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

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Current Student
Joined: 31 Aug 2007
Posts: 368

Kudos [?]: 163 [0], given: 1

When an integer Q is divided by 5, it leaves 1 as a [#permalink]

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29 Nov 2007, 20:00
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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.

Kudos [?]: 163 [0], given: 1

Manager
Joined: 01 Oct 2007
Posts: 86

Kudos [?]: 28 [0], given: 0

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29 Nov 2007, 20:22
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.

Kudos [?]: 28 [0], given: 0

Senior Manager
Joined: 09 Oct 2007
Posts: 463

Kudos [?]: 54 [0], given: 1

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29 Nov 2007, 22:32
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.

Kudos [?]: 54 [0], given: 1

Current Student
Joined: 31 Aug 2007
Posts: 368

Kudos [?]: 163 [0], given: 1

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30 Nov 2007, 10:32
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?

Kudos [?]: 163 [0], given: 1

CIO
Joined: 09 Mar 2003
Posts: 461

Kudos [?]: 71 [0], given: 0

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30 Nov 2007, 10:47
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.

Kudos [?]: 71 [0], given: 0

30 Nov 2007, 10:47
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