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I did it in a more rudimentary fashion, but bunuel's explanation is outstanding.

Bunuel
shrive555
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 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.
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Bunuel
shrive555
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 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.


This is a great explanation.

Thanks
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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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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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Bunuel
shrive555
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 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
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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
Bunuel
shrive555
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 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
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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
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
Bunuel


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
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tirbah
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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shrive555
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

Given: When positive integer k is divided by 1869, the remainder is 102.

Asked: What is the remainder when k is divided by 89?

1869 = 89 * 21

k = 1869m + 102; where m is an integer
k = 89*21m + 89 + 13 = 89*(21m+1) + 13 = 89n + 13; where n = 21 m + 1

Remainder when k is divided by 89 is 13

IMO C
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shrive555
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

k clearly could be equal to 102, which gives a remainder of 13 when divided by 89, so 13 must be the answer (since the question can only have one correct answer, you must also get 13 for any other valid value of k). There is no need to divide 1869 by 89.
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Easiest way to do this:
1. k = 1869x + 102
2. k = 89x + R

(1) k=1869(0)+102 --->k=102
(2) 102 = 89x + R
102 = 89(1) +R
102-89 = R
R=13
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shrive555
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
Solution:

Since 1869/89 = 21 (that is, 1869 is divisible by 89), the remainder when k is divided by 89 is the same as the remainder when 102 is divided by 89. Since 102/89 = 1 R 13, the remainder is 13.

Answer: C
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A fast way to do this is 102/89 = 1 and something. Something is the remainder, which is a 102-89=13
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