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# When positive integer k is divided by 1869, the remainder is

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When positive integer k is divided by 1869, the remainder is  [#permalink]

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Updated on: 24 Sep 2013, 14:57
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82% (01:03) correct 18% (01:52) wrong based on 250 sessions

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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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10 Dec 2010, 11:54
5
3
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 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.

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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 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.

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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 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.

This is a great explanation.

Thanks
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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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10 Dec 2010, 13:26
2
3
shrive555 wrote:
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?

x+2 is divisible by 10 --> $$x+2=10n$$ for some positive integer $$n$$

$$x^2+4x+9=(x+2)^2+5=(10n)^2+5$$ so this expression divided by 10 yields remainder of 5.

Questions on remainders:
Theory: compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html

PS:
remainder-101074.html
remainder-problem-92629.html
number-properties-question-from-qr-2nd-edition-ps-96030.html
remainder-when-k-96127.html
ps-0-to-50-inclusive-remainder-76984.html
good-problem-90442.html
remainder-of-89470.html
number-system-60282.html
remainder-problem-88102.html

DS:
remainder-problem-101740.html
remainder-101663.html
ds-gcd-of-numbers-101360.html
data-sufficiency-with-remainder-98529.html
sum-of-remainders-99943.html
ds8-93971.html
need-solution-98567.html
gmat-prep-ds-remainder-96366.html
gmat-prep-ds-93364.html
ds-from-gmatprep-96712.html
remainder-problem-divisible-by-86839.html
gmat-prep-2-remainder-86155.html
remainder-94472.html
remainder-problem-84967.html

Hope it helps.
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24 Sep 2013, 09:01
2
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  [#permalink]

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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 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.

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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Re: When positive integer k is divided by 1869, the remainder is  [#permalink]

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01 Jan 2015, 12:12
1
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 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.

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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Re: When positive integer k is divided by 1869, the remainder is  [#permalink]

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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 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.

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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Re: When positive integer k is divided by 1869, the remainder is  [#permalink]

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02 Jan 2015, 00:31
1
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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Re: When positive integer k is divided by 1869, the remainder is  [#permalink]

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