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1) A number N when devided by 4, 7 gives 1, 4 as remainders

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1) A number N when devided by 4, 7 gives 1, 4 as remainders [#permalink] New post 05 Sep 2003, 04:14
1) A number N when devided by 4, 7 gives 1, 4 as remainders respectively. Find the minimum possible value of N.

2) A number N when divided by 3,4,7 gives 2,1,4 as remainders respectively. Find remainder when N is divided by 84? (slightly tougher).
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 [#permalink] New post 05 Sep 2003, 06:32
1. I get n=25

2. 53 gives a remainder of 2 for 3, 1 for 4 and 4 for 7
but 53/83 gives no remainder so I guess i made a mistake.
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 [#permalink] New post 05 Sep 2003, 07:41
rich28 wrote:
1. I get n=25

2. 53 gives a remainder of 2 for 3, 1 for 4 and 4 for 7
but 53/83 gives no remainder so I guess i made a mistake.


For 2, sure it does. the answer would just be 0 remainder 53
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 [#permalink] New post 05 Sep 2003, 07:44
I guess it does. Did you take the GMAT this week? If yes, How was it? I'm going for round two in Bethesda in 3 weeks.

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 [#permalink] New post 06 Sep 2003, 01:33
Great Rich...
ur answers are correct
1) N=25
2) remainder will be 53. When 53 is divided by 84 remainder is 53.
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 [#permalink] New post 06 Sep 2003, 12:07
rich28 wrote:
1. I get n=25

2. 53 gives a remainder of 2 for 3, 1 for 4 and 4 for 7
but 53/83 gives no remainder so I guess i made a mistake.


how did you find 53?
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 [#permalink] New post 08 Sep 2003, 03:28
I counted multiples until I got one that fit the problem. I hope there is an easier way, my way took 2 1/2 minutes.
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 [#permalink] New post 26 Jul 2004, 12:19
What is the best way to solve the second one?
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 [#permalink] New post 26 Jul 2004, 23:01
Ans to Q1. The approach is, let the number be N. We have

N=d1q1+r1 and N=d2q2+r2. d1 and d2 are divisors. q1 & q2 are quotients and r1&7 r2 are remainders. Its given that

N=4q1+1 and N=7q2+4. Where q1 and q2 are quotients. 4 and 7 are divisors and 1 and 4 are remainders.

Substituting q1=1,2,3,4,5, and 6, we have, N=5 or 9 or 13 or 17 or 21 or 25
substituting q2=1,2,3, we have N=11 or 18 or 25. So N=25 is common to both divisors 4 and 7. So the answer is 25.

For Q2: We can solve in two ways.
Method 1:

Apply one of the popular rules.
The rule says, if we have three divisors (d1, d2 and d3) and three remainders (r1, r2 and r3), then the complete remainder is given by the formula d1d2r3 + d1r2 + r1.

Note that 84 is a product of 3, 4 and 7.
Here we have d1=3, d2=4, d3=7 and r1=2, r2=1 and r3=4. Applying the rule we have
48 + 3 + 2 = 53. Hence the answer is 53

Alternate Method 2: (easier one - need not remember any formula and is a technique based approach. Dont ask me to prove it. But works all the time)

We know N= 3q1 + 2 or 4q2 + 1 or 7q3 + 4.

let q3=0, we then have 7q3+4 = 4.
substitute 4 as q2. We then have 4q2 + 1 = 17.
substitute 17 as q1. We then have 3q1 + 2 = 53. So N= 53. 53 when divided by 84 leave remainder 53.

I will make it even better by extrapolating what has been done above.

Let q3=1, we have 7q3+4 = 11
substitute 11 as q2. We then have 4q2 + 1 = 45.
substitute 45 as q1. We then have 3q1 + 2 = 137. So N= 137. 137 when divided by 84 leave remainder 53.
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 [#permalink] New post 28 Jul 2004, 11:37
venksune, thanks for your detailed explanation!
It's the first time i see this method for solving this kind of problems.

About the second problem:

We know N= 3q1 + 2 or 4q2 + 1 or 7q3 + 4.

If i begin with q1=0, i get N=2.
So q2=2 -> N=9.
Now q3=9 -> 9*7 + 4 = 67.

So that doesn't seem to work.
So i don't think i understand this method...

Can you please explain how it works ?
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 [#permalink] New post 28 Jul 2004, 17:10
Dookie,
You start with q3=0, NOT q1=0. Start with the last quotient, not the first one.
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 [#permalink] New post 29 Jul 2004, 09:30
I don't understand. why does it matter which one i calculate first ?
The order they give in the question is just random.

Isn't it ?
:?:
  [#permalink] 29 Jul 2004, 09:30
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1) A number N when devided by 4, 7 gives 1, 4 as remainders

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