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Unit's digit of the product

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What is the unit's digit [#permalink] New post 06 May 2013, 11:14
What is the unit's digit of \(21^3\)x\(21^2\)x\(34^7\)x\(46^8\)x\(77^8\)?
A. 4
B. 8
C. 6
D. 2
E. 3
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Unit's digit of the product [#permalink] New post 06 May 2013, 11:20
If the unit's digit in the product (47n x 729 x 345 x 343) is 5, what is the maximum no. of values that n may take?
A. 9
B. 3
C. 7
D. 5
E. 4
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Re: What is the unit's digit [#permalink] New post 06 May 2013, 11:24
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The question can be solved using patterns: we have to find the unit digit of each term and them multiply them.

21 raised by any number will have 1 as last digit. (\(21*21=xx1\) every time)

34 follows this pattern (I'll consider only the unit digit)
\(34^1=4\)
\(34^2=6\)
\(34^3=4\)
\(34^4=6\)
So \(34^7\) (odd) \(= 4\).

46 follows this pattern (every exp will end with a 6)
\(46^1=6\)
\(46^2=6\)
\(46^3=6\)
So \(46^8 = 6\)

77 follows this pattern
\(77^1=7\)
\(77^2=9\)
\(77^3=3\)
\(77^4=1\) and then repeats
So \(77^8=1\)

Finally 1*1*4*6*1=24
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If a,b,c,d are consecutive odd numbers, then [#permalink] New post 06 May 2013, 11:24
If a,b,c,d are consecutive odd numbers, then \((a^2 + b^2 + c^2 + d^2)\) is always divisible by:-
A. 5
B. 7
C. 3
D. 4
E. 6
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When x is divided by 6, remainder obtained is 3. Find the [#permalink] New post 06 May 2013, 11:27
When x is divided by 6, remainder obtained is 3. Find the remainder when \(x^4 + x^3 + x^2 + x + 1\) is divided by 6.

A. 3
B. 4
C. 1
D. 5
E. 2
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Last edited by GMATtracted on 06 May 2013, 11:53, edited 1 time in total.
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Re: Unit's digit of the product [#permalink] New post 06 May 2013, 11:30
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Consider only the last digits: \(9*5*3=135\). Now 135*n must finish with a 5 (and n can be \(0,1,...,9\))
135*0 will not end with a 5
135*1 will end with a 5 (5*1=5)
135*2 will not end with a 5 (5*2=10)

Only the ODD value of n mantain \(5\) as the last digit, so n can be any odd value : \(1,3,5,7,9\). Five possible values
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Re: Unit's digit of the product [#permalink] New post 06 May 2013, 11:55
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Thanks Zarrolou for the explanation. This is my explanation:-
One very interesting thing about problems concerning unit's digit is that we can eliminate every other digit no matter how large the number is.
Thus, the expression would eventually reduce to finding the unit's digit for the expression:- \(1^3\) x \(1^2\) x \(4^7\) x \(6^8\) x \(7^8\)
Now the end digit for various indices of 4 can be written as:-
\(4^1\)------4
\(4^2\)------6
\(4^3\)------4
\(4^4\)------6
We clearly see that 4 and 6 are repeating in nature. Such a nature is called cyclicity and in case of number 4 the cyclicity is '2'
In the expression the power of 4 is 7 which is of the form \((2n+1)\).
Hence the unit's digit of \(4^7\) is 4.
Similarly the cyclicity for nos. 6 and 7 can be written down as shown below:-
\(6^1\)-----6
\(6^2\)-----6
We observe that 6 to the power of any integer will eventually lead to a value having its unit digit 6.
Thus the unit's digit for \(6^8\) would be 6.
For number 7:-
\(7^1\)-----7
\(7^2\)-----9
\(7^3\)-----3
\(7^4\)-----1
\(7^5\)-----7
Clearly then the cyclicity of 7 is 4.
And since the power of 7 in the expression is 8(which is of the form 4n) , therefore the unit's digit in this case would be 1
Thus, clearly the unit's digit of the resultant expression would be (1 x 1 x 4 x 6 x 1)=24.
4 is the right answer. :)
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Re: When x is divided by 6, remainder obtained is 3. Find the [#permalink] New post 06 May 2013, 12:00
GMATtracted wrote:
When x is divided by 6, remainder obtained is 3. Find the remainder when \(x^4 + x^3 + x^2 + x + 1\) is divided by 6.

A. 3
B. 4
C. 1
D. 5
E. 2


When x is divided by 6 the remainder is 3. Take x=3 (remaider of 3/6 is 3)

\(3^4 + 3^3 + 3^2 + 3 + 1\).It requires a lil bit of old fashion math, but it never hurt anybody

\(81+27+9+3+1=121\)
Reminder of \(\frac{121}{6}\) is 1

or:
\(3(27+9+3+1)+1=121\)
The first part is 3*(even>2) => divisible by 6. The remainder is 1

PS: your explanation of the above question works just fine
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Re: If a,b,c,d are consecutive odd numbers, then [#permalink] New post 06 May 2013, 12:15
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GMATtracted wrote:
If a,b,c,d are consecutive odd numbers, then \((a^2 + b^2 + c^2 + d^2)\) is always divisible by:-
A. 5
B. 7
C. 3
D. 4
E. 6


Take "smart numbers":
a=-3 b=-1 c=1 d=3
20 is divisible by 5 and 4, two options remain

a=-1 b=1 c=3 d=5
36, is divisible by 3,4 and 6

So the only common divisor is 4
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Re: Unit's digit of the product [#permalink] New post 07 May 2013, 05:14
Really nice explanation Zarrolou. Your method is the MOST suitable one for exams like GMAT. Taking examples of nos.(like 3 in this case) always helps reaching an answer faster.
Here is a more general explanation:-
From the question we can say that x will be of the form (6k+3), where k is an integer.
Therefore, the expression would be:-
\((6k+3)^4 + (6k+3)^3 + (6k+3)^2 + (6k+3) + 1\)
What would the remainder be if the above expression is divided by 6?
If we expand each bracketed expression given above, every individual term of the expanded expression would contain '6k' except the following:-
\(3^4, 3^3, 3^2, 3 and 1\)
So, the question reduces to----->"What would the remainder be if \((3^4+3^3+3^2+3+1)\) is divided by 6?"
Now, if we observe carefully any power of 3(\(like 3^2, 3^3, 3^4.....or 3^{1000} as the case may be\)) when divided by 6 always leaves a remainder 3.
Therefore, \((3^4+3^3+3^2+3+1)\) can be written as:-(6m+3)+(6n+3)+(6p+3)+(6q+3)+1=6(m+n+p+q+2)+1.Hence 1 is the required remainder
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Re: Unit's digit of the product   [#permalink] 07 May 2013, 05:14
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