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Bunuel
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The units digit of (44^91)*(73^37) is: 05 Mar 2015, 10:02
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

25% (medium)

Question Stats:

74% (01:17) correct 26% (01:24) wrong based on 349 sessions

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The units digit of (44^91)*(73^37) is:

(A) 2
(B) 4
(C) 6
(D) 8
(E) 0
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A
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The units digit of (44^91)*(73^37) is: 05 Mar 2015, 13:03
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IMO it should be 2 (A)
this is how I would have done in exam :
power repetitive of 4..
4^1 = 4
4^2 = 6
4^3 = 4
4^4 = 4

this means remainders of power ending with 1 gives unit digit 4
remainders of power ending with 2 gives unit digit 6
91/4 gives remainder of 3... hence unit digit left is 4

3^1=3
3^2=9
3^3=7
3^4=1
3^5=3
3^6=9
3^7=7

this means remainders of power ending with 1 gives unit digit
remainders of power ending with 2 gives unit digit 9
this means remainders of power ending with 3 gives unit digit 7
remainders of power ending with 4 gives unit digit 1
repetitive nature is 3
37/3 gives remainder 1
hence unit digit is 3
4*3=12
unit digit is 2


Kudos please if my solution is right and having appropriate method
Thanks
Celestial


Bunuel
The units digit of (44^91)*(73^37) is:

(A) 2
(B) 4
(C) 6
(D) 8
(E) 0

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KS15
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The units digit of (44^91)*(73^37) is: 07 Mar 2015, 00:34
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Bunuel
The units digit of (44^91)*(73^37) is:

(A) 2
(B) 4
(C) 6
(D) 8
(E) 0

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+1 for A. Last digit of 44^91=4 and last digit of 73^37=3. 4*3=12
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The units digit of (44^91)*(73^37) is: 08 Mar 2015, 15:06
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Bunuel
The units digit of (44^91)*(73^37) is:

(A) 2
(B) 4
(C) 6
(D) 8
(E) 0

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MAGOOSH OFFICIAL SOLUTION:

We have to figure out each piece separately, and then multiply them. The powers of 4 are particularly easy.

4^1 has a units digit of 4
4^2 has a units digit of 6 (e.g. 4*4 = 16)
4^3 has a units digit of 4 (e.g. 4*6 = 24)
4^4 has a units digit of 6
4^5 has a units digit of 4
4^6 has a units digit of 6

Four to any odd power will have a units digit of 4. Thus, any number with a units digit of four, raised to an odd power, will also have a units digit of 4. The first factor, 44^91, has a units digit of 4.

Now, the base in the second factor ends in a 3 (we can ignore the tens digit). Here is the pattern for powers of three.

3^1 has a units digit of 3
3^2 has a units digit of 9
3^3 has a units digit of 7 (e.g. 3*9 = 27)
3^4 has a units digit of 1 (e.g. 3*7 = 21)
3^5 has a units digit of 3
3^6 has a units digit of 9
3^7 has a units digit of 7
3^8 has a units digit of 1

The period is 4. This means, 3 to the power of any multiple of 4 will have a units digit of 1.

3^36 has a units digit of 1
3^37 has a units digit of 3

Thus, any number with a units digit of 7, when raised to the power of 37, will have a units digit of 3. The second factor, 73^37, has a units digit of 3.

Of course, 4*3 = 12, so any number with a units digit of 4 times any number with a units digit of 3 will yield a product with a units digit of 2.

Answer = A
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The units digit of (44^91)*(73^37) is: 10 Apr 2025, 11:21
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