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1^1+2^2+3^3+...+10^10 is divided by 5. What is the

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1^1+2^2+3^3+...+10^10 is divided by 5. What is the [#permalink] New post 17 Jan 2004, 01:35
1^1+2^2+3^3+...+10^10 is divided by 5. What is the remainder?

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

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 [#permalink] New post 17 Jan 2004, 02:55
1+4+7+6+5+6+3+6+9+0

47.

2.

While there's undoubtedly a more elegant way, brute force was quick enough.
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 [#permalink] New post 17 Jan 2004, 03:52
B) 2 Stoolfi is it 47 or 37 becoz I am getting 37 but anyways the remainder is 2 . 1+4+7+6+5+9+4+1+0 = 37/5 = 2
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 [#permalink] New post 17 Jan 2004, 20:53
stoolfi wrote:
1+4+7+6+5+6+3+6+9+0

47.

2.

While there's undoubtedly a more elegant way, brute force was quick enough.


Brute force is fine for this problem so long as you understand the principle behind what you are doing. Good job.

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 [#permalink] New post 31 Dec 2004, 15:05
can you explain what are you doing here?
i get a remainder of 4? i dont understand how you got 7 for 3^3?

rakesh1239 wrote:
B) 2 Stoolfi is it 47 or 37 becoz I am getting 37 but anyways the remainder is 2 . 1+4+7+6+5+9+4+1+0 = 37/5 = 2
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 [#permalink] New post 31 Dec 2004, 15:29
fresinha12 wrote:
can you explain what are you doing here?
i get a remainder of 4? i dont understand how you got 7 for 3^3?

rakesh1239 wrote:
B) 2 Stoolfi is it 47 or 37 becoz I am getting 37 but anyways the remainder is 2 . 1+4+7+6+5+9+4+1+0 = 37/5 = 2

3^3 = 27
All you care for is the units digit and you see the pattern.

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 [#permalink] New post 31 Dec 2004, 15:47
Paul..u r right...i am just concerned that how I could do this in 2-3 minutes? I took me about 5-6 minutes, and i still made a mistake, is there a short cut to these unit digit problems,



Paul wrote:
fresinha12 wrote:
can you explain what are you doing here?
i get a remainder of 4? i dont understand how you got 7 for 3^3?

rakesh1239 wrote:
B) 2 Stoolfi is it 47 or 37 becoz I am getting 37 but anyways the remainder is 2 . 1+4+7+6+5+9+4+1+0 = 37/5 = 2

3^3 = 27
All you care for is the units digit and you see the pattern.
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 [#permalink] New post 31 Dec 2004, 15:57
With practice, a problem like this should take you about 1 min. You will see, practice makes perfect. Developing a familiarity with many different type of GMAT problems will make you find the right way to approach it more quickly. The solving part should only take a minute though :wink:

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 [#permalink] New post 31 Dec 2004, 16:20
yeah you are right...but i usually make mistakes when I have a lot of calculations to do....one other number whos digit is hard to calculate is 7...
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 [#permalink] New post 31 Dec 2004, 18:54
AkamaiBrah wrote:
stoolfi wrote:
1+4+7+6+5+6+3+6+9+0

47.

2.

While there's undoubtedly a more elegant way, brute force was quick enough.


Brute force is fine for this problem so long as you understand the principle behind what you are doing. Good job.


AkamaiBrah and Paul,

this is really a good question. is there any other short-cut way of solving problems like this?


MA

Last edited by MA on 31 Dec 2004, 18:55, edited 1 time in total.
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 [#permalink] New post 31 Dec 2004, 19:12
I believe there is another method explained by another member before but I would stick to the pattern method. It really does not take much time and with practice, a problem like this could be done within 1 min, never mind 2 minutes.

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 [#permalink] New post 31 Dec 2004, 21:46
I would look for a pattern in the last digit (without doing the multiplications for large n say 7^7 or 9^9):

1^1 = 1
2^(2, 4, 6, or 8) = 4, 6, 6, or 6 resp.
3^3 = 7
5^5 = 5
7^7 = 3
9^9 = 9
10^10 = 0

Sum = 47; remainder 2.

Brute force will also do for this problem .
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pattern for units digits [#permalink] New post 02 Jan 2005, 03:26
Hi

The units digits of numbers ending with the following numbers have a cyclicity of 4

numbers ending in 2, 3, 7 and 8

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64 and so on. i.e., every 4th power has the same units digit. The same holds for 3, 7 and 8.

The units digits of numbers ending in 4 and 9 (both are perfect squares) has a cyclicity of 2

4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256 and so on.

The units digits of numbers ending in 0, 1, 5, 6 all end in 0, 1, 5, and 6 respectively.

this should help solve the above problem quickly

Manish
pattern for units digits   [#permalink] 02 Jan 2005, 03:26
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