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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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12 Jun 2017, 01:32
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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

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

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12 Jun 2017, 03:17
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5
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...
##### General Discussion
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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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12 Jun 2017, 03:32
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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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

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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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12 Jun 2017, 07:12
2
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

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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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=(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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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=(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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12 Jun 2017, 13:38
1
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,
gracie
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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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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.

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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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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 , 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..

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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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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  [#permalink]

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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  [#permalink]

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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

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Re: When positive integer y is divided by 7, the remainder is 2. When y is &nbs [#permalink] 04 Nov 2018, 23:24
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