Problem Solving

goyalsau wrote:I am not able to recall the technique by which by which we solve such problems........ Please Help Guys........


if n is the smallest number that leaves respective remainders of 4, 6 and 9 when divided successively by 13, 11 and 15.

shovan85 wrote:

goyalsau wrote:I am not able to recall the technique by which by which we solve such problems........ Please Help Guys........


if n is the smallest number that leaves respective remainders of 4, 6 and 9 when divided successively by 13, 11 and 15.

The n can be written as,

n = 13k + 4
n = 11k' + 6
n = 15k" + 9

So, in order to get the number we have to fullfil the above mentioned numbers one by one....

The number with 15 divided 9 as remainder = 15x+9
The number with 11 divided 6 as remainder = 11*(15x+9) + 6
The number with 13 divided 4 as remainder = 13*[11*(15x+9) + 6] +4 = 2145x +1369 (Thus the general form of the required number)

In order to get the lowest put x= 1 thus 3514

goyalsau wrote:I am not able to recall the technique by which by which we solve such problems........ Please Help Guys........


if n is the smallest number that leaves respective remainders of 4, 6 and 9 when divided successively by 13, 11 and 15.

Anurag@Gurome wrote: Note: I believe the original question belongs to CAT and it looks something like "Find the sum of the remainders obtained when a number n is divided by 9 and 7 successively, if n is the smallest number that leaves respective remainders of 4, 6 and 9 when divided successively by 13, 11 and 15."
Am I right Saurabh?

Anurag@Gurome wrote:

goyalsau wrote:if n is the smallest number that leaves respective remainders of 4, 6 and 9 when divided successively by 13, 11 and 15.

The number should be written in the following ways,

• 1. n = 11a + 6
  2. n = 13b + 4
  3. n = 15c + 9

Where a, b, and c are non-negative integers.

Our job is to combine all these three to get a single expression for n, from which we can easily determine the smallest possible value.

Combining first two relation we can say (13b + 4) gives a remainder of 6 when divided by 11. Now, (13b + 4) = 11b + (2b + 4). Hence (2b + 4) also gives a remainder of 6 when divided by 11. Which again implies 2b will give a remainder of (6 - 4) = 2 when divided by 11. Hence b will give a remainder of 1 when divided by 11. Thus, b can be written as (11x + 1) for some non-negative integer x.

Anurag@Gurome wrote: Now we have to combine this with the third relation. Which clearly implies (143x + 17) will give a remainder of 9 when divided by 15. Now, (143x + 17) = (135x + 8x + 15 + 2) = 15*(9x + 1) + (8x + 2). Hence (8x + 2) also gives a remainder of 9 when divided by 15. Which again implies 8x will give a remainder of (9 - 2) = 7 when divided by 15. Thus, 8x must be of the form (15y + 7) for some non-negative integer y. Note that this means 8 divides (15y + 7) = (8y + 7y + 7) = 8y + 7(y + 1). Again this implies 8 divides 7(y + 1). Now as 8 doesn't divide 7, 8 must divide (y + 1). Thus y) must be of the form (8z + 7) for some non-negative integer z. Hence, 8x = (15y + 7) = 15*(8z + 7) + 7 = (120z + 112) = 8*(15z + 14). hence x is of the form (15z + 14) for some non-negative integer z.

Hence, n = 13*(11x + 1) + 4 = (143x + 17) = 143*(15z + 14) + 17 = (2145z + 2019)

Hence minimum possible value of n is 2019.

GMATGuruNY wrote:

goyalsau wrote:I am not able to recall the technique by which by which we solve such problems........ Please Help Guys........


if n is the smallest number that leaves respective remainders of 4, 6 and 9 when divided successively by 13, 11 and 15.

I would simply try the answer choices -- starting with the smallest -- until I found one that satisfies all the conditions given.

Find the sum of the remainders obtained when a number n is divided by 9 and 7 successively, if n is the smallest number that leaves respective remainders of 4 and 6 when divided successively by 13 and 11.