Problem Solving

Md.Nazrul Islam wrote:when positive integer x is divided by 5, the remainder is 3,and when x is divided by 7 the remainder is 4. when positive integer y is divided by 5,the remainder is 3,and when y is divided by 7 the remainder is 5.if x>y, what must be a factor of x-y ?

Anurag@Gurome wrote:

Md.Nazrul Islam wrote:when positive integer x is divided by 5, the remainder is 3,and when x is divided by 7 the remainder is 4. when positive integer y is divided by 5,the remainder is 3,and when y is divided by 7 the remainder is 5.if x>y, what must be a factor of x-y ?

When a positive integer x is divided by 5, the remainder is 3 implies x = 5a + 3, here x can be 3, 8, 13, 18, 23, ...
When x is divided by 7 the remainder is 4 implies x = 7b + 4, here x can be 4, 11, 18, 25, ...
Now divisor will be the least common multiple (LCM) of above two divisors 5 and 7, which is 35.
Now 18 is the first common integer in the above two series, 3, 8, 13, 18, 23, ... and 4, 11, 18, 25, ...
So, x = 35c + 18
Similarly, y = 35d + 18, say.

Then x - y = 35c + 18 - (35d + 18) = 35(c - d), which implies that x - y should be a multiple of 35.

killer1387 wrote: @Anurag
the bold face part above is wrong

it should be

y=35d + 33

x-y= 35(c-d)+ (18-33)=35(c-d)-15=5{c-d-3}

it should be 5.

Also the example can be

x=53;y=33 where x>y
x-y=20 which is a multiple of 5 NOT 35.


Pls explain if i am going wrong somewhere.

Pharo wrote:

killer1387 wrote: @Anurag
the bold face part above is wrong

it should be

y=35d + 33

x-y= 35(c-d)+ (18-33)=35(c-d)-15=5{c-d-3}

it should be 5.

Also the example can be

x=53;y=33 where x>y
x-y=20 which is a multiple of 5 NOT 35.


Pls explain if i am going wrong somewhere.

You are right man. Anurag made a mistake with 18 there. y = 35d + 33 is the correct formula. However, you made a factoring mistake as well

x-y = 35 (c-d) + 15;
x-y = 5 (7(c-d) - 3);

The answer remains unchanged though; it's still 5