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Senior Manager
Joined: 22 Aug 2003
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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, 05:14
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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).



Manager
Joined: 02 Jul 2003
Posts: 58

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.



Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland

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



Manager
Joined: 02 Jul 2003
Posts: 58

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



Senior Manager
Joined: 22 Aug 2003
Posts: 257
Location: Bangalore

Great Rich...
ur answers are correct
1) N=25
2) remainder will be 53. When 53 is divided by 84 remainder is 53.
Vicks



SVP
Joined: 03 Feb 2003
Posts: 1604

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?



Manager
Joined: 02 Jul 2003
Posts: 58

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.



Senior Manager
Joined: 19 May 2004
Posts: 291

What is the best way to solve the second one?



Director
Joined: 16 Jun 2004
Posts: 891

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.



Senior Manager
Joined: 19 May 2004
Posts: 291

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 ?



Director
Joined: 16 Jun 2004
Posts: 891

Dookie,
You start with q3=0, NOT q1=0. Start with the last quotient, not the first one.



Senior Manager
Joined: 19 May 2004
Posts: 291

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 ?










