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When positive integer y is divided by 7, the remainder is 2. When y is

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When positive integer y is divided by 7, the remainder is 2. When y is  [#permalink]

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New post 12 Jun 2017, 02:32
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When positive integer y is divided by 7, the remainder is 2. When y is  [#permalink]

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New post 12 Jun 2017, 04:17
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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  [#permalink]

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New post 12 Jun 2017, 04:32
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Is it E?

Given:
y=7x+2 and
y=11x+3.

From this it follows that (y-3) 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  [#permalink]

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New post 12 Jun 2017, 07:19
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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  [#permalink]

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New post 12 Jun 2017, 08:12
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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 ... s-part-ii/
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When positive integer y is divided by 7, the remainder is 2. When y is  [#permalink]

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New post 12 Jun 2017, 10:24
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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=(y-2)/7-(y-3)/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  [#permalink]

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New post 12 Jun 2017, 13:24
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=(y-2)/7-(y-3)/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  [#permalink]

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New post 12 Jun 2017, 14:38
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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=(y-2)/7-(y-3)/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,
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Re: When positive integer y is divided by 7, the remainder is 2. When y is  [#permalink]

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New post 14 Jun 2017, 16:17
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  [#permalink]

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New post 15 Jun 2017, 04:32
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 , y-2 = 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  [#permalink]

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New post 15 Jun 2017, 07:18
[/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  [#permalink]

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New post 15 Jun 2017, 07:35
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  [#permalink]

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New post 05 Nov 2018, 00:24
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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Re: When positive integer y is divided by 7, the remainder is 2. When y is  [#permalink]

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New post 04 Sep 2019, 00:05
Let me share a shortcut that I found to calculate the ranges of dividend for a given remainder when quotient is not given.
For any unknown dividend with a remainder, the first value of several possible dividends is the given remainder itself.
Next set of possible dividends are calculated by simple consecutive addition of the divisor to the preceding dividends.
In this case, first dividend for first equation is the remainder 2 itself
Second dividend - 2+7=9
third - 9+7=16
4th - 16+7=23
Dividend ranges for first equation - 2 9 16 13 30 37 44 51 58
Dividend ranges for second equation - 3 14 25 36 47 58

58 is the common dividend which is divisible by both 7 and 11. This is the number you are looking for.
58=5+8=13
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Re: When positive integer y is divided by 7, the remainder is 2. When y is   [#permalink] 04 Sep 2019, 00:05
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