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

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)

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)

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

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

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..

[/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.

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

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

Posted from my mobile device

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

Please help me understand why you're taking the difference of the quotients.

Alternate Solution (for fast calculations):-

According to the question, Y is defined in 2 ways -
7a+2 & 11b+3

Since both 7a+2 & 11b +3 should equal the same number y

Break down 11b+3 in form of first (7a+2),

(7b+2) + (4b+1), Now this 4b +1 must be divisible by 7

b=1, 4b+1 = 5
b=2. 4b +1 = 9
b=3, 4b +1 = 13
b= 4, 4b+1 = 17
b= 5, 4b+1 = 21 (Divisible by 7)

put b=7 in 11b +3, you get 58
now add the digits 5+8 to get 13 (Answer choice E)

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