Pattern finding of last few digit of a^b

Question

The question is:Find the last two digits of 9^2013.(No calculators allowed to solve this question.)

Options: (A)01 (B)29 (C)41 (D)81 (E)89

From this point,I will show my working,

9^2=81(Last Digit 1) 9^3=1x9=9 (Now only focusing on ONLY last digit) 9^4=9(Last Digit of the previous answer)x9(The next power)=1 (Last Digit)

Noticing a pattern that as the powers increases,the pattern goes 9,1,9,1,9,1,9,1,9 so the repeated pattern is 1 and 9.So I took

2013÷2=1006R(Remainder)1,where R1 is the number 9.So we know the digit is 9.

This eliminates the option (A),(C) and (D),leaving with (B) and (E)

I attempted to find the 2md last digit number,but to no avail as the number increases,no pattern can be formed.Any ideas how?(The solution doesn't have to use concept of patterns)

Edited working error.

Hint: $9^{10}\equiv 01\pmod{100}$. And $9^3\equiv 29\pmod{100}$. Conclusion left to the reader ... — String, Mar 23 '15 at 16:05
@String Oh sh*t,I haven't learned mod or modulus yet,out of my syllabus.I'll learn it myself now :P — ministic2001, Mar 23 '15 at 16:10
OK, all you need to know is that the two last digits of $9^n$ repeats each time $n$ is increased by $10$. — String, Mar 23 '15 at 16:14

score 3 · Accepted Answer · answered Mar 24 '15 at 01:59

3

You can do the same thing:

$$\underline{09}\to81\to29\to61\to49\to41\to69\to21\to89\to01\to\underline{09}\to\ldots$$

The cycle length is $10$, and $9^{2013}=9^{10\cdot201+3}$, so ...

Alternatively, you can notice that $9^{10}$ ends in $01$, and the product of numbers ending in $01$ ends in $01$, so $9^{2010}=\left(9^{10}\right)^{201}$ ends in $01$.

answered Mar 24 '15 at 01:59

Brian M. Scott

616,228

Mysterious thing is,I can take 9^2010 and the last 2 digit would be 09.And now were left with 9^3.So therefore,9^2010x9^3=09×29=261,as the last 2 digit to be 61.But the thing is,there isn't an opitoon called "61" and second of all,odd powers last digit should end with the digit "9". – ministic2001 Mar 26 '15 at 07:29
@ministic2001: No, the last two digits of $9^{2010}$ are $01$; you’re off by $1$ in the cycle. – Brian M. Scott Mar 26 '15 at 07:31
Ahh right,its 01.So the answer to 9^2013 should be 9^10^201 x 9^3=01×29=29 as the last 2 digits rights? – ministic2001 Mar 26 '15 at 07:35
@ministic2001: That’s right. – Brian M. Scott Mar 26 '15 at 07:35
Thank you really much Brian! – ministic2001 Mar 26 '15 at 07:36
@ministic2001: You’re very welcome! – Brian M. Scott Mar 26 '15 at 07:45

Pattern finding of last few digit of a^b

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