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9827^9826 divided by 5 has a remainder of 1) 2 2) 4 3)

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9827^9826 divided by 5 has a remainder of 1) 2 2) 4 3) [#permalink] New post 18 Nov 2003, 09:52
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9827^9826 divided by 5 has a remainder of

1) 2
2) 4
3) 3
4) 1
Director
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 [#permalink] New post 18 Nov 2003, 10:00
7*7 is 49
keep the 9, *7 --63,
keep the 3, *7 --21
keep the 1, *7 --7
keep the 7, *7 --49
Keep the 9, *7 --63



Any number ending in 3, divided by 5, yields a remainder of 3.
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 [#permalink] New post 18 Nov 2003, 10:53
wonder_gmat--

If you were multiplying the two figures, you would be correct...
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 [#permalink] New post 18 Nov 2003, 11:01
Oh I divided by a wrong number. nevermind.

My new guess is 4
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 [#permalink] New post 18 Nov 2003, 11:02
wonder_gmat, you are correct. I used seven sevens, when I meant to use six sevens.

D'oh!
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 [#permalink] New post 18 Nov 2003, 11:22
I will :2gunfire: :( myself for it. Thanks for correcting it guys..
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 [#permalink] New post 19 Nov 2003, 06:52
can you explain S. why you use 5 sevens? thanks.
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 [#permalink] New post 19 Nov 2003, 07:22
I can explain how I approached... here it goes..

9827^9826

7^1 unit digit = 7
7^2 unit digit = 9
7^3 unit digit = 3
7^4 unit digit = 1
7^5 unit digit = 7
......

9826/7 = 1403*7 +5

9827^1403*7 . 9827^5 = unit digit 7 + unit digit 7
divide by 5 ...remainder = 2+2 = 4

this seems lot of work, but you can solve it in <40sec. I could..!
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 [#permalink] New post 19 Nov 2003, 07:47
allycat wrote:
can you explain S. why you use 5 sevens? thanks.


I used seven sevens, before I start listing them line-by line, i use two of them immediately above the line items in my post.

As I stated, this was wrong, as I should have used six of them!
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 [#permalink] New post 19 Nov 2003, 10:18
Hi Stoolfi,
i like your approach in solving this problem. please explain WHY you use 7 (unit digit) and raise to the 6th power (or using 6 sevens). i'm not sure the logic behind it. where do you get 6 from (the total number of digits, which is 8 in this case? or is there a special rule/formula?) thanks again. D
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 [#permalink] New post 19 Nov 2003, 11:05
Well, allycat, the more I think about this problem, the less assured I am about my approach being correct. In fact, it's wrong.

I will eat crow and wipe the egg off of my face. I should know better.

The longer (and probably correct) way:

7^1 leaves remainder of 2
7^2 leaves remainder of 4
7^3 leaves remainder of 3
7^4 leaves remainder of 1
7^5 leaves remainder of 2
7^6 leaves remainder of 4
7^7 leaves remainder of 3
7^8 leaves remainder of 1

If the pattern continues, any multiple of four (other than zero) will leave a remainder of 1.

Any multiple of two that is not a multiple of four will leave a remainder of 4.

You can split 9826 in half, and wind up with an integer only once. So it is a multiple of two, but not four. So the remainder will be four.

I am humbled by wonder_gmat.
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 [#permalink] New post 19 Nov 2003, 11:24
Hey S. your new approach makes perfect sense to me. thank you.

i like the previous approach you used because it was so simple. hence, i thought you knew some shortcuts. perhaps you can tell me what you were thinking when you decided to raise 7 to the 6th power -- which i think may work as well 7^6 (raise the unit digit by the unit digit of the exponent)

Was this what you were thinking?
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 [#permalink] New post 19 Nov 2003, 13:45
Hallo guys, i came up with something but i need your opinion:
((9825+1)+1)^9826 so it seems to me that the remainder should be 2, since 9825 is exactly divisible by 5,please give me your advise :?
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 [#permalink] New post 19 Nov 2003, 13:56
ooooohhhh... !!

BG,

consider: (a+b)^2 = a^2 + b^2 + 2ab . not a^2 + b^2

(9825+2)^9826 will be eq. to 9825^9826 + 2^9826 + something.....

Last edited by dj on 20 Nov 2003, 06:38, edited 1 time in total.
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 [#permalink] New post 20 Nov 2003, 05:46
stoolfi,
Thanks. You approach is perfectly fine too. In fact, mine is very similiar to yours.
------

Here's how I did it:

You only have to focus on the unit digit, 7 in this case, of the actual number. You raise 7 to power by first few integers, i.e. 1, 2, 3, 4, 5, 6. You don't even need to multiply out whole the number, you can just get away with finding the unit digit. i.e. 7, 9, 3, 1, 7, 9, etc. You can see that the patter in the unit digit place takes place and that the patter repeats every four steps.

So, I divided 9826 by 4 and got remainder 2. That tells me that there four complete patterns plus 2 extra steps. Now I know that raising 7 to 9824 has to leave 1 as the unit digit. I just go two steps further and I get 9 in the unit digit. Just divide 9 by 5 and you get remainder 4!

I don't know if you care for the timing but I could solve this problem in 50 sec using this trick. It could be done faster but I wanted to double check my answer. The trick is pretty straight forward, I suppose.
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 [#permalink] New post 24 Nov 2003, 15:22
stoolfi wrote:
7*7 is 49
keep the 9, *7 --63,
keep the 3, *7 --21
keep the 1, *7 --7
keep the 7, *7 --49
Keep the 9, *7 --63



Any number ending in 3, divided by 5, yields a remainder of 3.




i did it this way also....but as you mentioned you used an extra 7, you should have stopped at the point when the units digit was 9 and thus if you divide anything with a units digit of nine by 5 you get a remainder of 4....so why did you switch your method stoolfi??
  [#permalink] 24 Nov 2003, 15:22
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