blueviper wrote:
EgmatQuantExpert wrote:
Solution
Given:• n is a positive integer greater than 6.
To find:• The value of remainder when n is divided by 6.
Statement-1: “When n is divided by 9, the remainder is 2. “
• n= 9a+2= 6a+3a+2
• n= 6*a+(3a+2)
o Hence, the remainder is 3a+2 whose values depends on the variable a.
Hence, Statement 1 alone is not sufficient to answer the question.
Statement-2: “When n is divided by 4, the remainder is 1. “
• n= 4a+1
o Again, the value of remainder depends on the variable a.
Statement 2 alone is not sufficient to answer the question.
Combining both the statements: From Statement 1: n= 9a+2
From Statement 1: n= 4a+1
By combining both the statements: n= 36k+29= 6*6k+6*4+5= 6*m+5
o Hence, the remainder is 5.
Hence, we can find the answer by combining both the statements.
Answer: Ccan you please elaborate how we got n= 36k+29 ?
Hello
I will try. N when divided by 9 gives a remainder of 2, thus N can be written as '9a + 2', where a is a non negative integer.
N when divided by 4 gives a remainder of 1, so N can also be written as '4b + 1', where b is also a non negative integer.
Since N is same, lets equate them. 9a+2 = 4b+1. Now we write one of the variables (out of a and b) in terms of other variable.
So 4b = 9a+1 or b = (9a+1)/4
Now with some trial, we have to find the first integer value of 'a' which will also give an integer value of 'b'. This will give us the least possible value of N (which satisfies both statements conditions).
We put a=1, 9a+1 = 10, not divisible by 4.
We put a=2, 9a+1 = 19, not divisible by 4.
We put a=3, 9a+1 = 28, this IS divisible by 4.
So least value of a=3, and least value of b=28/4 = 7.
So the least value of N becomes either '9*3+2' or '4*7+2' which is 29.
Once we have the least value, a general form of N is given by the following formula:
N = least number obtained + (LCM of divisors)*K , where K is a non negative integer.
So the least number is 29 and LCM of divisors (9 and 4 here) is 36. So by the above formula
N = 29 + 36K