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Two numbers when divided by a divisor leave reminders of 248 [#permalink]
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23 Nov 2011, 09:14
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Two numbers when divided by a divisor leave reminders of 248 and 372 respectively. The reminder obtained when the sum of the numbers is divided by the same divisor is 68. Find the divisor. A. 276 B. 552 C. 414 D. 1104 E. 2202
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Re: Numbers #3 [#permalink]
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23 Nov 2011, 09:57
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Fun question.
Say the two numbers are x and y, and divisor is a.
x divided by a leaves a remainder of 248. This means that x = a*N + 248, where N is the integer result of the division. y divided by a leaves a remainder of 372. This means that x = a*K + 372, where K is the integer result of the division.
x+y divided by a leaves a remainder of 68. This means that x = a*M + 68, where M is the integer result of the division.
From definitions above:
x+y = (a*N + 248) + (a*K + 372) = a*(N+K) + 620.
a*(N+K) + 620 = a*M + 68 552 = a*(MNK)
We know that M, N, and K are integers and that a must be at least 373 (to leave a 372 remainder). The only possible value for (MNK) is 1.
Therefore, a = 552. B.



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Re: Numbers #3 [#permalink]
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23 Nov 2011, 10:58
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cleetus wrote: Two numbers when divided by a divisor leave reminders of 248 and 372 respectively. The reminder obtained when the sum of the numbers is divided by the same divisor is 68. Find the divisor. A) 276 B) 552 C) 414 D 1104 E) 2202 Thanks kostyan5. My approach is similar to that of urs. This is how i did it. Let the 2 numbers be X and Y; Let D= Divisor X = D*N+248 , N = Quotient got when X is divided by divisor R Y = D*K+372 , K = Quotient got when Y is divided by divisor R X+Y = (D*N+248) + (D*K+372) = D(N+K)+620 = D(N+K+552/D)+68 As N+K+552/D must be an integer, D must be a factor of 552. As any divisor is greater than the reminder, D>372 So D=552 Answe B
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Re: Numbers #3 [#permalink]
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30 Jan 2012, 21:55
kostyan5 wrote: Fun question.
Say the two numbers are x and y, and divisor is a.
x divided by a leaves a remainder of 248. This means that x = a*N + 248, where N is the integer result of the division. y divided by a leaves a remainder of 372. This means that x = a*K + 372, where K is the integer result of the division.
x+y divided by a leaves a remainder of 68. This means that x = a*M + 68, where M is the integer result of the division.
From definitions above:
x+y = (a*N + 248) + (a*K + 372) = a*(N+K) + 620.
a*(N+K) + 620 = a*M + 68 552 = a*(MNK)
We know that M, N, and K are integers and that a must be at least 373 (to leave a 372 remainder). The only possible value for (MNK) is 1.
Therefore, a = 552. B. ok, "a" must be at least 373, but then why not 414 instead of 552? Thanks!



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Re: Numbers #3 [#permalink]
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31 Jan 2012, 02:21
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Saurajm wrote: kostyan5 wrote: Fun question.
Say the two numbers are x and y, and divisor is a.
x divided by a leaves a remainder of 248. This means that x = a*N + 248, where N is the integer result of the division. y divided by a leaves a remainder of 372. This means that x = a*K + 372, where K is the integer result of the division.
x+y divided by a leaves a remainder of 68. This means that x = a*M + 68, where M is the integer result of the division.
From definitions above:
x+y = (a*N + 248) + (a*K + 372) = a*(N+K) + 620.
a*(N+K) + 620 = a*M + 68 552 = a*(MNK)
We know that M, N, and K are integers and that a must be at least 373 (to leave a 372 remainder). The only possible value for (MNK) is 1.
Therefore, a = 552. B. ok, "a" must be at least 373, but then why not 414 instead of 552? Thanks! If we follow kostyan5's way we get 552=a*(MNK) > (MNK)=integer=552/a, no other value from the answer choices will yield an integer for this expression except 552 and 276, and as a>372 then a=552. Hope it's clear.
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Re: Two numbers when divided by a divisor leave reminders of 248 [#permalink]
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31 Jan 2012, 03:24
Once we know that a=552/ (MNK),
Can we say that a < or = 552. And since 372 is one of the remainders (eliminates A. 276) the only possibility is 552 itself.



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Re: Numbers #3 [#permalink]
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11 May 2012, 20:33
kostyan5 wrote: Fun question.
Say the two numbers are x and y, and divisor is a.
x divided by a leaves a remainder of 248. This means that x = a*N + 248, where N is the integer result of the division. y divided by a leaves a remainder of 372. This means that x = a*K + 372, where K is the integer result of the division.
x+y divided by a leaves a remainder of 68. This means that x = a*M + 68, where M is the integer result of the division.
From definitions above:
x+y = (a*N + 248) + (a*K + 372) = a*(N+K) + 620.
a*(N+K) + 620 = a*M + 68 552 = a*(MNK)
We know that M, N, and K are integers and that a must be at least 373 (to leave a 372 remainder). The only possible value for (MNK) is 1.
Therefore, a = 552. B. Why did you decide the a must be at least 373 and not 248? That's the other remainder.



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Re: Numbers #3 [#permalink]
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12 May 2012, 02:43
elaskova wrote: kostyan5 wrote: Fun question.
Say the two numbers are x and y, and divisor is a.
x divided by a leaves a remainder of 248. This means that x = a*N + 248, where N is the integer result of the division. y divided by a leaves a remainder of 372. This means that x = a*K + 372, where K is the integer result of the division.
x+y divided by a leaves a remainder of 68. This means that x = a*M + 68, where M is the integer result of the division.
From definitions above:
x+y = (a*N + 248) + (a*K + 372) = a*(N+K) + 620.
a*(N+K) + 620 = a*M + 68 552 = a*(MNK)
We know that M, N, and K are integers and that a must be at least 373 (to leave a 372 remainder). The only possible value for (MNK) is 1.
Therefore, a = 552. B. Why did you decide the a must be at least 373 and not 248? That's the other remainder. Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is nonnegative integer and always less than divisor).So, the divisor mus be greater than both remainders, which means that a>372. Also check this: twonumberswhendividedbyadivisorleaveremindersof123645.html#p1036863Hope it helps.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Numbers #3 [#permalink]
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06 Oct 2012, 06:32
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A rule to solve all similar problems 
If two numbers, say a & b, are divided by the same divisor (d) leaving remainders r1 & r2.
Then the remainder (R), when Sum (a+b) / d = (r1+r2)  d. Note  If R becomes negative, then R = (r1+r2) only.
Hence Solution to the above problem 
d = 68, r1 = 248, r2 = 372 so Remainder R when Sum (a+b) / 68 = (248+372)  68 = 620  68 = 552
Note  Difference (ab) is exactly divisible by the same divisor (d).
Hope it helps.



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Re: Two numbers when divided by a divisor leave reminders of 248 [#permalink]
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17 Dec 2013, 15:33
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cleetus wrote: Two numbers when divided by a divisor leave reminders of 248 and 372 respectively. The reminder obtained when the sum of the numbers is divided by the same divisor is 68. Find the divisor.
A. 276 B. 552 C. 414 D. 1104 E. 2202 Easy, why 68 if the sum of the remainders is 248+372=620? Cause the divisor is eating the other part. Then the divisor is 62068=552 Answer is A Cheers! J



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Re: Two numbers when divided by a divisor leave reminders of 248 [#permalink]
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01 May 2014, 03:05
jlgdr wrote: cleetus wrote: Two numbers when divided by a divisor leave reminders of 248 and 372 respectively. The reminder obtained when the sum of the numbers is divided by the same divisor is 68. Find the divisor.
A. 276 B. 552 C. 414 D. 1104 E. 2202 Easy, why 68 if the sum of the remainders is 248+372=620? Cause the divisor is eating the other part. Then the divisor is 62068=552 Answer is A Cheers! J you mean B !!. Answer given is correct but option marked is incorrect. This was the smartest approach of the lot and I used the same.. If 620 is equating to 68 , what was the remaining amount ( 620  68). This added one unit to divisor



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Re: Two numbers when divided by a divisor leave reminders of 248 [#permalink]
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12 Jul 2016, 04:27
friends... I have a question ... Why (MNK) has to be one.... In other words why the answer is 552 and not 1104...



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