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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]

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05 Sep 2003, 04:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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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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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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Sept 3rd

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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.

Rich
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06 Sep 2003, 01:33
Great Rich...
1) N=25
2) remainder will be 53. When 53 is divided by 84 remainder is 53.
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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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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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26 Jul 2004, 12:19
What is the best way to solve the second one?
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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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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.

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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28 Jul 2004, 17:10
Dookie,
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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 ?
29 Jul 2004, 09:30
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