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Re: Manhattan Remainder Problem [#permalink]
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13 Sep 2012, 04:59
Jp27 wrote: Bunuel wrote: To elaborate more.
Suppose we are told that: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?
The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ... The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...
The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).
So we should derive general formula (based on both statements) that will give us only valid values of \(n\).
How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.
Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).
Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).
Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...
Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps. This is awesome I was searching for this logic for a long time. Just one question  you said, i quote, Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\). Is there a quick way to come with this value without listing the nos out. sometime the common nos in both the series would come after 6 or 7 term hence sometimes consumes time. Is there a quicker way? My sincere thanks. cheers In some cases we can use some other approaches, though I think simple listing is the easiest way.
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Re: Manhattan Remainder Problem [#permalink]
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15 Sep 2012, 21:32
Bunuel wrote: In some cases we can use some other approaches, though I think simple listing is the easiest way. Thanks for your response. Yes that easy. Sorry this might be a silly questions.... I have a basic doubt here Lets assume p = 9 n = 11 and what is the reminder when pn / 15? so 99 / 15 give me a reminder of 9 but if i split the denominator for ex 9 * 11 / 3 * 5 = 9/3 * 11 / 5 => I can always mutiply reminders as long as i can correct the excess so here 9/3 reminder 0 11/5 reminder 1 = 0 * 1 = 0 = reminder. So i should never split the denominator or the numertator when calculating for root or can I? something is wrong here! pls help.



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Re: Manhattan Remainder Problem [#permalink]
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03 Nov 2012, 01:46
Bunuel wrote: To elaborate more.
Suppose we are told that: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?
The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ... The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...
The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).
So we should derive general formula (based on both statements) that will give us only valid values of \(n\).
How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.
Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).
Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).
Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...
Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps. Sorry , i didn't understand the last statement written marked in red... instead of 12 if we have other values like 14 or 20 then how remainder would vary.. just lil but curious to understand the gravity



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Re: Manhattan Remainder Problem [#permalink]
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03 Nov 2012, 01:50
breakit wrote: Bunuel wrote: To elaborate more.
Suppose we are told that: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?
The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ... The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...
The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).
So we should derive general formula (based on both statements) that will give us only valid values of \(n\).
How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.
Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).
Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).
Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...
Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps. Sorry , i didn't understand the last statement written marked in red... instead of 12 if we have other values like 14 or 20 then how remainder would vary.. just lil but curious to understand the gravity \(n=24k+10=12(2k)+10\) > \(n\) can be: 10, 34, 58, ... \(n\) divided by 12 will give us the reminder of 10. As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.
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Re: Manhattan Remainder Problem [#permalink]
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03 Nov 2012, 14:14
Bunuel wrote: To elaborate more.
Suppose we are told that: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?
The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ... The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...
The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).
So we should derive general formula (based on both statements) that will give us only valid values of \(n\).
How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.
Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).
Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).
Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...
Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps. Hi Bunuel, I had a quick question with this explanation: Do we have to find the LCM? I just multiplied 6 x 8 and got 48 => n = 48k + 10 which also leads to a remainder of 10. My question is is finding the LCM necessary?



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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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16 Dec 2012, 14:22
n=24k+10=12(12k)+10 > n can be: 10, 34, 58, ... n divided by 12 will give us the reminder of 10.
As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.
Bunuel  Can you please explain this? how does 24k+10 = 12(12k)+10
and can you help me to visualize how you would divide 24k+10 by 12? thanks!



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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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16 Dec 2012, 23:29
jmuduke08 wrote: n=24k+10=12(12k)+10 > n can be: 10, 34, 58, ... n divided by 12 will give us the reminder of 10.
As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.
Bunuel  Can you please explain this? how does 24k+10 = 12(12k)+10
and can you help me to visualize how you would divide 24k+10 by 12? thanks! It's n=24k+10=12*2k+10, not n=24k+10=12*12k+10.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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18 Dec 2012, 01:40
Ans: since n is greater than 30 we check for the number which gives a remainder of 4 after dividing by 6 and 3 after dividing by 5 , the number comes out to 58. So it will give a remainder of 28 after dividing by 30. Answer (E).
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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05 Feb 2014, 23:41
Listing the integers to obtain a common value takes time. Can this be done some other way ?



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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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13 Apr 2014, 22:37
Maybe I did something wrong, or I just got lucky, so please inform me. I also had 28 as the solution, by just plugging in the values. a. 3 / 6 = 0 r 6 INCORRECT b. 12 / 6 = 2 r 0 INCORRECT c. 18 / 6 = 3 r 0 INCORRECT d. 22 / 6 = 3 r 4 & 22 / 5 = 4 r 2 INCORRECT e. 28 / 6 = 4 r 4 & 28 / 5 = 4 r 3 CORRECTSupersimple math, so there must be something wrong here, haha.



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Re: Manhattan Remainder Problem [#permalink]
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18 May 2014, 12:55
Bunuel wrote: To elaborate more.
Suppose we are told that: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?
The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ... The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...
The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).
So we should derive general formula (based on both statements) that will give us only valid values of \(n\).
How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.
Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).
Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).
Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...
Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12). Hope it helps. Hi Bunuel, All of this makes sense but I would like to challenge the last statement highlighted above. In this case, we can obviously write n=12(2k)+10 as you highlighted in the later posts. If for some reason, let's say it asked for a number that wasn't divisible by 24, wouldn't that make the equation n=24k+10 invalid? Meaning, it asked what is the remainder that n leaves after division by 11? Additionally, what if it asked "what is the remainder that n leaves after division by 48. Could we still apply the same logic and say 10? Thanks!



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Re: Manhattan Remainder Problem [#permalink]
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19 May 2014, 08:00
russ9 wrote: Bunuel wrote: To elaborate more.
Suppose we are told that: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?
The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ... The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...
The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).
So we should derive general formula (based on both statements) that will give us only valid values of \(n\).
How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.
Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).
Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).
Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...
Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12). Hope it helps. Hi Bunuel, All of this makes sense but I would like to challenge the last statement highlighted above. In this case, we can obviously write n=12(2k)+10 as you highlighted in the later posts. If for some reason, let's say it asked for a number that wasn't divisible by 24, wouldn't that make the equation n=24k+10 invalid? Meaning, it asked what is the remainder that n leaves after division by 11? Additionally, what if it asked "what is the remainder that n leaves after division by 48. Could we still apply the same logic and say 10? Thanks! If the question were what is the remainder when n is divided by 11, then the answer would be "cannot be determined". The same if the question asked about the remainder when n is divided by 48. See, according to this general formula valid values of \(n\) are: 10, 34, 58, ... These values give different remainders upon division by 11, or 48.
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Re: Manhattan Remainder Problem [#permalink]
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19 May 2014, 17:55
Bunuel wrote: russ9 wrote: Bunuel wrote: To elaborate more.
Suppose we are told that: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?
The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ... The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...
The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).
So we should derive general formula (based on both statements) that will give us only valid values of \(n\).
How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.
Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).
Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).
Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...
Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12). Hope it helps. Hi Bunuel, All of this makes sense but I would like to challenge the last statement highlighted above. In this case, we can obviously write n=12(2k)+10 as you highlighted in the later posts. If for some reason, let's say it asked for a number that wasn't divisible by 24, wouldn't that make the equation n=24k+10 invalid? Meaning, it asked what is the remainder that n leaves after division by 11? Additionally, what if it asked "what is the remainder that n leaves after division by 48. Could we still apply the same logic and say 10? Thanks! If the question were what is the remainder when n is divided by 11, then the answer would be "cannot be determined". The same if the question asked about the remainder when n is divided by 48. See, according to this general formula valid values of \(n\) are: 10, 34, 58, ... These values give different remainders upon division by 11, or 48. Makes complete sense. Thanks.



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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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31 May 2014, 14:12
Hi, Probably a silly question.... But can some one explain why Remainder r would be the first common integer in the two patterns n=6p+4 n=5q+3
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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15 Aug 2014, 22:17
jmuduke08 wrote: n=24k+10=12(12k)+10 > n can be: 10, 34, 58, ... n divided by 12 will give us the reminder of 10.
As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.
Bunuel  Can you please explain this? how does 24k+10 = 12(12k)+10
and can you help me to visualize how you would divide 24k+10 by 12? thanks! Is this a sub600 level question?



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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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18 Aug 2014, 02:33
alphonsa wrote: jmuduke08 wrote: n=24k+10=12(12k)+10 > n can be: 10, 34, 58, ... n divided by 12 will give us the reminder of 10.
As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.
Bunuel  Can you please explain this? how does 24k+10 = 12(12k)+10
and can you help me to visualize how you would divide 24k+10 by 12? thanks! Is this a sub600 level question? No. It is 650700 level question. It seems quite doable only because we have done the concept of remainders in great detail. For a newbie, this problem can be quite challenging.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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01 Sep 2014, 00:28
Bunuel wrote: bchekuri wrote: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30? (A) 3 (B) 12 (C) 18 (D) 22 (E) 28
How to approach this Problem? Positive integer n leaves a remainder of 4 after division by 6 > \(n=6p+4\) > 4, 10, 16, 22, 28, ... Positive integer n leaves a remainder of 3 after division by 5 > \(n=5q+3\) > 3, 8, 13, 18, 23, 28, ... \(n=30k+28\)  we have 30 as lcm of 5 and 6 is 30 and we have 28 as the first common integer in the above patterns is 28. Hence remainder when positive integer n is divided by 30 is 28. Answer: E. P.S. n>30 is a redundant information. What will be the answer of they ask us what the remainder is if n is divided by something other than 30, lets say 11 or 12. any random number which is not a factor of 30.



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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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01 Sep 2014, 00:41
rohansangari wrote: Bunuel wrote: bchekuri wrote: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30? (A) 3 (B) 12 (C) 18 (D) 22 (E) 28
How to approach this Problem? Positive integer n leaves a remainder of 4 after division by 6 > \(n=6p+4\) > 4, 10, 16, 22, 28, ... Positive integer n leaves a remainder of 3 after division by 5 > \(n=5q+3\) > 3, 8, 13, 18, 23, 28, ... \(n=30k+28\)  we have 30 as lcm of 5 and 6 is 30 and we have 28 as the first common integer in the above patterns is 28. Hence remainder when positive integer n is divided by 30 is 28. Answer: E. P.S. n>30 is a redundant information. What will be the answer of they ask us what the remainder is if n is divided by something other than 30, lets say 11 or 12. any random number which is not a factor of 30. In this case we wouldn't be able to determine the remainder. For example, if we were asked to find the remainder of n divided by 11, then we would get different answers: if n = 28, then remainder would be 6 but if n = 58, then the remainder would be 3.
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Re: Positive integer n leaves a remainder of 4 after division by
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01 Sep 2014, 00:41



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