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19 May 2004, 11:09
Kudos
Bookmarks
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?
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?
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20 May 2004, 04:20
Kudos
Bookmarks
solve this one too..
20 May 2004, 04:48
Kudos
Bookmarks
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
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
