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The units digit of 35^87 + 93^46 is:

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The units digit of 35^87 + 93^46 is: [#permalink]

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The units digit of (35)^(87) + (93)^(46) is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0


For strategies on tough units-digit questions, as well as a complete explanation to this problem, see:
http://magoosh.com/gmat/2013/gmat-quant ... questions/

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Re: The units digit of 35^87 + 93^46 is: [#permalink]

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Step 1: (35)^(87) can be broken up into (5)^87 x (7)^87. Each power of 7 ends in a units digit of either a 7,9,3 or 1. Each power of 5 ends in a 5. When you multiply 5 by any odd number you will end up with a units digit of 5.

Step 2: (93)^(46) can be broken up into (3)^46 x (31)^46. Each power of 31 ends in a units digit of 1. Each power of 3 ends in a units digit of either 3,9,7 and 1. Since there is a pattern here where every 4th power of 3 ends in a units digit of 3, the 46th power of 3 would end in a units digit of 9. When you multiply 1 by 9 you end up with a units digit of 9.

Therefore, the units digit we are looking for is 5 + 9 = 14. Units digit will be 4. Answer B.
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Re: The units digit of 35^87 + 93^46 is: [#permalink]

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mikemcgarry wrote:
The units digit of (35)^(87) + (93)^(46) is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0


For strategies on tough units-digit questions, as well as a complete explanation to this problem, see:
http://magoosh.com/gmat/2013/gmat-quant ... questions/

Mike :-)

It is quite intuitive to go for a basic two step approach for this problem.

When dealing with 35^87, we can apply a simple concept here. 5 raised to any power > 0 must have 5 as its units digit.
When dealing with 93^46, we can apply the concept of cyclicity. Since the cyclicity of 3 is 4, so units digit of 93^46 is equivalent to teh units digit of 3^2 i.e. 9.
On adding these 2 digits i.e. 9 and 5, we get 14 of which the units digit is 4.
Will be curious to know how others deal with such questions.
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Re: The units digit of 35^87 + 93^46 is: [#permalink]

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mikemcgarry wrote:
The units digit of (35)^(87) + (93)^(46) is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0


For strategies on tough units-digit questions, as well as a complete explanation to this problem, see:
http://magoosh.com/gmat/2013/gmat-quant ... questions/

Mike :-)


Easier approach would be:

The units digit of (35)^(87) is the same as the units digit of 5^(87). 5 in ANY positive integer power has the units digit of 5.

The units digit of (93)^(46) is the same as the units digit of 3^(46)=9^23. 9 in odd power has the units digit of 9.

5 + 9 = 14.

Answer: B.
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Re: The units digit of 35^87 + 93^46 is: [#permalink]

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New post 30 Oct 2013, 01:23
Marcab wrote:
mikemcgarry wrote:
The units digit of (35)^(87) + (93)^(46) is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0


For strategies on tough units-digit questions, as well as a complete explanation to this problem, see:
http://magoosh.com/gmat/2013/gmat-quant ... questions/

Mike :-)

It is quite intuitive to go for a basic two step approach for this problem.

When dealing with 35^87, we can apply a simple concept here. 5 raised to any power > 0 must have 5 as its units digit.
When dealing with 93^46, we can apply the concept of cyclicity. Since the cyclicity of 3 is 4, so units digit of 93^46 is equivalent to teh units digit of 3^2 i.e. 9.
On adding these 2 digits i.e. 9 and 5, we get 14 of which the units digit is 4.
Will be curious to know how others deal with such questions.



Good question, I have the same question in my mind.

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Re: The units digit of 35^87 + 93^46 is: [#permalink]

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New post 28 Mar 2014, 10:36
Bunuel wrote:
mikemcgarry wrote:
The units digit of (35)^(87) + (93)^(46) is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0


For strategies on tough units-digit questions, as well as a complete explanation to this problem, see:
http://magoosh.com/gmat/2013/gmat-quant ... questions/

Mike :-)


Easier approach would be:

The units digit of (35)^(87) is the same as the units digit of 5^(87). 5 in ANY positive integer power has the units digit of 5.

The units digit of (93)^(46) is the same as the units digit of 3^(46)=9^23. 9 in odd power has the units digit of 9.

5 + 9 = 14.

Answer: B.



Thanks Bunuel. "9 in odd power has units digit of 9". Missed that simple logic... :-D :-D
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The units digit of 35^87+93^46 is: [#permalink]

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Re: The units digit of 35^87+93^46 is: [#permalink]

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New post 05 Mar 2015, 13:06
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IMO it should be 8 (D)

this is how I would have done in exam :
power repetitive of 5..
5^1 = 5
5^2 = 5
5^3 = 5
5^4 = 5

this means remainders of power ending with 1 gives unit digit 5
remainders of power ending with 2 gives unit digit 5
87/5 gives remainder of 2... hence unit digit left is 5

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
46/3 gives remainder 1
hence unit digit is 3
5+3=8
unit digit is 8

Kudos please if my solution is right and having appropriate method
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Bunuel wrote:
The units digit of 35^87 + 93^46 is:

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


Kudos for a correct solution.
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The units digit of 35^87+93^46 is: [#permalink]

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I think the answer is B. This is how I approached it..

35^87 + 93^46

Looking at 35^87 and its unit digit (5)- you know this number will end with a units digit 5 because that's how the cycle works with 5's (5^1=5, 5^2=25, 5^3-125 etc..)
Looking at 93^46 - you can establish a pattern with the 3's as I've demonstrated below...

3^1=3
3^2=9
3^3=27
3^4=81
3^5=243 (this is where the cycle starts to repeat)

So the cycle is in 4. 46/4 leaves a remainder of 2 so you know this unit digit will end in 9.

5 + 9= 14, thus units digit is 4 and answer is (B)
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Re: The units digit of 35^87+93^46 is: [#permalink]

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New post 05 Mar 2015, 14:23
Hey!
Just saw my error! Apologies and no excuses why it happened.. but it happened.
Will be more careful posting next time
Thanks
Celestial


soniasawhney wrote:
I think the answer is B. This is how I approached it..

35^87 + 93^46

Looking at 35^87 and its unit digit (5)- you know this number will end with a units digit 5 because that's how the cycle works with 5's (5^1=5, 5^2=25, 5^3-125 etc..)
Looking at 93^46 - you can establish a pattern with the 3's as I've demonstrated below...

3^1=3
3^2=9
3^3=27
3^4=81
3^5=243 (this is where the cycle starts to repeat)

So the cycle is in 4. 46/4 leaves a remainder of 2 so you know this unit digit will end in 9.

5 + 9= 14, thus units digit is 4 and answer is (B)

The units digit of (35)^(87) is the same as the units digit of 5^(87). 5 in ANY positive integer power has the units digit of 5.

The units digit of (93)^(46) is the same as the units digit of 3^(46)=9^23. 9 in odd power has the units digit of 9.

5 + 9 = 14.

Answer: B.
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Re: The units digit of 35^87+93^46 is: [#permalink]

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I will also go with B

The units digit of a number raised to the 5th power will always be 5.

For the units digit of 93^46 we only need the units digit of 3^46.
This is the cyclicity of 3:
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = ..1 After that we will get a units digit of 3, so wee stop. The cyclicity of 3 is 4. So, 46/4 leaves a remainder of 2, so we are looking at the second power, which gives a units digit of 9.

Adding the two units digits we get: 5+9 = 14, so the units digit should be 4. ANS B
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Re: The units digit of 35^87+93^46 is: [#permalink]

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Bunuel wrote:
The units digit of 35^87 + 93^46 is:

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


Kudos for a correct solution.


all numbers follow a repetitive cycle after every 4th power, although few repeat even before that for example 6,5,0,1...
5 to any power will give last digit as 5...
3 follows a repetitive cycle after a multiple of 4...
3^1=3..
3^2=9..
3^3=7..
3^4=1..
3^5=3.. and so on..
now 46=4*11+2.. power of multiple of 4+2=9..
so last digit =5+9=14.. last digit 4..
ans B
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Re: The units digit of 35^87+93^46 is: [#permalink]

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Bunuel wrote:
The units digit of 35^87 + 93^46 is:

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


Kudos for a correct solution.


+1 for B. Last digit of 35^87=5 and last digit of 93^46=9
9+5=14.
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Re: The units digit of 35^87+93^46 is: [#permalink]

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I think the ans should be (B).
Anything with unit digit 5 and any power will give me unit digit of 5.
For no with unit digit 3, we will have to check the cyclicity of 3. It's 4. So we divide 46 by 4 and the remainder is 2.
Hence, 3^2 is 9.
Thus, the net unit digit is 5+9=14 i.e. 4. (B).
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Re: The units digit of 35^87+93^46 is: [#permalink]

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New post 08 Mar 2015, 15:02
Bunuel wrote:
The units digit of 35^87 + 93^46 is:

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


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

We have to figure out each piece separately, and then add them. The first piece is remarkably easy — any power of anything ending in 5 always has a units digit of 5. So the first term has a units digit of 5. Done.

The second term takes a little more work. We can ignore the tens digit, and just treat this base as 3. Here is the units digit patter for the powers of 3.
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.

3^44 has a units digit of 1
3^45 has a units digit of 3
3^46 has a units digit of 9

Therefore, the second term has a units digit of 9.

Of course 5 + 9 = 14, so something with a units digit of 5 plus something with a units digit of 9 will have a units digit of 4.

Answer = B
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Re: The units digit of 35^87 + 93^46 is: [#permalink]

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New post 16 Mar 2017, 13:21
Quote:
The units digit of (35)^(87) + (93)^(46) is:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0



Since we only need to determine the units digit of (35)^(87) + (93)^(46), we can rewrite the question as 5^87 + 3^46.

Let’s first determine the units digit of 5^87. Since 5 raised to any positive integer power will always have a units digit of 5, 5^87 has a units digit of 5.

Next we can determine the units digit of 3^46. We can evaluate the pattern of the units digits of 3^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 3. When writing out the pattern, notice that we are ONLY concerned with the units digit of 3 raised to a power.

3^1= 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

As we can see from the above, the pattern of the units digit of any power of 3 repeats every 4 exponents. The pattern is 3–9–7–1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. Thus:

3^44 has a units digit of 1, 3^45 has a units digit of 3, and thus 3^46 has a units digit of 9.

Since the sum of the units digits of 5^87 and 3^46 is 5 + 9 = 14, (35)^(87) + (93)^(46) has a units digit of 4.

Answer: B
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Re: The units digit of 35^87 + 93^46 is: [#permalink]

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New post 16 Mar 2017, 13:37
I used the following logic:

the units digit of any number that end in 5 is 5. And then to find the units digit of 3 to the 46 power I used the following:
3^1=3
3^2=9
3^3=27
3^4= 81
So the cycles for 3 are: 3-9-7-1 for every exponent of base 3 divisible by 4 the units digit will be 1 because 3^4 has a unit digit of 1. Let´s start from 3^44 the units digit is 1, 3^45 units digit will be 3 and 3^46 units digit is 9.
Then the sum of the units digit is 5+9=14 which is the units digit of 4.
Hope this is a clear explanation and you don´t need to memorize the powers of any number.
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Re: The units digit of 35^87 + 93^46 is: [#permalink]

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New post 17 Mar 2017, 02:52
(35)^(87) + (93)^(46)
for 35 ^87 unit digit will be 5

3^4 ends with 1. therefore 3^44 = (3^4)^11 will end with 1. Multiplying it with left 3^2 = 9 we get the unit digit for 93^46 as 9

adding 5+9 we get unit digit as 4
Re: The units digit of 35^87 + 93^46 is:   [#permalink] 17 Mar 2017, 02:52
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