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saurya_s
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Dookie
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saurya_s
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Is the answer 8?
A weird method to explain. If OA matches, I will surely explain.
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twixt
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I spent 5 minutes on this one and and did not find any trick.
Btw Dookie could you please explain how you obtained 4 as a units digit ?
Maybe there is something wrong with 642 which contains the prime 107...
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Dookie
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I was looking for the units digit.
Noticed that the units digit changes in a sequence of 4:
2,4,8,6,2,4,8,6,2,4,8,6....
642/4 = 160 + Remainder 2.
So the units digit should be "4".
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twixt
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I ve got the trick for the units digit... just a serie 2 4 8 6 2 4 8 6...
I think there is the same, more complex, approach to conduct for the tens digit... I com back with the answer ;-)
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twixt
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Finally I obtain 4 as a units digit and 8 for the tens digit.

For units, 2 <=> 2^5 642=2+5*128
For tens, 2 <=> 2^20 642=2+20*32
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myfrankenstein
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I have a longish approach to solving this problem that involves successive breakdowns.
642 = 2*3*107
2 ^ 642 = (60+4) ^ 107 = (60 + 4) ^ 107.
(Now the critical part) Considering a binomial expansion in all binomial terms the only term that will affect the ten's digit or add to it shall be 4^107. The rest have 0 at ten's digit.
4^107 = (4^2)*(4^105) = (16) * [( 1020 + 4 ) ^ 21] [ i am comfortable with big multiples ..they could be smaller as well]
Need to consider only 16 * (4 ^21) = 16* 4 * (4^20)
= (60+4) * (1020 + 4 )^4
Again consider only 4 * (4^4 )
4* (250 + 6) => last 2 digits r like 24

We can say for sure that 4 is the unit's digit and 2 the ten's digit. We can't be say anything about the hundred's digit as we eliminated only considering the ten's digit.
Instead of 60 + 4 yu can use 1020+ 4 or any other combination as well. This involves using binomial expansion to eliminate non considerants.
Unit's digit : 4
Ten's digit : 2
Time taken 130 seconds.
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venksune
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Just a quick tip to calculate the unit digits of a crazy exponential number as in the case here.
Dookie has explained well the pattern method. There is a shorter trickier method.

Case1: When 2,3, 7 or 8 are the digits in the unit place of the base number, then divide the power of that number by 4 and replace the power of the remainder 1,2,3, or 0. In the problem under discussion, we will divide 642 by 4 (as 2 is the unit digit of the base number). We have remainder as 2. So, the unit digit of 2^642 is 2^2 = 4.

Case 2: When 0,1, 5 or 6 are the digits in the unit place of the base number, then for all values of the power, the number has the same digit in the unit place. Say. 5^322 will be 5.

Case 3: When 4 is in the unit place of the base number, then the unit digit will contain 4 or 6 according to odd or even power respectively.

Case 4: When 9 is in the unit palce of the base number then the unit digit will contain 9 or 1 in the unit place according to the odd power or even power respectivey. (ex. 9^1=9, 9^2 =81, 9^3=729...)
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The repetition is based on recognizing the pattern and works only for the unit's digit.
You r better off understanding the pattern and analysis rather than the rule.
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venksune
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Franky,
The rule is a derivative of the pattern approach and analysis. Yes, its only for the units digits.
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saurya_s
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Franky, your approach is right. What is the basis of selecting 4^107 term. Terms affecting tens digit should have only one zero (terms which won't afect tens digit are those with at least two zeros at end) I guess on that basis you should also consider the term where 60 occurs one. Can you please explain it what I am missing.
S
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2^642=18249762470488780874564686422801165299572914028994239722316770071597100668834709546023651245269485599114569238294377629242754818885501751993010645278888856753007978697441059800331496768986415104

that elimanates the idea that the tenth digit is 2.


Note: I didn't calculate it myself. (found the answer in 5 sec using Mathematica)
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saurya_s
Franky, your approach is right. What is the basis of selecting 4^107 term. Terms affecting tens digit should have only one zero (terms which won't afect tens digit are those with at least two zeros at end) I guess on that basis you should also consider the term where 60 occurs one. Can you please explain it what I am missing.
S

You r right about the hundred's digit consideration. This would make it too tedious then.
I made a mistake on ten's digit .
We can get the answer using the approach above...only that it will take too long :(
Either there is another shortcut or guessing is the best way forward.



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