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Exponents

by lj » Sat Feb 12, 2011 5:18 pm
Could someone please advise on how the solve the problem below ? Thanks

Q: what is the units digit of (13)^4 (17)^2 (29)^3

a. 9
b. 7
c. 5
d. 3
e. 1
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by Night reader » Sat Feb 12, 2011 6:48 pm
lj wrote:Could someone please advise on how the solve the problem below ? Thanks

Q: what is the units digit of (13)^4 (17)^2 (29)^3

a. 9
b. 7
c. 5
d. 3
e. 1
(1)*(9)*(9)=(1)
IOM E
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by lj » Sun Feb 13, 2011 4:58 pm
Night reader wrote:
lj wrote:Could someone please advise on how the solve the problem below ? Thanks

Q: what is the units digit of (13)^4 (17)^2 (29)^3

a. 9
b. 7
c. 5
d. 3
e. 1
(1)*(9)*(9)=(1)
IOM E
Thanks for the help. How did you get those numbers ? Cant seem to figure that part out .
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by Ashish8 » Sun Feb 13, 2011 5:43 pm
I also agree with Night Reader E

3 * 3 * 3 * 3 = units digit of 1

7 * 7 = units digit of 9

9 * 9 * 9 = units digits of 9


We just need to multiply the units digits cause anything else doesn't affect the product of the units digit.


9 * 9 * 1 = units digit of 1.
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by Anurag@Gurome » Mon Feb 14, 2011 5:44 am
lj wrote:What is the units digit of (13)^4 (17)^2 (29)^3 ?

a. 9
b. 7
c. 5
d. 3
e. 1
Unit's digit of (13^4)*(17^2)*(29^3) will be same as the unit's digit of the product of unit's digits of (13^4), (17^2), and (29^3).

Now, unit's digit of
  • (13^4) = unit's digit of (3^4) = 1
    (17^2) = unit's digit of (7^2) = 9
    (29^3) = unit's digit of (9^3) = 9
Hence, required unit's digit = unit's digit of the product of 1, 9, and 9 = unit's digit of 81 = 1
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