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When positive integer y is divided by 7, the remainder is 2. When y is
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12 Jun 2017, 01:32
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When positive integer y is divided by 7, the remainder is 2. When y is
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12 Jun 2017, 03:17
Bunuel wrote: When positive integer y is divided by 7, the remainder is 2. When y is divided by 11, the remainder is 3. What is the sum of the digits of the smallest possible value that meets the definition for y?
A. 9 B. 10 C. 11 D. 12 E. 13 Required number is in form; \(y = 7k + 2\)  (k is the quotient) The same number when divided by 11 leaves remainder 3. ie; dividing \((7k + 2)\) by \(11\) has remainder \(3\). Subtracting this remainder \(3\) from the the number \((7k +2)\) gives number which would be divisible by \(11\). Therefore; \(7k + 2  3 = (7k 1)\) is divisible by \(11\). Try out values \(0, 1, 2\) ... for k. \(k = 8\) \(7 *8 1 = 55\) \(\frac{55}{11} = 5\)
When \(k = 8\), number is divisible by \(11\). Substitute \(k = 8\) in original number form \((7k +2)\) to get the number. \(7 * 8 +2 = 56 + 2 = 58.\) Sum of digits of \(58 = 13\). Answer E...




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Re: When positive integer y is divided by 7, the remainder is 2. When y is
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12 Jun 2017, 03:32
Is it E?
Given: y=7x+2 and y=11x+3.
From this it follows that (y3) should be divisible by 11.
Substituting the first equation for y, we get \(\frac{(7x+2)3}{11}\) should be an integer.
When x=8, it is the smallest number where the above equation divisible by 11.
Therefore y=7x+2=7(8)+2=58 Sum of digits is 13 (E)



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Re: When positive integer y is divided by 7, the remainder is 2. When y is
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12 Jun 2017, 06:19
Bunuel wrote: When positive integer y is divided by 7, the remainder is 2. When y is divided by 11, the remainder is 3. What is the sum of the digits of the smallest possible value that meets the definition for y?
A. 9 B. 10 C. 11 D. 12 E. 13 When positive integer y is divided by 7, the remainder is 2 i.e. y may be {2, 9, 16, 23, 30, 37, 44, 51, 58... etc} When positive integer y is divided by 11, the remainder is 3 i.e. y may be {3, 14, 25, 36, 47, 58, 69, 80, 91, 102... etc} First Common value = 58 Sum of digits = 5+8 = 13 answer: Option E
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Re: When positive integer y is divided by 7, the remainder is 2. When y is
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12 Jun 2017, 07:12
Bunuel wrote: When positive integer y is divided by 7, the remainder is 2. When y is divided by 11, the remainder is 3. What is the sum of the digits of the smallest possible value that meets the definition for y?
A. 9 B. 10 C. 11 D. 12 E. 13 y = 7a + 2 y = 11b + 3 Look for the first such value by hit and trial. If b = 1, y = 14 but it is not of the form 7a + 2 If b = 2, y = 25 but it is not of the form 7a + 2 If b = 3, y = 36 but it is not of the form 7a + 2 If b = 4, y = 45 but it is not of the form 7a+2 If b = 5, y = 58 = 7*8 + 2 So 58 is first such value. The sum of its digits is 5+8 = 13 Answer (E) For more on this, check: https://www.veritasprep.com/blog/2011/0 ... unraveled/https://www.veritasprep.com/blog/2011/0 ... emainders/https://www.veritasprep.com/blog/2011/0 ... spartii/
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When positive integer y is divided by 7, the remainder is 2. When y is
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12 Jun 2017, 09:24
Bunuel wrote: When positive integer y is divided by 7, the remainder is 2. When y is divided by 11, the remainder is 3. What is the sum of the digits of the smallest possible value that meets the definition for y?
A. 9 B. 10 C. 11 D. 12 E. 13 let x=difference between y/7 and y/11 quotients x=(y2)/7(y3)/11 y=(77x+1)/4 3 is the least value of x giving a multiple of 4 y=(77*3+1)/4=58 5+8=13 E



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Re: When positive integer y is divided by 7, the remainder is 2. When y is
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12 Jun 2017, 12:24
gracie wrote: Bunuel wrote: When positive integer y is divided by 7, the remainder is 2. When y is divided by 11, the remainder is 3. What is the sum of the digits of the smallest possible value that meets the definition for y?
A. 9 B. 10 C. 11 D. 12 E. 13 let x=difference between y/7 and y/11 quotients x=(y2)/7(y3)/11 y=(77x+1)/4 3 is the least value of x giving a multiple of 4 y=(77*3+1)/4=58 5+8=13 E I reached the same answer (E) by listing down all possible values of y for y=7q+2 and for y=11k+3. The first common value was 58. Your approach seems straightforward but I could not understand it. Can you help explain? How did you get the value of x here as 3 ? What does "3 is the least value of x giving a multiple of 4 " mean?



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When positive integer y is divided by 7, the remainder is 2. When y is
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12 Jun 2017, 13:38
anupama000 wrote: gracie wrote: Bunuel wrote: When positive integer y is divided by 7, the remainder is 2. When y is divided by 11, the remainder is 3. What is the sum of the digits of the smallest possible value that meets the definition for y?
A. 9 B. 10 C. 11 D. 12 E. 13 let x=difference between y/7 and y/11 quotients x=(y2)/7(y3)/11 y=(77x+1)/4 3 is the least value of x giving a multiple of 4 y=(77*3+1)/4=58 5+8=13 E I reached the same answer (E) by listing down all possible values of y for y=7q+2 and for y=11k+3. The first common value was 58. Your approach seems straightforward but I could not understand it. Can you help explain? How did you get the value of x here as 3 ? What does "3 is the least value of x giving a multiple of 4 " mean? hi anupama, looking at the equation y=(77x+1)/4, it's clear that, if y is an integer, then the dividend, 77x+1, must be a multiple of the divisor, 4 testing possible values of 0, 1, and 2 for x, none of these work 3 is the least value of x that gives 77x+1 divisibility by 4 I hope this helps, gracie



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Re: When positive integer y is divided by 7, the remainder is 2. When y is
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14 Jun 2017, 15:17
Bunuel wrote: When positive integer y is divided by 7, the remainder is 2. When y is divided by 11, the remainder is 3. What is the sum of the digits of the smallest possible value that meets the definition for y?
A. 9 B. 10 C. 11 D. 12 E. 13 We are given that when positive integer y is divided by 7, the remainder is 2, and that when y is divided by 11, the remainder is 3. Let’s first determine the values of y that produce a remainder of 2 when divided by 7: y could be: 2, 9, 16, 23, 30, 37, 44, 51, 58, ... Next let’s determine the values of y that produce a remainder of 3 when divided by 11: y could be: 3, 14, 25, 36, 47, 58, ... Thus, we see that the smallest value is 58 and the sum of the digits of 58 is 13. Answer: E
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Re: When positive integer y is divided by 7, the remainder is 2. When y is
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15 Jun 2017, 03:32
sashiim20 wrote: Bunuel wrote: When positive integer y is divided by 7, the remainder is 2. When y is divided by 11, the remainder is 3. What is the sum of the digits of the smallest possible value that meets the definition for y?
A. 9 B. 10 C. 11 D. 12 E. 13 Required number is in form; \(y = 7k + 2\)  (k is the quotient) The same number when divided by 11 leaves remainder 3. ie; dividing \((7k + 2)\) by \(11\) has remainder \(3\). Subtracting this remainder \(3\) from the the number \((7k +2)\) gives number which would be divisible by \(11\). Therefore; \(7k + 2  3 = (7k 1)\) is divisible by \(11\). Try out values \(0, 1, 2\) ... for k. \(k = 8\) \(7 *8 1 = 55\) \(\frac{55}{11} = 5\)
When \(k = 8\), number is divisible by \(11\). Substitute \(k = 8\) in original number form \((7k +2)\) to get the number. \(7 * 8 +2 = 56 + 2 = 58.\) Sum of digits of \(58 = 13\). Answer E... sashiim20... Your approach to this problem is very nice. But to make it shorter Lets say y = 11 k + 3 So , y2 = 11k +3 2 = 11k+1 is divisible by 7 Now we check by putting the values of k as 0, 1, 2...... k=0, 11k+1 = 1 k=1, 11k+1 = 12 k=2, 11k+1 = 23 k=3, 11k+1 = 34 k=4, 11k+1 = 45 k=5, 11k+1 = 56 So y = 11k+3 = 58...... If we take y w.r.t bigger number, we can solve in little less time thatn if we would have considered y=7k+2.. Answer E..
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Re: When positive integer y is divided by 7, the remainder is 2. When y is
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15 Jun 2017, 06:18
[/quote]I reached the same answer (E) by listing down all possible values of y for y=7q+2 and for y=11k+3. The first common value was 58. Your approach seems straightforward but I could not understand it. Can you help explain? How did you get the value of x here as 3 ? What does "3 is the least value of x giving a multiple of 4 " mean?[/quote]
hi anupama, looking at the equation y=(77x+1)/4, it's clear that, if y is an integer, then the dividend, 77x+1, must be a multiple of the divisor, 4 testing possible values of 0, 1, and 2 for x, none of these work 3 is the least value of x that gives 77x+1 divisibility by 4 I hope this helps, gracie[/quote]
Thanks gracie. Its clear now. I think I was reading it wrong as it wasn't obvious to me immediately that x needs to be an integer( as it is the difference in quotients). This is a much cleaner and faster approach I must say.



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Re: When positive integer y is divided by 7, the remainder is 2. When y is
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15 Jun 2017, 06:35
Bunuel wrote: When positive integer y is divided by 7, the remainder is 2. When y is divided by 11, the remainder is 3. What is the sum of the digits of the smallest possible value that meets the definition for y?
A. 9 B. 10 C. 11 D. 12 E. 13 Possible values of y = { 9 , 16 , 23 , 30 , 37 , 44 , 51 , 58................} Possible values of x = { 14, 25, 36 , 47 , 58................} Sum of the digits is 5 + 8 = 13 Thus, answer will be (E) 13
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Re: When positive integer y is divided by 7, the remainder is 2. When y is
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04 Nov 2018, 23:24
First case Y=7p+2 Y=11q+3 7p=11q+1(1)
7*8=11*5+1=56
So, p+q=8+5=13 Answer:E
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