Always Divisible By 17

Number Theory Level 2

$x$ and $y$ are integers such that $(2x+3y)$ is divisible by 17.

For which of the following values of $k$ , must $9x + ky$ always be divisible by 17?

5 7 9 3

10 solutions

Shashvat Shukla
Apr 13, 2014

Here is how I solved it. Start with $17|(2x+3y)$ Then clearly $17|(20x+30y)$ Since $17|(17x+17y)\Rightarrow17|(20x+30y)-(17x+17y)\Rightarrow17|(3x+13y)$ Then clearly $17|(9x+39y)$ Since $17|34y\Rightarrow17|(9x+39y)-34y\Rightarrow17|(9x+5y)\Rightarrow\boxed{k=5}$

2x+ 3y is divisible by 17 and obviously 17x+17y is divisible by 17.If two nos are divisible by ano. Then their sum or difference will also be divisible by thst no.What do we have to subtract from 17x to get 9x?8x.Therefore the term 2x + 3y has to be multiplied by a no. which makes 2x equal to 8x.this is 4.Hence 3y becomes 12y.8x+12y is a multiple of 2x+3y and hence also divisicle by 17.Subtracting 8x+12y from 17x+17y gives 9x+5y.Hence k=5.

Amit Chopra - 7 years, 1 month ago

haha, I was awestruck by the explanations found, here, well that was a nice problem. I guessed the solution (x,y) to be (1,5) and compared it with the given values of K... Only 5 satisfied it... :) well good work, Mr. shukla.. ;)

Aravind M - 7 years, 1 month ago

thanks :)

Shashvat Shukla - 7 years, 1 month ago

same here

Pratik Kulkarni - 7 years, 1 month ago

done it exactly you did it.

MOHD FARAZ - 7 years, 1 month ago

???? not understanding :-(

Tania 123 - 7 years, 1 month ago

Which part?

Shashvat Shukla - 7 years, 1 month ago

all of it haha....please explain thoroughly step by step ? thanks

Zack Yeung - 7 years, 1 month ago

(17x + 17y) is always divisible by 17 for integer values x and y.... Also, it is given that (2x + 3y) is divisible by 17.... Now, algebraically manipulate to get coefficient of x as 9.. Hence, subtract 4(2x + 3y) from (17x + 17y).. Coefficient of y = 5.

Rahul Basu - 7 years, 1 month ago

nice

Max B - 7 years, 1 month ago

if x=1, then y=5, if x=2 then y=10 and so on, so solve it for k, it will be k=5. start solving the problem from easy and small numbers.

Ilyas Raza - 7 years, 1 month ago

pls explain this problem i can't understand

Izzie Stephanie - 7 years, 1 month ago

from the eqn 2x+3y is divisible by 17 when x=1,y=5 then when comparing it with 2nd eqn 9x+ky is divisible by 17 when x=1,y=5 and k value will be 5 then only 9x+ky will be divisible by 17

Sara Rahman - 7 years, 1 month ago

I don't understand could you please explain clearly?

Zack Yeung - 7 years, 1 month ago

Shashvat Shukla - 7 years, 1 month ago

what does | means?? sorry I am slow learner haha

Zack Yeung - 7 years, 1 month ago

Wow :D

JohnDonnie Celestre - 7 years, 1 month ago

Raviteja Deekonda
Apr 11, 2014

TAKE SMALLEST MULTIPLE OF 17 i.e 17 itself, 2x+3y=17, let x=4, y=3. substitute the values of x and y in 9x+ky=multiple of 17, substitute k as any value from options and check

I don't see why there has to be a fixed value for k. The question should probably make it clear that x or y are non-zero integers because if either x or y are zero, any value of k will satisfy the equation. For example, x=17 and y=0. This satisfies the first equation which is 2x + 3y = (2 * 17 + 3 * 0) = 34 which is divisible by 17. And therefore for 9x + ky, any value of k would always give a constant value which is always divisible by 17.

Similarly for x= 0 and y=17, (2x + 3y) would give 51 which is divisible by 17. And as for 9x + ky, since 9x = 0, ky (k * 17) will always give a multiple of 17 irrespective of the value of k.

Vishwa Shah - 7 years, 2 months ago

You have to find a value of $k$ which satisfies the conditions for ALL ordered pairs $(x,y)$ , not only a few select ones.

Siddhartha Srivastava - 7 years, 2 months ago

This is obviously a Diophantine equation.

Joeie Christian Santana - 7 years, 2 months ago

Agreed. However, it would be clearer if mentioned because it could be simply a question of substitution

Vishwa Shah - 7 years, 2 months ago

Sanjay Balaji
Apr 7, 2014

4 (2x + 3y) + (9x + 5y) = 17 (x + y) If 2x + 3y = 17n, then 9x + 5y = 17 ((x + y) - 4n). Hence divisible by17

hey can u plz explain in detail .

Anushree Pandey - 7 years, 2 months ago

very nice

Vijaysekhar Chellaboina - 7 years, 1 month ago

Kvsnlr Kvsnlr
Apr 14, 2014

We can solve the problem by trail and error method first assume x to be 7 and y to be 1 so that 2x+3y=17 which is the least number divisible by 17 thensubstitute x value and y value in eqn we get 63+k×1as the numberequate it to 68 bcz it is nearest to 68 which is multiple of 17 tjen weget 5 as answer

Vijaysekhar Chellaboina
Apr 20, 2014

To show that $k=5$ is a solution note that $2x+3y=17a$ for some $a$ . Then $18x+10y=18x+27y-17y=9(2x+3y)-17y=17(9a-y).$ Hence, $18x+10y$ is divisible by $17$ which implies that $9x+5y$ is divisible by $17$ .

Mursid A
Apr 19, 2014

Muhammad Arslan
Apr 19, 2014

Easy question;)

Bibhuti` Bhushan
Apr 18, 2014

multiple of 17

Manoj K Sharma
Apr 18, 2014

Just choose a random value for x say x=7, so for 2x+3y to be divisible by 17,y value must be 3 now.. so we hav x and y values,nogiven:9x+ky,substitue x&y values,it will give us (63+k)|17... now fr this to be divisible,the value of k shud be 5 only!

Aravind M
Apr 17, 2014

haha, I was awestruck by the explanations found, here, well that was a nice problem. I guessed the solution (x,y) to be (1,5) and compared it with the given values of K... Only 5 satisfied it... :)

Always Divisible By 17

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