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# This one is from a Kaplan test I took : What is the

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This one is from a Kaplan test I took : What is the [#permalink]

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12 Aug 2007, 19:29
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This one is from a Kaplan test I took :

What is the result of adding the units digit of 2002^2002 and 7007^7007?
A) 1
B) 4
C) 5
D) 7
E) 9

OA is D .Can someone explain how to do this.
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Re: PS question - number properties [#permalink]

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12 Aug 2007, 19:41
Kiski wrote:
This one is from a Kaplan test I took :

What is the result of adding the units digit of 2002^2002 and 7007^7007?
A) 1
B) 4
C) 5
D) 7
E) 9

OA is D .Can someone explain how to do this.

Since we are only looking at the units digit:
2002^2002
2 * 2 = 4
we know 2002^2002 ends with 4

7007^7007
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 343*7 ends in 1
7^5 = 1 * 7 = ends with 7 <-- here is where the sequence repeats
7^6 = 7 * 7 = ends with 9
7^7 = 9 * 7 = ends with 3
we know 7007^7007 ends with 3

thus adding the two unit digits: 4+3 = 7 (D)
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Re: PS question - number properties [#permalink]

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12 Aug 2007, 20:22
Kiski wrote:
This one is from a Kaplan test I took :

What is the result of adding the units digit of 2002^2002 and 7007^7007?
A) 1
B) 4
C) 5
D) 7
E) 9

OA is D .Can someone explain how to do this.

2^1=2
2^2=4
2^3=8
2^4=16

2^5=32
2^6=64
2^7=128
2^8=256

see the pattren. in every 4 powers, the pattren repeats.
so 2002^2002 has 4 in unit digit.

similarly:
7^1=7
7^2=49
7^3=343
7^4=2401
7^4=16807

now here to the repeats after every 4 powers and 7007^7007 has 3 in unit digit.

therfore the sum is 7.
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12 Aug 2007, 21:52
Just thought I should mention/ask: Once you have the pattern, you divide the exponent by the number in the sequence (four for both 2002 and 7007 in this case) and then your remainder determines what the actual digit is...correct?

So, in this case:

2
4
8
6
-------> cycle is 4, so 2002 / 4 = 500 r2, so the digit is the second number in the sequence: 4

7
9
3
1
-------> cycle is also 4, so 7007 / 4 = 1751 r3, so the digit is the third number in the sequence: 3
12 Aug 2007, 21:52
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