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# m09 32

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CIO
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03 Sep 2010, 06:10
The OA is NOT B, but D. See Bunuel's explanation closely:
m09-74730.html#p685866
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Manager
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05 Sep 2010, 09:23
obvious D. No need to debate on answer. A quick press of the calculator reviewed 7625597484987. Last digit is 7.
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07 Sep 2010, 22:24
Its D.

3 to the power 3 to the power 3 translates into 3 to the power 27. Cyclicity is 3, 9, 7, 1 and to degree 4. So 3 to the power 27 goes for 3rd value in the cycle -7.
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08 Sep 2011, 05:48
alexgmd wrote:
easy "B"

ans is D not B.

look for above explanation.
if still you have any doubt, i can try to clarify them.
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17 Sep 2011, 09:00
Nice question. Thanks.
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07 Sep 2012, 05:54
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gijoedude wrote:

Step 1: 3^3= 27.

Step 2: Therefore 3^(3^3)=3^27. = 3*3*3*.......3 (27 3s).

Step 3: We know 3*3*3 = 27

Step 4: Hence step 2 can be simplified as 27*27*...27 (9 27s)

Step 5: We know 27*27*27 ends in 3 (multiply it out. No need to do the entire multiplication. Only the last digits matter).

Step 6: Hence we end up abc3*abc3*abc3 = a number ending with 7 (again only the last digits matter)

Step 7: Voila!

This is way too complicated !
Every digit has a pattern that repeats. See the chart given below. Some repeat in multiples of 4, others repeat in multiples of 1 and 2.
Key takeaway is that EVERY POWER REPEATS !

Powers of 2 - 2,4,8,6,,2,4,8,6
Powers of 3- 3,9,7,1,3,9,7,1
Powers of 4 - 4,6,4,6,4,6
Powers of 5 - 5,5,5,5,5,5,5,
Powers of 6 - 6,6,6,6,6,6,6,
Powers of 7 - 7,9,3,1,7,..
Powers of 8 - 8,4,2,6,8
Powers of 9 - 9,1,9,1

They key is to find the pattern for reptition. A units digit for a power of 3 repeats every 4 terms.
The units digit for 3 ^ 27 ==> Hence the term that corresponds to Remainder of (27 divided by 4) = 3rd term = 7 is the answer

To stretch this a lil bit more
2 ^ 100 = Remainder of (100 divided by 4) = Remainder = 0 , ie 4th term

A word of caution - Focus on the word remainder.
The division only tells us the POSITION of the term in the sequence and NOT the repeating term itself.
2^50 =-> REMAINDER OF (50 divided by 4) = 2ND TERM in the sequence, hence 4
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10 Sep 2012, 01:58
Exponentiation is not associative. Addition and multiplication are. For example, (2 + 3) + 4 = 2 + (3 + 4) = 9 and (2·3)·4 = 2·(3·4) = 24, but 23 to the 4 is 84 or 4,096, whereas 2 to the 34 is 281 or 2,417,851,639,229,258,349,412,352. Without parentheses to modify the order of calculation, by convention the order is top-down, not bottom-up:

$$b^{p^q} = b^{(p^q)} != (b^p)^q = b^{(p \c*q)} = b^{p \c*q}$$
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07 Sep 2013, 05:22
this works for units digit only.
for any integer a
a ^ (4n+x) = a^x...

3^27 = 3^(4*6+3) = 3^3 =27...

so units digit is 7....
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25 Aug 2014, 01:43
The resultant of the power of the number 3 repeats the unit place's digit after every 4th power, i.e.
3^1=3 --> 3
3^2=9 --> 9
3^3=27 --> 7
3^4=81 --> 1
3^5=243 --> 3
3^6=729 --> 9
3^7=2187 --> 7
3^8=6561 --> 1
... and so on.

Given this pattern, the 25th - 28th power of 3 will also have the unit's place in the same order,i.e. 3,9,7,1 resulting in the 27th power ending with the digit 7. Moreover, this way you could find out for any power of 3.

Hope this clarifies some doubts.
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26 Aug 2014, 04:37
In order to find the last digit of 3^{3^3} , we need to the no 3's cycle

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243

Now you see the last digit when 3^5 is again 3. Therefore the distinct last digits that 3 raised to any integer power will take are 3,9,7,1 respectively in the same order

now back to the qn: It can be expressed as

3^27

now divide the 27/4(cyclicity) and find out the remainder. If remainder is 0 then the last unit digit for a no raised to certain power will be the last digit in its cycle.

In this case, the remainder is 3. So the last digit shall be

3, 9 ,7 ,1

Hope this helps.
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31 Aug 2014, 11:21
3 raised to power 4k-1 (k>=1) would always leave a remainder with 10 as 7 or give a units digit as 7...

3^3(4*1 -1) = 27
3^7(4*2 -1) = 2187

and so on... thus 3^(3^3) = 3^27 = 3^(28-1) which would give 7 as the last digit
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Re: m09 32   [#permalink] 31 Aug 2014, 11:21

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# m09 32

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