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# 1) What is the units digit of 9^231? 2) What is units digit

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1) What is the units digit of 9^231? 2) What is units digit [#permalink]  19 May 2004, 10:09
1) What is the units digit of 9^231?

2) What is units digit of 7^231 * 2^28?

3) What is units digit of 6^31 * 7^7 * 3^35?

4) What is the remainder when 6^31 * 7^7 * 3^35 is divided by 11?
CEO
Joined: 15 Aug 2003
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solve this one too..
Senior Manager
Joined: 11 May 2004
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1) What is the units digit of 9^231?

2) What is units digit of 7^231 * 2^28?

3) What is units digit of 6^31 * 7^7 * 3^35?

4) What is the remainder when 6^31 * 7^7 * 3^35 is divided by 11?

ANS:

FOR ANY DIGIT "N"

N^(4*X+1) WILL HAVE UNIT DIGIT SAME AS N WHERE X IS A WHOLE NUMBER.

BASED ON THIS RULE

1) What is the units digit of 9^231?

LET US FIND

231 = 57*4 + 3

SO UNIT DIGIT OF 9^231 IS SAME AS 9 ^3

WHICH IS 9 IN THIS CASE

2) What is units digit of 7^231 * 2^28?

SAME RULE APPLIES

UNIT DIGIT IS 6*1 = 6

3) What is units digit of 6^31 * 7^7 * 3^35?

6*6*7

UNIT DIGIT = 2

4) What is the remainder when 6^31 * 7^7 * 3^35 is divided by 11?

FOR EVERY FACTOR TRY TO FIND A NUMBER CLOSE TO 11

6^31 * 7^7 * 3^35

= 6* (6^2) ^ 15 * 7*(7^2)^3 *3^2*(3^3)^11

= 6*(33+3)^15*7*(44+5)^3*9*(22+5)^11

REMAINDER WILL BE

= 6* 3^15 * 5 ^3 * 9 * 5^11

BY FOLLOWING SAME PROCESS UNTIL WE FIND A NUMBER < 11

= 2 * 3^18 * 5^14

= 2* 5^6* 3^7

= 2* 3^2* 3* 5^2
= 2* 5^3

= 2*5* 3 = 30 = 22+ 8
= 8

BUT THIS METHOD IS VERY CUMBERSOME..

ANY ONE HAVE BETTER METHOD??

SAMMY
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Joined: 31 Mar 2004
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Location: texas
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hey sammy,

i think you got the process but i think you made some typos in number 2 and 3..

number two shd be 3 *1 instead of 6*1 and the unit digit shd be 3

also in number 3,
it shd be 6*9*7 instead of 6*6*7
and so the unit digit shd be 8
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