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When positive integer n is divided by 13, the remainder is 2. When n

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When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


n=13Q1+2
n=8Q2+5

13Q1+2=8Q2+5
13Q1=8Q2+3

When we list the multiples of 13, only when n=91 can we have a remainder of 3 when n is divided by 8.
OR
We will get a multiple of 13 only when Q2=11

Answer: B
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


the equation that can be formed is 13x+2=8y+5..
13x-3=8y...
as we can see x can take only odd values as the RHS will always be even..
Also x can take values till 13 as 13*14>180..
now we have to substitue x as 1,3,5,7,9,11,13...
once we find 7 fitting in , any other value need not be checked as every 4th value will give us answer so next value will be 15..
ans 1.. B
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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AmoyV wrote:
Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


n=13Q1+2
n=8Q2+5

13Q1+2=8Q2+5
13Q1=8Q2+3

When we list the multiples of 13, only when n=91 can we have a remainder of 3 when n is divided by 8.
OR
We will get a multiple of 13 only when Q2=11

Answer: B


hi
you do not have to check all multiples of 13.. the multiple has to be odd so Q1 has to be odd.. so check for only 1,3,5,7,9.. .. may save some time
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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New post 02 Apr 2015, 06:59
chetan2u wrote:
AmoyV wrote:
Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


n=13Q1+2
n=8Q2+5

13Q1+2=8Q2+5
13Q1=8Q2+3

When we list the multiples of 13, only when n=91 can we have a remainder of 3 when n is divided by 8.
OR
We will get a multiple of 13 only when Q2=11

Answer: B


hi
you do not have to check all multiples of 13.. the multiple has to be odd so Q1 has to be odd.. so check for only 1,3,5,7,9.. .. may save some time


Thanks for pointing that out.
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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chetan2u wrote:
AmoyV wrote:
Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


n=13Q1+2
n=8Q2+5

13Q1+2=8Q2+5
13Q1=8Q2+3

When we list the multiples of 13, only when n=91 can we have a remainder of 3 when n is divided by 8.
OR
We will get a multiple of 13 only when Q2=11

Answer: B


hi
you do not have to check all multiples of 13.. the multiple has to be odd so Q1 has to be odd.. so check for only 1,3,5,7,9.. .. may save some time



And we can save even more time:
all numbers that will be multiple to 13 and 8+3 will be in periods 104 (13 * 8 = 104 )
so if our first number is 91 than next will be 91 + 104 = 195 and next 195 + 104 = 299 ans so on.

And we know that we have diapason from 1 to 180
so 180 - 104 = 76
if we have one number that will be divisible on 13 and 8 + 3
and that number will be less that 76 than we have 2 numbers
if we don't have such number before 76 than we have only one number in this diapason.

for example we can solve the same task but change 8 + 3 on 8 + 4
the first number that will be divisible on 13 and 8 + 4 it's 52
this number before 76 so we have two numbers before 180
second number is 52 + 104 = 156

This is not very useful on this task but if we will have diapson not 1..180 but 1...800 for example
it'll be very useful shortcut.
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


\(N = 13q+2\) (possible values: 2,15,28,41,54,67,80,93,106....
\(N = 8z+5\) (possible values: 5,13,21,29,37,45,53,61,69,77,85,93

Therefore,\(N = (13*8)k+93\)
\(N = 104k+93\) (possible values:93,197)

Clearly only one value is possible that is less than 180.

Answer:B
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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New post 03 Apr 2015, 09:07
chetan2u wrote:
AmoyV wrote:
Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


n=13Q1+2
n=8Q2+5

13Q1+2=8Q2+5
13Q1=8Q2+3

When we list the multiples of 13, only when n=91 can we have a remainder of 3 when n is divided by 8.
OR
We will get a multiple of 13 only when Q2=11

Answer: B


hi
you do not have to check all multiples of 13.. the multiple has to be odd so Q1 has to be odd.. so check for only 1,3,5,7,9.. .. may save some time


That saves a ton of time. Is it because 8y+3 will always be odd ?
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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New post 03 Apr 2015, 17:30
ak1802 wrote:
chetan2u wrote:

hi
you do not have to check all multiples of 13.. the multiple has to be odd so Q1 has to be odd.. so check for only 1,3,5,7,9.. .. may save some time


That saves a ton of time. Is it because 8y+3 will always be odd ?



hi ak1802,
you are correct, 8y+3 will always be odd irrespective of the value of y... and that is why we check for odd values 1,3,5,7..
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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New post 05 Apr 2015, 00:43
Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


The number should be of the form 13k+2 leaving a remainder of 5 when divided by 8.
Calculating different mutilples of 13 i.e 13,26,39,52,65,78,91
Only 91 satisfies i.e 91+2=93 and 93 leaves a remainder of 5 when divided by 8.
Answer=B
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

The LCM of 8 and 13 is 104. Hence there cannot be more than 2 such values less than 180. Options (D) and (E) are out of the window for sure.

The number n should be of the following two forms:
n = 8a + 5
n = 13b + 2

In a given bunch of numbers, there will be many more numbers of the form (8a + 5) and fewer of the form (13b + 2) so let’s start with a number of the form (13b + 2).
If b = 0, n = 2. Is it of the form (8a + 5)? No. n/8 gives a remainder of 2, not 5.
If b = 1, n = 15. Is it of the form (8a + 5)? No. n/8 gives a remainder of 7, not 5.
If b = 2, n = 28. Is it of the form (8a + 5)? No. n/8 gives a remainder of 4, not 5.
If b = 3, n = 41. Is it of the form (8a + 5)? No. n/8 gives a remainder of 1, not 5.
If b = 4, n = 54. Is it of the form (8a + 5)? No. n/8 gives a remainder of 6, not 5.
If b = 5, n = 67. Is it of the form (8a + 5)? No. n/8 gives a remainder of 3, not 5.
If b = 6, n = 80. Is it of the form (8a + 5)? No. n/8 gives a remainder of 0, not 5.
If b = 7, n = 93. Is it of the form (8a + 5)? Yes! n/8 gives a remainder of 5.

The smallest value of n is 93. The next value of n = 93 + 104 = 197 i.e. greater than 180. Hence there is just one value of n less than 180.

Answer: B.
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When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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Bunuel wrote:
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


Kudos for a correct solution.


Possible Values of n will be

FOR SET 13 ={ 15 , 28 , 41 , 54 , 67 , 80 , 93 , 106 , 119 , 132, 145, 158 , 171 }

FOR SET 8 = { 13 , 21 , 29 , 37 , 45 , 53, 61 , 69 , 77, 85 ............. Go no further }

For set 8 there is a patter to observe the numbers are all ODD numbers , but for set 13 the alternate numbers are Odd Numbers....

So, now check the odd numbers in set 13 which when divided by 8 leaves remainder 5 { Here start from 93 as you have already checked numbers upto 85 and no number is common }

93/8 = 11*8 + 5 { This is our number }

Check further ( only the odd numbers ) you cant find any ODD number....


Hence we have only the number 93 which is common to both SET 13 and SET 8 , so this is our number !!

Answer will be only one possible number and that is 93
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When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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New post 31 Aug 2016, 16:12
When positive integer n is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

n=13q+2
n=8p+5
13q+2=8p+5➡
13q-3=8p
the least value of q that will make 13q-3 a multiple of 8=7
13*7+2=93
93 is least value of n
next highest value of n=93+13*8=197, which is>180
thus, 93 is the only value of n<180
B

Last edited by gracie on 29 Nov 2017, 19:05, edited 2 times in total.
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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New post 04 Dec 2016, 01:38
Great Question.
Here is what i did =>
n=13k+2
n=8k+5
combining the two equations above we can say that n=104k+93
n=> 93,93+104,93+2*104,...
Clearly there is only one value of n<180.
Hence B

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When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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New post 29 Nov 2017, 00:00
I kind of noticed a minor shortcut with after reading the Divisibility Applied to Remainders post

Quote:
Now, let’s come back to our original question. We have a number n which when divided by 3 gives a remainder 1 and when divided by 7 gives a remainder 5. We can say the number is of the form:

n = 3a + 1
and
n = 7b + 5

...

Can we say that the remainder in both the cases is (-2) since we need another 2 to make complete groups of 3 and 7? When n is divided by 3 and the remainder obtained is 1, it is the same as saying the remainder is -2. n is 1 more than a multiple of 3 which means it is 2 less than the next multiple of 3. Therefore, we can say n = 3x – 2 and n = 7y – 2.

Now this is exactly like the situation we discussed above. When we divide n by 21, remainder will be -2


Taking the same principle we see that 13a + 2 = 8b + 5

However, if we add 2 to b, we're now an additional 11 units away from (8b + 5), as 5 - (8*2) = -11
Similarly, if we take add 1 to a, we're now an additional 11 units away from (13a + 2), as 2 - (13*1) = -11

So (13[a+1] - 11) = (8[b+2] - 11) or (13x - 11) = (8y - 11). The LCM(13, 8) is 104, so amongst the potential values of n it must follow the form (104q - 11).

Since n is a positive integer, q must be > 0

With q=1, n=93
With q=2, n=197: greater than our upper limit for the problem, leaving only 1 number that fits the question's criteria.
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Re: When positive integer n is divided by 13, the remainder is 2. When n [#permalink]

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New post 30 Dec 2017, 16:20
Trying over 5 numbers until finding the one that works seems inefficient. Here's what I did...

From the given info, we have these 2 equations:
(1) n = 13x + 2
(2) n = 8y + 5

Note that the above are equivalent to:
(1*) n = 13x - 11
(2*) n = 8y - 3

Thus, we have
- 13x - 11 = 8y - 3
- 13x = 8y + 8
- 13x = 8(y+1)
So,
- (y+1) has to be a multiple of 13, and
- x has to be a multiple of 8

The smallest multiple of 13 is 13 (obviously), so if we substitute y = 12 into (2*), then we get n = 93, which is the smallest value of n.
Similarly, if we substitute x = 8 into (1*), then we get the same answer, n = 93.

You can calculate the LCD of 8 and 11 (which is 104), and see that the next value of n is 197 (93 + 104).
Or, since we found the smallest value of n by using x = 8 or y = 12, we can find the next value by using x = 16 (the next multiple of 8) or y = 25 (which would make y+1 the next multiple of 13). If we substitute x = 16 into (1*), we get n = 197. Similarly, if we substitute y = 25 into (2*), we get the same n = 197.
Re: When positive integer n is divided by 13, the remainder is 2. When n   [#permalink] 30 Dec 2017, 16:20
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