Hi Sunita123,
Thanks for replying. Actually I have come across such situations many a times when I am not able to figure out that a particular no. is a factor of a another particular number. In order to find factors of a particular number I first try to see if it is divisible by 2,3,5,6,7 etc as checking divisibility with them is easier and once a number is not divisible by any of these I do not know what to do.
for example there is another question -
Question: If K is a positive integer, is (2^k) - 1 a prime number?
Statement 1: K is a prime number
Statement 2: K has exactly two positive divisors.
Here basically both statements convey the same thing so answer is either E or D.
I tried some prime number values for K to see if (2^k) - 1 is prime or not.....all values 3,5,7 gives the value of (2^k) - 1 as prime no. and when I checked with K=11 then -
2^11 - 1 = 2047....
I checked the divisibility of 2047 with all the numbers e.g. 2,3,5,7,13,19 etc and thought that it must be a prime number and as all the prime values of K resulted in prime number for the value of (2^k) - 1 I thought that answer should be D but the correct answer is E and it came out that 2047 is not a prime number and is divisible by 23 in 89 times. (2047 = 23*89)
So what I was trying to ask is - and now in the below question I missed to see that 1869 is 21*89. So I am not sure if something is wrong with my approach or if I am doing something wrong somewhere?
Please let me know if you have another perspective to deal with such questions. Many thanks.
Regards
sunita123 wrote:
how to find out that 89 will go into 1869 at 21 times
if you just divide 1869 with 89 , you will get 21
tirbah wrote:
Bunuel wrote:
Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).
So, positive integer k is divided by 1869, the remainder is 102 --> \(k=1,869q+102\). Now, 1,869 itself is divisible by 89: 1,869=89*21, so \(k=1,869q+102=89*21q+89+13=89(21q+1)+13\) --> first term (89(21q+1)) is clearly divisible by 89 and the second term 13 divided by 89 yields remainder of 13.
Answer: C.
Hi Bunuel,
Thanks for the wonderful solution to the problem however how to find out that 89 will go into 1869 at 21 times......I mean while trying to solve this question I thought if 1869 is divisible by 89 however after trying 4-5 multiples of 89 I gave up....is there a way to be able to see that? Thanks in advance.
Regards