If [(243)^x]*[(463)^y] = n, where x and y are positive integers, what is the units digit of n?
(1) x + y = 7
(2) x = 4
OA A
Exponent problem - what is the units digit of n?
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- logitech
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243 = 3^5tritrantran wrote:If [(243)^x]*[(463)^y] = n, where x and y are positive integers, what is the units digit of n?
(1) x + y = 7
(2) x = 4
OA A
463 ends with a 3. So we have to know how many times we will multiply 3's at the end of each numbers.
1) 7 times - SUF
2) we dont know Y - INSUF
Choose (a)
LGTCH
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3^1 = 3tritrantran wrote:I'm still not seeing it. Is there a rule for numbers ending in 3?
3^2 = 9
3^3 = 7
3^4 = 1
3^5 = 3
3^6 = 9
3^7 = 7
3^8 = 1
So the unit digits can be 1, 3, 7 or 9 depending on the power of 3.
LGTCH
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Most exponent problems of this type deal with what's called cyclicity of units digit.
As Logitech mentioned in his post above it cylces as follows for 3
3,9,7,1,3,9,7,1,3,9,7,1.......
Stmt I
For 3 it will be 3,9,7,1,3,9,7,1,3,9,7,1.......
Since x+y=7
the units digit of the product will be 7 no matter how u distribute 7 between x and y(x=4 and y=3 or x=6 and y=1 etc...)
SUFFICIENT
Stmt II
x=4
If y=1 then the units digit of the product will be 3
3,9,7,1,3,9,7,1,3,9,7,1.......
If y =2 then the units digit of the product will be 9
3,9,7,1,3,9,7,1,3,9,7,1.......
Any of the 4 values from 3,9,7,1(abive I have given 2 examples) possible
Cannot get to one defnite answer for the units digit
INSUFFICIENT
A)
Hope this helps also!
[/b]
As Logitech mentioned in his post above it cylces as follows for 3
3,9,7,1,3,9,7,1,3,9,7,1.......
Stmt I
For 3 it will be 3,9,7,1,3,9,7,1,3,9,7,1.......
Since x+y=7
the units digit of the product will be 7 no matter how u distribute 7 between x and y(x=4 and y=3 or x=6 and y=1 etc...)
SUFFICIENT
Stmt II
x=4
If y=1 then the units digit of the product will be 3
3,9,7,1,3,9,7,1,3,9,7,1.......
If y =2 then the units digit of the product will be 9
3,9,7,1,3,9,7,1,3,9,7,1.......
Any of the 4 values from 3,9,7,1(abive I have given 2 examples) possible
Cannot get to one defnite answer for the units digit
INSUFFICIENT
A)
Hope this helps also!
[/b]
- rohit_gmat
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Took me ages to understand this one...
(please correct me if my logic is wrong here)...
Since we are concerned only abt the unit's digits, lets focus on those only...
243 & 463 both have 3 in the units place...
to solve for the units digit's value, we can boil the equation down to :
(3^x)(3^y) = last few digits of n ....
so we have x+y to solve for... since we know the cyclic form of 3 to the power smths = 1 3 9 7 1 3 9 7 ...
Stmnt 1) x + y = 7
gives us exactly what we are looking for (x+y)
SUFF
Stmnt 2) x =4
y could be ANY other +ve integer on earth...
INSUFF
So it should be A....
(please correct me if my logic is wrong here)...
Since we are concerned only abt the unit's digits, lets focus on those only...
243 & 463 both have 3 in the units place...
to solve for the units digit's value, we can boil the equation down to :
(3^x)(3^y) = last few digits of n ....
so we have x+y to solve for... since we know the cyclic form of 3 to the power smths = 1 3 9 7 1 3 9 7 ...
Stmnt 1) x + y = 7
gives us exactly what we are looking for (x+y)
SUFF
Stmnt 2) x =4
y could be ANY other +ve integer on earth...
INSUFF
So it should be A....