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When positive integer k is divided by 1869, the remainder is
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Updated on: 24 Sep 2013, 14:57
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When positive integer k is divided by 1869, the remainder is 102. What is the remainder when k is divided by 89? A. 0 B. 1 C. 13 D. 23 E. 51
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Originally posted by shrive555 on 10 Dec 2010, 11:20.
Last edited by Bunuel on 24 Sep 2013, 14:57, edited 1 time in total.
Renamed the topic and edited the question.




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Re: remainder
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10 Dec 2010, 11:54
shrive555 wrote: When positive integer k is divided by 1869, the remainder is 102. What is the remainder when k is divided by 89?
0 1 13 23 51
How to approach remainder question ? THanks 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 nonnegative 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.
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Re: remainder
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10 Dec 2010, 12:10
I did it in a more rudimentary fashion, but bunuel's explanation is outstanding. Bunuel wrote: shrive555 wrote: When positive integer k is divided by 1869, the remainder is 102. What is the remainder when k is divided by 89?
0 1 13 23 51
How to approach remainder question ? THanks 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 nonnegative 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.



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Re: remainder
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10 Dec 2010, 12:18
Bunuel wrote: shrive555 wrote: When positive integer k is divided by 1869, the remainder is 102. What is the remainder when k is divided by 89?
0 1 13 23 51
How to approach remainder question ? THanks 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 nonnegative 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. This is a great explanation. Thanks
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Re: remainder
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10 Dec 2010, 12:38
Thanks B. one more question, If remainder is Zero or if we have any algebraic expression. The concept would be the same ? For example : If x is a positive integer and x+2 is divisible by 10, what is the remainder when x2+4x+9 is divided by 10?
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Re: remainder
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10 Dec 2010, 13:26



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Re: remainder
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24 Sep 2013, 09:01
K= 1869A +102
K/89 => {1869A + 102}/89
Since 1869 is perfectly divisible by 89
REM(K/89) = REM (102/89) = 13



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Re: When positive integer k is divided by 1869, the remainder is
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01 Jan 2015, 11:06
Bunuel wrote: shrive555 wrote: When positive integer k is divided by 1869, the remainder is 102. What is the remainder when k is divided by 89?
0 1 13 23 51
How to approach remainder question ? THanks 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 nonnegative 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 45 multiples of 89 I gave up....is there a way to be able to see that? Thanks in advance. Regards



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Re: When positive integer k is divided by 1869, the remainder is
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01 Jan 2015, 12:12
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: shrive555 wrote: When positive integer k is divided by 1869, the remainder is 102. What is the remainder when k is divided by 89?
0 1 13 23 51
How to approach remainder question ? THanks 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 nonnegative 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 45 multiples of 89 I gave up....is there a way to be able to see that? Thanks in advance. Regards
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Re: When positive integer k is divided by 1869, the remainder is
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01 Jan 2015, 23:15
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 nonnegative 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 45 multiples of 89 I gave up....is there a way to be able to see that? Thanks in advance. Regards



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Re: When positive integer k is divided by 1869, the remainder is
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02 Jan 2015, 00:31
tirbah wrote: 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 
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.
Right! So use the same approach: 1869 Not divisible by 2. Divisible by 3 since 1+8+6+9 = 24 which is divisible by 3. Divide by 3: 1869 = 3*623 Now 623 is not divisible by 2, 3 and 5. Try dividing by 7. 623 = 7*89 So 1869 = 3*7*89 89 is a prime number so no more factors are possible 1869 = 21*89 Also, another way  faster I might add  would be to start with 89 and see if it is a factor. Your approach depends on whether 89 is a factor of 1869 or not. Divide 1869 by 89. You get 21. So you know 1869 = 21*89
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