What is the units digit of 9^19 - 7^15? : Problem Solving (PS)

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07 Oct 2019, 22:20

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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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08 Oct 2019, 02:53

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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

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08 Oct 2019, 03:52

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Quote:

What is the units digit of \(9^{19}−7^{15}\)?

A. 2
B. 4
C. 5
D. 6
E. 7

\(units:9^{19}…cycles(9)=[9,1]=2…remainder:19/2=1…units:1st=[9]\); 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)

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08 Oct 2019, 04:56

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Ans D.. Use cyclicity concept

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08 Oct 2019, 05:11

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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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08 Oct 2019, 12:33

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.....9-.....3=....6
Hence, option D

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08 Oct 2019, 13:49

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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

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11 Oct 2019, 15:54

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Bunuel

Competition Mode Question

What is the units digit of \(9^{19} - 7^{15}\)?

A. 2
B. 4
C. 5
D. 6
E. 7

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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22 Jan 2020, 00:05

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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

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