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regarding the reminders problems

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Re: regarding the reminders problems [#permalink]

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New post 31 Mar 2011, 06:54
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ammulujnt wrote:
when a number n is divided by 5 yeilds reminder 1 and when divided by 7 yeilds reminder 3.what is the least possible value of K so that K+n is divisible by 35.Please propose a simple solution for this

Ans: 4


First, understand the theory behind it. Then you can get to the answer within seconds.

Let us say I have a number n which is divisible by 5 and by 7. We all agree that it will be divisible by 35, the LCM of 5 and 7.

Now, if I have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 1, we can say the number is of the form
n = 5a + 1 e.g. 5 + 1, 10 + 1, 15 + 1, 20 + 1, 25 + 1, 30 + 1, 35 + 1 etc
and
n = 7b + 1 e.g. 7 + 1, 14 + 1, 21 + 1, 28 + 1, 35 + 1 etc
So when it is divided by the LCM, 35, it will give 1 as remainder (as is apparent above)

Next, if I have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 3, we can say the number is of the form
n = 5a + 1
and
n = 7b + 3
Now, the only thing you should try to understand here is that when n is divided by 5 and if I say the remainder is 1, it is the same as saying the remainder is -4. e.g. When 6 is divided by 5, remainder is 1 because it is 1 more than a multiple of 5. I can also say it is 4 less than the next multiple of 5, can't I? 6 is one more than 5, but 4 less than 10.
Therefore, we can say n = 5x - 4 and n = 7y - 4 (A remainder of 3 when divided by 7 is the same as getting a remainder of -4)
Now this question is exactly like the question above. So when you divide n by 35, remainder will be -4 i.e. n will be 4 less than a multiple of 35. So you must add 4 to n to make it a multiple of 35

A trickier version is: If I have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 5, what is the remainder when n is divided by 35?
n = 5a + 1 = 5x - 4
n = 7b + 5 = 7y -2
Nothing common! Now, I will need to check for the smallest such number.
I put b = 1. n = 12. Is it of the form 5a + 1? No.
Put b = 2. n = 19. Is it of the form 5a + 1? No.
Put b = 3. n = 26. Is it of the form 5a + 1? Yes.
When 26 is divided by 5, it gives a remainder of 1. When it is divided by 7, it gives a remainder if 5.
Next such number will be 35 + 26.
Next will be 35*2 + 26
and so on...
The remainder when n is divided by 35 will be 26 (or we can say it will be -9). If we want to find the number that must be added to n to make it divisible by 35, that number will be 9.
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Re: regarding the reminders problems [#permalink]

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New post 31 Mar 2011, 20:13
n is divided by 5 yeilds reminder 1 ---> n is short by 4
n is divided by 7 yeilds reminder 3 ----> n is short by 4
I think n is short by 4. So add 4 to make a whole i.e. a number divisible by 35.
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Re: regarding the reminders problems [#permalink]

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New post 04 Apr 2011, 19:47
ammulujnt wrote:
when a number n is divided by 5 yeilds reminder 1 and when divided by 7 yeilds reminder 3.what is the least possible value of K so that K+n is divisible by 35.Please propose a simple solution for this

Ans: 4


Good question!

It would be great if you can post the answer choices and remove the text "Ans: 4" so that people can try out this question.
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Re: regarding the reminders problems [#permalink]

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New post 20 Mar 2017, 11:58
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Re: regarding the reminders problems   [#permalink] 20 Mar 2017, 11:58
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