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1,2,4,7,11,16,22,29,37,46,56,… Fin the remainder when the 2008th number in the above pattern is divided by 5.


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Hi. Just saw this and first thought is it may be too hard for P5 to know sum of consecutive series if not doing Math Olympiad. Initially, I’m not sure why the below pattern occurs but the “why” is not critical for deducing the pattern to solving the question. (The “why” is deduced at footnote for interest only.)

1, 2, 4, 7, 11,     16, 22, 29, 37, 46,    

56, 67, 79, 92, 106    

121, 137, 154, 172, 191….

(I have doubts at first on the remainder pattern too, that is why I did another 2 sets. My approach was to solve for the answer first and then, since I do not have exam time constraints, to figure why the pattern exists.)

remainder when divide by 5:

1, 2, 4, 2 1      1, 2, 4 , 2 1            

1, 2, 4, 2,1,    

1,2,4,2,1   …..

five in a set

2008th number will be 2005th + 3rd number in a set. 

Remainder is 4 for 2008th term. 

*Really for interest only:

We start with first number “1” inside a black box. We know we keep adding consecutive numbers into the box to derive a number. So add 1, then add 2, then add 3…… . Without knowing what the ending number is, we can deduce the remainder (when divide by 5) for first few numbers easily. So let’s start:

First number “1” inside th box, remainder is 1 when divided by 5.

Add 1. Since first remainder is 1, the new remainder is 2.

Add 2. Since last remainder is 2, the new remainder is 4.

Add 3. Since last remainder is 4, new remainder is 2.

Add 4. Since last remainder is 2, new remainder 1.

Add 5. No effect on remainder. Same remainder 1.

Add 6. It is just like adding 1 (because the divisor is 5). Likewise, adding 7,8,9,10 has same effect on remainder as adding 2,3,4,5 respectively. So the pattern resets every 5 numbers and holds true throughout the series. 

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P5 should have already learn how to add  1 + 2 + 3 + …….

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