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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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If you shut your door to all errors, truth will be shut out.

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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It is beyond a doubt that all our knowledge that begins with experience.

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.
_________________

If you shut your door to all errors, truth will be shut out.

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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If you shut your door to all errors, truth will be shut out.

gmatclubot

Re: Unit's digit of the product
[#permalink]
07 May 2013, 05:14

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