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When an integer Q is divided by 5, it leaves 1 as a

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When an integer Q is divided by 5, it leaves 1 as a [#permalink] New post 29 Nov 2007, 19:00
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When an integer Q is divided by 5, it leaves 1 as a remainder. When Q is divided by 7, it leaves 3 as a remainder. Find the smallest positive Q. Then find the next smallest positive Q.
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Re: PS division [#permalink] New post 29 Nov 2007, 19:22
young_gun wrote:
When an integer Q is divided by 5, it leaves 1 as a remainder. When Q is divided by 7, it leaves 3 as a remainder. Find the smallest positive Q. Then find the next smallest positive Q.


Q = 5k + 1 and 7t + 3, where k & t are positive integers.

So 5k + 1 = 7t + 3

5k = 7t + 2

k = (7t + 2)/5. We want values of t such that 7t + 2 is divisible by 5. The lowest is 4; the next lowest is 9.

If t = 4, Q = 31.

If t = 9, Q = 66.
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 [#permalink] New post 29 Nov 2007, 21:32
I like johnrb's method. Mine wasn;t as fancy.

Did 2 lists. First, the column on the right with multiples of 7. Then a column on the right, adding 3 to the number on the left:

7 10
14 17
21 24
28 31
35 38
42 45
49 52
56 59
63 66
70 73

Then went though it to find a mutliple of five+1. Whole thing didn't take a minute to do.
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Re: PS division [#permalink] New post 30 Nov 2007, 09:32
johnrb wrote:
young_gun wrote:
When an integer Q is divided by 5, it leaves 1 as a remainder. When Q is divided by 7, it leaves 3 as a remainder. Find the smallest positive Q. Then find the next smallest positive Q.


Q = 5k + 1 and 7t + 3, where k & t are positive integers.

So 5k + 1 = 7t + 3

5k = 7t + 2

k = (7t + 2)/5. We want values of t such that 7t + 2 is divisible by 5. The lowest is 4; the next lowest is 9.

If t = 4, Q = 31.

If t = 9, Q = 66.


agreed, this method is nice...how can you quickly find values of t such that 7t + 2 is divisible by 5?
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 [#permalink] New post 30 Nov 2007, 09:47
asdert wrote:
I like johnrb's method. Mine wasn;t as fancy.

Did 2 lists. First, the column on the right with multiples of 7. Then a column on the right, adding 3 to the number on the left:

7 10
14 17
21 24
28 31
35 38
42 45
49 52
56 59
63 66
70 73

Then went though it to find a mutliple of five+1. Whole thing didn't take a minute to do.


Personally, when it comes to remainders, simpler is better. Lots of algebra looks great and sophisticated, but often is confusing and time consuming. I think this method, though not "as fancy", yields the right answer nearly immediately.

Note that asdert started with multiples of 7. That's the right move. S/He could have started with 5, but then there are just more numbers to write out on the list. So start with the higher number, then make the other number conform.
  [#permalink] 30 Nov 2007, 09:47
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When an integer Q is divided by 5, it leaves 1 as a

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