Tens digit of 36^10

36 = 6²

So 36¹� = 6²�.

Now look for a pattern to the tens digits of powers of 6.

6¹ = 6 Tens Digit: 0
6² = 36 Tens Digit: 3
6³ = 216 Tens Digit 1
6� = --96 Tens Digit 9
6� = --76 Tens Digit 7
6� = ---56 Tens Digit 5
6� = ----36 Tens Digit 3

OK. Now we are back to 3, and the pattern will repeat from here.

3 1 7 9 5 3 1 7 9 5

There are five numbers in the pattern, and to get from 6� to 6²� we need fourteen powers of 6.

Getting fourteen powers requires two times then entire pattern, and then the first four again.

So starting with 6� we have the following tens digits.

3 1 7 9 5 3 1 7 9 5 3 1 7 9

The tens digit of 6²� is 9.

The correct answer is A.

Marty Murray wrote:36 = 6²

So 36¹� = 6²�

Now look for a pattern to the tens digits of powers of 6.

6¹ = 6 Tens Digit: 0
6² = 36 Tens Digit: 3
6³ = 216 Tens Digit 1
6� = --96 Tens Digit 9
6� = --76 Tens Digit 7
6� = ---56 Tens Digit 5
6� = ----36 Tens Digit 3

OK. Now we are back to 3, and the pattern will repeat from here.

3 1 7 9 5 3 1 7 9 5

There are five numbers in the pattern, and to get from 6� to 6²� we need fourteen powers of 6.

Getting fourteen powers requires two times then entire pattern, and then the first four again.

So starting with 6� we have the following tens digits.

3 1 7 9 5 3 1 7 9 5 3 1 7 9

The tens digit of 6²� is 9.

The correct answer is A.

Thanks for your reply.
How about this approach?

36¹� = 6²�

STEP 1 : Find periodicity of ten's digit - Clearly its 5

STEP 2 : 20/5 = 4

STEP 3 : 6²� will have same tens digit as 6^4

STEP 4 :6^4 = 1296

Ten's digit is 9 ; option A ?

No, that does not work.

Using 20/5 = 4, then what you would have, if the tens digit of 6¹ were the first member of the pattern, which is not the case, would be four times the entire pattern.

31795 31795 31795 31795

So the tens digit of 6²� would be 5.

However the pattern starts at 6².

So getting to 6²� takes 19 powers.

Using your method would indicate that the answer is based on 19/5, and that the tens digit of 6²� is the same as the tens digit of 6¹�/� or 6³⋅�.

You could play more with the logic of what you said to make more clear to yourself that that method would not work.

Marty Murray wrote:No, that does not work.

Using 20/5 = 4, then what you would have, if the tens digit of 6¹ were the first member of the pattern, which is not the case, would be four times the entire pattern.

31795 31795 31795 31795

So the tens digit of 6²� would be 5.

However the pattern starts at 6².

So getting to 6²� takes 19 powers.

Using your method would indicate that the answer is based on 19/5, and that the tens digit of 6²� is the same as the tens digit of 6¹�/� or 6³⋅�.

You could play more with the logic of what you said to make more clear to yourself that that method would not work.

So does it mean that this method would only work if the cyclicity starts from first? (in this case 6^1)

Actually, I did try this with another number and it worked.

Ten's digit of 6^17

17/Mod 5 = 2

6^2 = 36 ...Hence this confusion

Regards,
MoM

Manonamission wrote:

Marty Murray wrote:No, that does not work.

Using 20/5 = 4, then what you would have, if the tens digit of 6¹ were the first member of the pattern, which is not the case, would be four times the entire pattern.

31795 31795 31795 31795

So the tens digit of 6²� would be 5.

However the pattern starts at 6².

So getting to 6²� takes 19 powers.

Using your method would indicate that the answer is based on 19/5, and that the tens digit of 6²� is the same as the tens digit of 6¹�/� or 6³⋅�.

You could play more with the logic of what you said to make more clear to yourself that that method would not work.

So does it mean that this method would only work if the cyclicity starts from first? (in this case 6^1)

Actually, I did try this with another number and it worked.

Ten's digit of 6^17

17/Mod 5 = 2

6^2 = 36 ...Hence this confusion

Regards,
MoM

The method does not work no matter where you start.

You are not dividing by 4 or 2. You are dividing by 5 in order to find how many times the pattern of 5 repeats and how many numbers come after the last full pattern of five.

To find which member of the pattern will be the digit in question, you have to look at the remainder, not the quotient.

For instance, 19/5 = 3 with remainder 4. So the digit in question will be the fourth member of the pattern.

Look at the logic of this, not at what seemed to work in a couple of cases.

Manonamission wrote: So does it mean that this method would only work if the cyclicity starts from first? (in this case 6^1)

Actually, I did try this with another number and it worked.

Ten's digit of 6^17

17/Mod 5 = 2

6^2 = 36 ...Hence this confusion

You're on the right track with modular arithmetic here. The question is essentially asking for 36¹� mod 100, so you want to find the cycle mod 100.

6¹ = 6 = 6 mod 100
6² = 36 = 36 mod 100
6³ = 216 = 16 mod 100
6� = 1296 = 96 mod 100
6� = 7776 = 76 mod 100
6� = 46656 = 56 mod 100

From here, the pattern will return to 36, meaning that the cycle is length 5, and 6¹ is an exception. That gives us 6� giving the following tens digits, with x any nonnegative integer:

If n = 5x + 2, the tens digit is 3;
If n = 5x + 3, the tens digit is 1;
If n = 5x + 4, the tens digit is 9;
If n = 5x + 5, the tens digit is 7;
If n = 5x + 1, with x necessarily greater than 0 in this case only, the tens digit is 5.

You want 6²�, so n = 20 = 5*3 + 5, meaning you must have a tens digit of 7.

EDIT: Added the bit in red.

Marty Murray wrote:
3 1 7 9 5 3 1 7 9 5

3 1 9 7 5! We're in Avon Lake, Ohio here, not Ty Ty, Georgia. (The latter does sound much more happening, though.)