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77% (01:12) correct
23%
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What is the units digit of \(9^{19} - 7^{15}\)?
A. 2
B. 4
C. 5
D. 6
E. 7
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Cyclic order of 9 raised to the power is [1,9]
Cyclic order of 9 raised to the power is [9,3,1,7]
So, the difference of the said expression is 2
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Cyclic order of 9 raised to the power is [9,3,1,7]
So, the difference of the said expression is 2
Posted from my mobile device
9^19 - 7^15
9^1 = 9 unit digit = 9
9^2 = 81 unit digit = 1
9^3 = 729 unit digit = 9
9^4= 6561 unit digit = 1
hence cyclisity of the powers of 9 is 2
7^1= 7 unit digit =7
7^2= 49 unit digit =9
7^3= 343 unit digit =3
7^4= 2401 unit digit =1
7^5=16807 unit digit =7
hence cyclisity of the powers of 7 is 4
9^19 - 7^15 = (9^18)*9 - (7^12)*7^3
=1*9 - 1*3 = 9-3 =6
Hence D.
9^1 = 9 unit digit = 9
9^2 = 81 unit digit = 1
9^3 = 729 unit digit = 9
9^4= 6561 unit digit = 1
hence cyclisity of the powers of 9 is 2
7^1= 7 unit digit =7
7^2= 49 unit digit =9
7^3= 343 unit digit =3
7^4= 2401 unit digit =1
7^5=16807 unit digit =7
hence cyclisity of the powers of 7 is 4
9^19 - 7^15 = (9^18)*9 - (7^12)*7^3
=1*9 - 1*3 = 9-3 =6
Hence D.
What is the units digit of \(9^{19}−7^{15}\)?
A. 2
B. 4
C. 5
D. 6
E. 7
For \(9^{19}\) unit digit would be 9 as odd power of 9 gives 9 as unit digit(even power gives one)
For \(7^{15}\) unit power would be 3 since 7 raised to power which is a multiple of 3 gives 3 as unit digit.
Hence unit place of \(9^{19}−7^{15}\) = Unit place of \(9^{19}\) - Unit place of \(7^{15}\)
= 9 - 3
= 6
Answer D.
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A. 2
B. 4
C. 5
D. 6
E. 7
For \(9^{19}\) unit digit would be 9 as odd power of 9 gives 9 as unit digit(even power gives one)
For \(7^{15}\) unit power would be 3 since 7 raised to power which is a multiple of 3 gives 3 as unit digit.
Hence unit place of \(9^{19}−7^{15}\) = Unit place of \(9^{19}\) - Unit place of \(7^{15}\)
= 9 - 3
= 6
Answer D.
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What is the units digit of 9^19 - 7^15?
9^1=9
9^2=81
9^3= _ _ 9
the pattern is _9,_1,_9,_1
19/2=8 R1
therefore the unit digit of 9^19 is 9
7^1=7
7^2=49
7^3=_ _ 3
7^4= _ _ _ 1
7^5= _ _ _ _ _ 7
the pattern is _7, _9,_3,_1
15/4=3 R3
therefore the unit digit of 7^15 is 3
9-3= 6
D
9^1=9
9^2=81
9^3= _ _ 9
the pattern is _9,_1,_9,_1
19/2=8 R1
therefore the unit digit of 9^19 is 9
7^1=7
7^2=49
7^3=_ _ 3
7^4= _ _ _ 1
7^5= _ _ _ _ _ 7
the pattern is _7, _9,_3,_1
15/4=3 R3
therefore the unit digit of 7^15 is 3
9-3= 6
D
Quote:
\(units:9^{19}…cycles(9)=[9,1]=2…remainder:19/2=1…units:1st=[9]\); or,
\(units:9^{19}=(3^2)^{19}=3^{38}…cycles(3)=[3,9,7,1]=4…remainder:38/4=2…units:2nd=[9]\)
\(units:7^{15}…cycles(7)=[7,9,3,1]=4…remainder:15/4=3…units:3rd=[3]\)
\(units:9^{19}−7^{15}=9-3=6\)
Answer (D)
Posts: 14
Ans D.. Use cyclicity concept
9^1=9
9^2=.1
...
These units digits are repeated in every two terms.
—> If the power of 9 is odd, units digit ends with 9.
—> if the power of 9 is even, units digit ends with 1.
——>Units digit of 9^19 ends with 9
7^1=7
7^2=.9
7^3=..3
7^4=..1
These units digits are repeated in every four terms.
—> units digit of 7^15 is the same as that of 7^3 —> ...3
—> ...9 —3=...6
The answer is D
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9^2=.1
...
These units digits are repeated in every two terms.
—> If the power of 9 is odd, units digit ends with 9.
—> if the power of 9 is even, units digit ends with 1.
——>Units digit of 9^19 ends with 9
7^1=7
7^2=.9
7^3=..3
7^4=..1
These units digits are repeated in every four terms.
—> units digit of 7^15 is the same as that of 7^3 —> ...3
—> ...9 —3=...6
The answer is D
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Posts: 346
.....9-.....3=....6
Hence, option D
Posted from my mobile device
Hence, option D
Posted from my mobile device
The powers of 9 have the following pattern:
9^1 = 9
9^2 = 81
9^3 = 729
9^4 = 6,561
9^5 = 59,049
As you can see, all even powers will have a units digit of 1 and odd powers will have a units digit of 9.
9^19, 19 is odd so units digit is 9.
Powers of 7 repeat itself after 4 transitions, for example:
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2,401
7^5 = 16,807
We see that after the fourth power the pattern starts over again, so:
7^15, we take 15 and divide it by four leaving us with 3, the units digit is the 3rd in the repeating list -> 3.
9-3 = 6
IMO
9^1 = 9
9^2 = 81
9^3 = 729
9^4 = 6,561
9^5 = 59,049
As you can see, all even powers will have a units digit of 1 and odd powers will have a units digit of 9.
9^19, 19 is odd so units digit is 9.
Powers of 7 repeat itself after 4 transitions, for example:
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2,401
7^5 = 16,807
We see that after the fourth power the pattern starts over again, so:
7^15, we take 15 and divide it by four leaving us with 3, the units digit is the 3rd in the repeating list -> 3.
9-3 = 6
IMO
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Expert reply
Posts: 22,737
Bunuel
The patter of units digits for a base of 9 is:
9^1 = 9
9^2 = 1
9^3 = 9
So, 9 raised to an odd power results in a units digit of 9.
The pattern of units digits for a base of 7 is:
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
Thus, we see that 7^4n (where n is an integer) has a units digit of 1. Thus, 7^16 has a units digit of 1, and 7^15 has a units digit of 3. Thus, the units digit of 9^19 - 7^15 is 9 - 3 = 6.
Answer: D
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Cyclicity Concept:
Pattern of 9^2: 9,81,729 --> unit digit pattern is (9,1,9,1..)
pattern is repeated after every second term so 19/2= quotient is 9 and remainder is 1 (R1)
Pattern of 7^2: 7,49, 343, 2401, 16807 --> unit digit pattern is (7,9,3,1,7...)
pattern is repeated after every fourth term so 15/4= quotient is 3 and remainder is 3 (R3)
We will match the remainders with pattern position:
R1's position is 1 in the pattern of 9 i.e ---> at 1 position there exists 9
R3's position is 3 in the pattern of 7 i.e ---> at 3 position there exists 3
so 9 -3 = 6
Pattern of 9^2: 9,81,729 --> unit digit pattern is (9,1,9,1..)
pattern is repeated after every second term so 19/2= quotient is 9 and remainder is 1 (R1)
Pattern of 7^2: 7,49, 343, 2401, 16807 --> unit digit pattern is (7,9,3,1,7...)
pattern is repeated after every fourth term so 15/4= quotient is 3 and remainder is 3 (R3)
We will match the remainders with pattern position:
R1's position is 1 in the pattern of 9 i.e ---> at 1 position there exists 9
R3's position is 3 in the pattern of 7 i.e ---> at 3 position there exists 3
so 9 -3 = 6
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