Unit digit-how to find out?

[Mahi111]

Can someone please tell me how to find of the unit digit of the following.

 

Find out the unit digit of 4^27 * 5^27 * 3^27?

 

Not only the above one,how to find unit digit of any value to any power?

 

Please some one explain me this concept.

 

Thx.

[moe]

5, 15, 25.... to any power, always unit digit=5, and same goes for 6

 

others follow easy patterns:

 

the easiest are 4 and 9:

 

4, 6, 4, 6.... and 9, 1...

 

2^1=2

2^2=4

2^3=8

2^4=16

2^5=32 (back to the start, unit = 2)

 

for 8 it's simmilar to 2:

8, 4, 2, 6

 

3 and 7 are simmilar too:

 

3, 9, 7, 1 and 7, 9, 3, 1

 

helps?![dance] [banana] [dance]

[moe]

and to find out the unit digit of 4^27 * 5^27 * 3^27... 4^27 will have unit = 4 and 5^27, unit = 5 ==> 4^27 * 5^27, unit = 0 ==> 4^27 * 5^27 * 3^27, unit = 0

 

is that the answ?

[Mahi111]

Thank you moe,Yes.u r answer is correct.

 

but I did not understand this part

 

and anything to the power of 2, mean that first I should check where it is repeating,then, that reaping one is the unit digit or else ,I strucked up here.

 

 

2^1=2

2^2=4

2^3=8

2^4=16

2^5=32 (back to the start, unit = 2)

 

 

for 8 it's simmilar to 2:

8, 4, 2, 6

 

3 and 7 are simmilar too:

 

3, 9, 7, 1 and 7, 9, 3, 1

 

I think I have not understand the concept perfectly,can you please help me by saying bit clearly.

[moe]

ok... lets see if i explain myself any better...:blush:

 

2^1=2

2^2=4

2^3=8

2^4=16

2^5=32 ==> same unit digit as 2 to the power of 1 (2^1=2)

2^6=64 ==> same unit digit as 2 to the power of 2 (2^2=4)

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..

...

units digit can be grouped into a sequence of 4 numbers that repeat themselves 2/4/8/6==> 2 will be the unit number for 2 to the power of 1, 5, 9, 13... and, in the same way, 4 will be the unit number for 2 to the power of 2, 6, 10, 14...

eg: what's the unit digit of 2^16? 16/4=4 no remainder, therefore, the units digit will be the last one in the series=6

 

(and this same sequence applies to the powers of numbers 12, 22, 32... 32^3 will have a unit digit of 8, the same as 22^3 and 2^3)

 

numbers 3, 7 and 8 also have sequences of 4 numbers:

for nº3: 3/9/7/1

eg:3^4 will have a unit digit of 1 (and 13^4 as well), 3^5 will have unit digit 3...

the sequences of 7 (7, 9, 3, 1) and 8 (8, 4, 2, 6) follow the same rules

[sruthi07]

explanation moe, and time saving too!!

thank you!

 

i had a diff approach, let me know whther im right

 

viz; a ^b * c^d * e ^f ; unit digit of this one will be

 

a^1* b^1* c^1...???

 

so in the problem above

5*4*3 has a unit digit = 0

[moe]

i'm sorry, but i'm not sure i get the method... if i understand well, in a shorter example, like 4^2*3^7, u would do 4*2*3*7? that wouldn't work on this new example...

[gre_girl]

hi moe

sory but i cant understand this concept.can u explain again

thanks in advance

regards

[whaw]

helo mahi111, u can solv for the units digit by getting the patern...here's how

 

4^1=4

4^2=6

4^3=4

4^4=6

 

So 27/2=13 remainder 1 therefore units digit is 4.

 

5^1=5

5^2=5

5^3=5...therefore the units digit is 5...and so on..

[taminkeu]

4^27 * 5^27 * 3^27 = (4*5*3)^27 = 60^27

unit of 60^1 = 0

unit of 60^2 = 0

unit of 60^3 = 0

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unit of 60^27= 0

[mystic87]

choices, choices, choices...

these are all great by the way. really makes you think about the properties of numbers.

[kobra]

1. 4^2 * 3^7 unit digit = (4*3) = 12 =2

 

2. 4^2 unit digit = 6

3^7 unit digit = 7

6*7 = 42

unit dig =2

[kobra]

isnt the unit digit of 3^27 = 7

 

3

power 1 = 3

2 = 9

3 = 27

4 = 81

3,9,7,1

27/4 = 6 R3

3rd down is 7.....

 

still it would be irrelevant since the unit of the first two terms multiplied end in 0......

[Sonal Bhardwaj]

Find out the unit digit of 4^27 * 5^27 * 3^27?

 

Hi. I have worked out the solution for your question.

 

In the question, we have 4,5,3 as Base.

 

First step is to find out the cyclicity of these numbers.

 

For 3:

Base ^ Exponent = Unit Value

3^1 = 3

3^2 = 9 (3x3)

3^3 = 7 (3x9)

3^4 = 1 (3x7)

3^5 = 3 (3x1) and so on.

Here, you will observe that the pattern for the unit's digits will repeat after 1, i.e. after counting a total digits of 4. Hence, the cyclicity of 3 =4.

 

For 4:

Base ^ Exponent = Unit Value

4^1 = 4

4^2 = 6 (4x4)

4^3 = 4 (4x6)

4^4 = 6 (4x4) and so on.

Here, you will observe that the pattern for the unit's digits will repeat after 6, i.e. after counting a total digits of 2. Hence, the cyclicity of 4 =2.

 

For 5:

Base ^ Exponent = Unit Value

5^1 = 5

5^2 = 5 (5x5)

In the case of 5, each time 5 will be multiplied by 5, unit's place will always have 5. The cyclic value is 1. 5^n will always have 5 in the unit's place.

 

Coming back to the ques, taking the first term, we have: 4^27

We are the find a number less than 27 which must be a fully divisible by 4's cyclic value, i.e. 2. The nearest value is 26.

4^27 = 4^26.4

Here we have to be very careful with the digits. In the cyclic table of 4, the cycle repeats itself after 6. Thus, 6 will be unit's digit for 4^26 as 26%2=0. Substituting 6 in place of 4^26, we have,

4^27 = 4^26.4 = 6.4 = 24

Again, the unit's place for 24 is 4. Set aside this value.

 

Similarly, 5^27 = 5^25.5^2 = 5.5^2 = 125

Again, the unit's place for 25 is 5. Set aside this value.

 

Similarly, 3^27 = 3^24.3^3

In the cyclic table of 3, the cycle repeats itself after 1. Thus, 1 will be unit's digit for 3^24 as 24%4=0. Substituting 1 in place of 3^24, we have,

=> 1.3^3 = 27

Again, the unit's digit for 27 is 7. Set aside this value too.

 

Combining the 3 values, we have, 4,5 and 7.

4x5x7 will give 140 which will have the unit's place = 0.

 

I hope you will understand this. This is the fastest way to solving such questions.

 

All the best !!!!!!

 

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Can someone please tell me how to find of the unit digit of the following.

 

Find out the unit digit of 4^27 * 5^27 * 3^27?

 

Not only the above one,how to find unit digit of any value to any power?

 

Please some one explain me this concept.

 

Thx.