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Cyclcity of 3 : 3,9,7,1
Cyclicity of 7 : 7,9,3,1
3^50 unit digit :9
7^52 unit digit : 1
9*1:9
Option E


BrentGMATPrepNow
What is the units digit of \((3^{50})(7^{52})\)?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

Challenge: Can you solve the question in your head?

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Archit3110
Cyclcity of 3 : 3,9,7,1
Cyclicity of 7 : 7,9,3,1
3^50 unit digit :9
7^52 unit digit : 1
9*1:9
Option E
Nice work!
Can you also spot a shortcut that helps us answer the question in 10 seconds?
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BrentGMATPrepNow
Another way is

3^50*7^50*7^2
For
3^50*7^50 unit digit is 1 and 7^2:unit digit 9
1*9;9 option E

BrentGMATPrepNow
Archit3110
Cyclcity of 3 : 3,9,7,1
Cyclicity of 7 : 7,9,3,1
3^50 unit digit :9
7^52 unit digit : 1
9*1:9
Option E
Nice work!
Can you also spot a shortcut that helps us answer the question in 10 seconds?

Posted from my mobile device
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BrentGMATPrepNow
What is the units digit of \((3^{50})(7^{52})\)?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

Challenge: Can you solve the question in your head?

Useful property: \((x^k)(y^k) = (xy)^k\)

First rewrite the given expression as follows: \((3^{50})(7^{52})=(3^{50})(7^{50})(7^{2})\)

Apply the property to get: \((3^{50})(7^{50})(7^{2}) = (3 \times 7)^{50}(7^{2}) \)

Evaluate: : \((3 \times 7)^{50}(7^{2})=(21)^{50}(7^{2}) \)

Since \(21^n\) has units digit \(1\) for ANY positive integer \(n\) we can write: \((21)^{50}(7^{2}) = (????1)(49) =\) ?????9

Answer: E
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