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# What is the units digit of the product (32^28) (33^47) (37^19)?

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Re: What is the units digit of the product (32^28) (33^47) (37^19)? [#permalink]
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That's a valid approach, amynicole.
There's another approach that's much faster. Can you spot it?

Hint:
notice that the largest exponent is equal to the sum of the other two exponents

Cheers,
Brent
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Re: What is the units digit of the product (32^28) (33^47) (37^19)? [#permalink]
GMATPrepNow
That's a valid approach, amynicole.
There's another approach that's much faster. Can you spot it?

Hint:
notice that the largest exponent is equal to the sum of the other two exponents

Cheers,
Brent

Hi Brent,

How does "notice that the largest exponent is equal to the sum of the other two exponents" Helps? All(2, 3, 7, 8) the 3 numbers have cycle of 4 repetitions.

Thanks
Nandish
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Re: What is the units digit of the product (32^28) (33^47) (37^19)? [#permalink]
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GMATPrepNow
That's a valid approach, amynicole.
There's another approach that's much faster. Can you spot it?

Hint:
notice that the largest exponent is equal to the sum of the other two exponents

Cheers,
Brent

I also solved it by using cycle and took me around 2:30 minutes
But this hint is genius

32^28 * 33^47 * 37^19 = (32^28 * 33^28 )*(33^19 * 37 ^19) == (32 *33)^28 * [Rest Does not matter] as 6 has cycle of 1 and unit digit will always be 6.

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Re: What is the units digit of the product (32^28) (33^47) (37^19)? [#permalink]
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Wow! This is cool. I did it the long way...hopefully I can remember this on the test

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Re: What is the units digit of the product (32^28) (33^47) (37^19)? [#permalink]
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GMATPrepNow
What is the units digit of the product (32^28)(33^47)(37^19)?

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

Cyclicity of number 2 = 4

So, $$32^{28}$$ will have units digit 6

Cyclicity of number 3 = 4

So, $$33^{47} = 33^{44+3}$$ will have units digit 7

Cyclicity of number 7 = 4

So, $$37^{19} = 37^{16+3}$$ will have units digit 3

Thus, the units digit of $$32^{28}*33^{47}*37^{19} = 6*7*3 = 6$$; Answer will be (D) 6
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Re: What is the units digit of the product (32^28) (33^47) (37^19)? [#permalink]
BrentGMATPrepNow
What is the units digit of the product $$(32^{28})(33^{47})(37^{19})$$?

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

Hint:
There’s a fast approach and a slow approach

*kudos for all correct solutions
$$2^4 = 6$$

So, $$32^{28} = 2^{140} = 2^{4*35} =$$Units digit $$6$$

$$3^4 = 1$$ & $$3^3 = 7$$

So, $$33^{47} = 33^{4*11+3} = 1*7 =$$ units digit $$7$$

$$7^4 = 1$$ & $$7^3 = 3$$

So, $$37^{19} = 37^{4*4+3} =$$ Units digit 3

Thus, units digit is $$6*7*3 = 6$$, Answer is (D)
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Re: What is the units digit of the product (32^28) (33^47) (37^19)? [#permalink]
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Re: What is the units digit of the product (32^28) (33^47) (37^19)? [#permalink]
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