1 are equal to 2 mod 3?" (so maybe it is by now clear that it's 1/3 of the numbers of this form...)

8 = (7*1 + 1) Mod 3 = 7*1 Mod 3 + 1 Mod 3 = 2 mod 3
15 = (7*2 + 1) Mod 3 = 2 Mod 3 + 1 Mod 3 = 0 mod 3
22 = (7*3 + 1) Mod 3 = 3 mod 3 + 1 Mod 3 = 1 mod 3
29 = 2 Mod 3
36 = 0 Mod 3
43 = 1 Mod 3
...
498 = (7*71 + 1) Mod 3 = 71 Mod 3 + 1 Mod 3 = 0 Mod 3

So there are 71 numbers less than 500 of the form 2 Mod 7 and 1 out of 3 is of the form 2 Mod 3 so it's just a matter of checking the endpoints and so the asnwer is 24 because we start with 2 Mod 3 and end with 0 mod 3.

That seems complicated and long-winded, but the principles can help you do these problems in no time (maybe there's a good review of modular arithmetic somewhere on the web).
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 [#permalink] New post 05 Nov 2006, 13:25
trivikram wrote:
7m + 1 =3n +2

7m = 3n+1

if n= 2,9,16......

Total 71 numbers


Nope (Check the numbers) 2 divided by 7 has remainder 2 (not 1), etc..
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 [#permalink] New post 05 Nov 2006, 16:32
joeydvivre wrote:
trivikram wrote:
7m + 1 =3n +2

7m = 3n+1

if n= 2,9,16......

Total 71 numbers


Nope (Check the numbers) 2 divided by 7 has remainder 2 (not 1), etc..


Sorry it should be 24 and not 71

I never mentioned that 2,7,16 leave a reminder when divided by 7 but they do form a pattern leaving a reminder 2 and substituting n in

7m = 3n+1 then 3n+1 is a multiple of 7

when n=2 m=1
when n=9 m=4
when n=16 m=7
....
.....
when n= 163 m=70

So there are 24 numbers like this....
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 [#permalink] New post 06 Nov 2006, 00:35
yes. there are 24.

n can be expressed as
n = 7m + 1, where m = 1, 4, 7, ..
then express m = 3x + 1, where x = 0, 1, 2, 3...

then n = 21x + 8. and 21x+8 < 500, u solve x <=23.

and x can be 0, so total no of x is 24. hence total no of n is 24.
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 [#permalink] New post 07 Nov 2006, 12:34
[quote="tennis_ball"]yes. there are 24.

n can be expressed as
n = 7m + 1, where m = 1, 4, 7, ..
then express m = 3x + 1, where x = 0, 1, 2, 3...

then n = 21x + 8. and 21x+8 < 500, u solve x <=23.

and x can be 0, so total no of x is 24. hence total no of n is 24.[/quote]

Where did you get the values of "m" from, and why did you make M a function of X.
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 [#permalink] New post 07 Nov 2006, 20:42
trivikram wrote:
joeydvivre wrote:
trivikram wrote:
7m + 1 =3n +2

7m = 3n+1

if n= 2,9,16......

Total 71 numbers


Nope (Check the numbers) 2 divided by 7 has remainder 2 (not 1), etc..


Sorry it should be 24 and not 71

I never mentioned that 2,7,16 leave a reminder when divided by 7 but they do form a pattern leaving a reminder 2 and substituting n in

7m = 3n+1 then 3n+1 is a multiple of 7

when n=2 m=1
when n=9 m=4
when n=16 m=7
....
.....
when n= 163 m=70

So there are 24 numbers like this....


What is X? how do we get that?
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Re: Number Properties [#permalink] New post 07 Nov 2006, 21:39
Paayal wrote:
x is a positive integer less than 500. When x is divided by 7, the
remainder is 1; when x is divided by 3, the remainder is 2. How many
such numbers are possible?


Vikram, 24 is answer for this question. What I was asking was what is 'x' in the question.

I agree..it is a series of numbers. But can you please post a few initial numbers in the series?
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 [#permalink] New post 07 Nov 2006, 22:18
x = 7m+1
x = 3n+2

7m+1 = 3n+2
7m = 3n+1

3n+1 must be a multiple of 7

3n+1 = 7, n = 2
3n+1 = 28, n = 9
3n+1 = 49, n = 16

etc... n keeps adding 7.

3n+2 < 500, so bigger number possible is 497 --> n = 165

So number of possible numbers = (165-2)/7 + 1 = 24
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Re: Number Properties [#permalink] New post 09 Nov 2006, 18:54
ak_idc wrote:
Paayal wrote:
x is a positive integer less than 500. When x is divided by 7, the
remainder is 1; when x is divided by 3, the remainder is 2. How many
such numbers are possible?


Vikram, 24 is answer for this question. What I was asking was what is 'x' in the question.

I agree..it is a series of numbers. But can you please post a few initial numbers in the series?


x is the # of "numbers" between 2 and <500 satisfying the given conditions....
Re: Number Properties   [#permalink] 09 Nov 2006, 18:54
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