Find the tens digit of 3^2007.?

Answer 1

we need to evaluate 3^ 2007 mod 100

we know 10 -1 = 9

so (10-1)^1003 = 9^1003

if we evaluate e last 2 terms of expression then we get using binomial theorem

1003 *10 * 1 -1 or 10030-1 or 10029

as other terms have 10^k where k >=2 so divisible by 100

so 9^1003 mod 100 = 29
or 3^2006 mod 100 = 29
or 3^2007 mod 100 = 29 *3 mod 100 = 87 mod 100

so last 2 digits are 87 and hence it is obvious to say 10's digit is 8

Answer 2

The "tens digit" will be ' 8.'

Here's how to find that out:

The first twenty one powers of 3 end in the following pairs of digits:

03, 09, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 07, 21, 63, 89, 67, 01, 03.

With the final ' 03 ,' above, we're back where we started. That means the last pair of digits goes through a cycle of twenty (20) pairs that then repeat. So, after 2000 terms, 100 cycles have been completed, and the "tens digit" in 3^2007 must be the same as the "tens digit" in the 7th term (87) in the original set of repeating pairs.

That digit is therefore ' 8.'

Live long and prosper.

P.S. FOUL! 'Nimish A' has removed his original argument (which was completely false) and substituted what is essentially mine. In the circles I move in, that is Plagiarism, pure and simple. That's why there are so many "thumbs down!"

Answer 3

Answer. 8

To solve any such problem you need to study the sequence. In this case, you need to study the sequence of the last 2 digits of 3^n and see when the cycle repeats itself.

So lets see the units and tens digits for 3^n, where n = 1, 2, 3...
3^0 (01)
3^1 (03)
3^2 (09)
3^3 (27)
3^4 (81)
3^5 2(43)
3^6 7(29)
3^7 21(87)
3^8 65(61)
3^9 196(83)
3^10 590(49)
3^11 1771(47)
3^12 5314(41)
3^13 15943(23)
3^14 47829(69)
3^15 143489(07)
3^16 430467(21)
3^17 1291401(63)
3^18 3874204(89)
3^19 11622614(67)
===
3^20 34867844(01)
3^21 104603532(03)
3^22 313810596(09)
3^23 941431788(27)

So, we see that the cycle is repeating itself after every 20 numbers i.e 3^1 and 3^21 and 3^41 will have the same last two digits.

So 3^2007 will have the smae last two digits as 3^7 i.e. 87. Hence, the tens digit for 3^2007 will be 8.

Answer 4

unit digit is 7..workin on the tens digit..i think we have to use

2007/3 ! + 2007/9! + 2007/ 27! ..so on( i dont kno why,but we did this method to find the tens digit of sum no. ) if u do by this method, i think u''ll get 4 as an answer

Answer 5

Nimish A has the right answer, but the wrong logic.
Try out 2^3006 according to his logic. It should be 8 according to him, but it is actually 2.

The following C program achieves this. It ouputs the last two digits.

try it out and verify the answer urself. It outputs the last two digits. the tens digit is 8.

#include

void main()
{
for( int i = 1, num = 1; i <= 2007; i++ )
num = ( ( num * 3 ) % 100 );
cout << num;
}

Modify the program to print the last three and last four digits.
You will find the hundreds digits repeating after every 100 powers and every 500 respectively. Thus the pattern follows a geometic progression!
ones digit - exponent repeats after 4 powers,
tens -after 20
hundreds -after 100
1000s - after 500
...

Answer 6

3.8225944275174390703263644921131 x 10^957

this means
the first ten digits would be

3822594427