the tens digit of 16^298 ....

**ukorov** · December 23rd 2009, 04:13 PM

Please someone tell me how to work out the TENS digit of the answer of $16^{298}$ WITHOUT using calculator.

I believe that it is somehow related to the principles such as $(x^a)^b = x^{ab}$

the original question also included the units digit but it should always be 6.

Thanks in advance.

**pickslides** · December 23rd 2009, 04:36 PM

$16^{298} = (2^4)^{298} = (2)^{4\times 298} =2^{1192}$

Now have a read of this article.

How to Find the Last Digits of a Positive Power of Two - Exploring Binary

**aidan** · December 23rd 2009, 05:19 PM

Originally Posted by ukorov

Please someone tell me how to work out the TENS digit of the answer of $16^{298}$ WITHOUT using calculator.

I believe that it is somehow related to the principles such as $(x^a)^b = x^{ab}$

the original question also included the units digit but it should always be 6.

Thanks in advance.

A few calculations (you should be able to multiply these easily w/o the abacus)

016 = 16^1
x16
===
_56 = 16^2 (16 times 16 = 256, but we only need the tens & ones digits)
x16
===
_96 = 16^3
x16
===
_36 = 16^4
x16
===
_76 = 16^5
x16
===
_16 = 16^6 (same as 16^1, exponent difference is 5)

$16^6 \equiv 16^1 \mod 100$

298/5 = 59.9
59*5 = 295

1+5=6
1+295=296

$16^1 \equiv 16^6 \equiv 16^{296} \mod 100$

&(add 2 to the exponents)

$16^3 \equiv 16^8 \equiv 16^{298} \mod 100$

see prior calculation above for the tens & ones digits for $16^3\mod 100$ which is the same as $16^{298}\mod 100$

.

**ukorov** · December 23rd 2009, 05:46 PM

Originally Posted by pickslides

$16^{298} = (2^4)^{298} = (2)^{4\times 298} =2^{1192}$

Now have a read of this article.

How to Find the Last Digits of a Positive Power of Two - Exploring Binary

$2^{1192}$
$\equiv (2^{22})^{54} \cdot 2^4$ (mod100)
$\equiv 4^{54} \cdot 2^4$ (mod100)
$\equiv 2^{108} \cdot 2^4$ (mod100)
$\equiv 2^{112}$ (mod100)
$\equiv (2^{22})^5 \cdot 2^2$ (mod100)
$\equiv 4^5 \cdot 2^2$ (mod100)
$\equiv 2^{10} \cdot 2^2$ (mod100)
$\equiv 2^{12}$ (mod100)
$\equiv 96$ (mod100)

**Bacterius** · December 23rd 2009, 07:20 PM

$16^{298} \equiv 16^{98} \pmod{100}$

$16^{98} = (2^{98})^4 = 2^{98 \times 4} = 2^{392}$

$2^{392} \equiv 2^{92} \pmod{100}$

Note that $92 = 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 1$

Thus $2^{92} = 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^1$

Note that $2^7 = 128$ (easily done mentally)
Thus $2^7 \equiv 28 \pmod{100}$

Thus $2^{92} \equiv 28^{13} \times 2 \pmod{100}$

Note that $28^{13} = 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^1$
You easily find out that $28^2 \equiv 84 \pmod{100}$

Thus, $28^{13} \equiv 84^{6} \times 28 \pmod{100}$

We get that $2^{92} \equiv 84^{6} \times 28 \times 2 \equiv 84^6 \times 56 \pmod{100}$

Note that $84^6 = 84^2 \times 84^2 \times 84^2$ .
You easily find out that $84^2 \equiv 56 \pmod{100}$ .

Thus, $2^{92} \equiv 56^{3} \times 56 \equiv 56^4 \pmod{100}$ .

Note that $56^4 = 56^2 \times 56^2$ .
You easily find out that $56^2 \equiv 36 \pmod{100}$ .

Thus, $56^4 \equiv 36^2 \pmod{100}$ .
You easily find out that $36^2 \equiv 96 \pmod{100}$ .

Thus, $56^4 \equiv 2^{92} \equiv 2^{392} \equiv 16^{98} \equiv 16^{298} \equiv 96 \pmod{100}$ .

Conclusion : the tens digit of $16^{298}$ is $9$ .

I know my method is long, boring and repetitive, but eh ? it gets the job done

**BabyMilo** · December 23rd 2009, 10:11 PM

Originally Posted by Bacterius

$16^{298} \equiv 16^{98} \pmod{100}$

$16^{98} = (2^{98})^4 = 2^{98 \times 4} = 2^{392}$

$2^{392} \equiv 2^{92} \pmod{100}$

Note that $92 = 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 1$

Thus $2^{92} = 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^7 \times 2^1$

Note that $2^7 = 128$ (easily done mentally)
Thus $2^7 \equiv 28 \pmod{100}$

Thus $2^{92} \equiv 28^{13} \times 2 \pmod{100}$

Note that $28^{13} = 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^2 \times 28^1$
You easily find out that $28^2 \equiv 84 \pmod{100}$

Thus, $28^{13} \equiv 84^{6} \times 28 \pmod{100}$

We get that $2^{92} \equiv 84^{6} \times 28 \times 2 \equiv 84^6 \times 56 \pmod{100}$

Note that $84^6 = 84^2 \times 84^2 \times 84^2$ .
You easily find out that $84^2 \equiv 56 \pmod{100}$ .

Thus, $2^{92} \equiv 56^{3} \times 56 \equiv 56^4 \pmod{100}$ .

Note that $56^4 = 56^2 \times 56^2$ .
You easily find out that $56^2 \equiv 36 \pmod{100}$ .

Thus, $56^4 \equiv 36^2 \pmod{100}$ .
You easily find out that $36^2 \equiv 96 \pmod{100}$ .

Thus, $56^4 \equiv 2^{92} \equiv 2^{392} \equiv 16^{98} \equiv 16^{298} \equiv 96 \pmod{100}$ .

Conclusion : the tenths digit of $16^{298}$ is $9$ .

I know my method is long, boring and repetitive, but eh ? it gets the job done

according to a computer the tenth digit is 7.

http://www.wolframalpha.com/input/?i=10th+digit+of+16^298

**Bacterius** · December 23rd 2009, 10:40 PM

Originally Posted by BabyMilo

according to a computer the tenthS digit is 7.

http://www.wolframalpha.com/input/?i=10th+digit+of+16^298

I mean the before-last digit, of course ... not the 10th digit from the left

$16^{298} =$ 67259685376506838351825694125205607357734081036908 94500677045631406539332516306882577142397357830536 43998864024685999522991598722297525821288734761044 09052127576185436156214378455780845781465327759636 97038693955559119408307906472347614286627358580697 16218116239821639793538226085261772687392217674908 12009984323096665122511297763735177027009515287152 894672896

I was talking about the red digit, (I believe it is the question) ...

**Soroban** · December 24th 2009, 05:25 AM

Hello, ukorov!

My method is similar to Aiden's . . .

Find the tens digit of the answer of $16^{298}$ without using calculator.

Note the last two digits of consecutive powers of 16 . . .

. . $\begin{array}{c|c}16^n & \text{ends in} \\ \hline 16^1 & 16 \\ 16^2 & 56 \\ 16^3 & 96 \\ 16^4 & 36 \\ 16^5 & 76 \\ 16^6 & 16 \\ \vdots & \vdots \end{array}$

We find that the last two digits have a 5-digit cycle: . $16, 56, 96, 36, 76$

. . Hence: . $16^{5n+1} \equiv 16 \text{ (mod 100)}$

Since $298 \:=\:[5(59) + 1] + 2$

. . $16^{298} \;\;\equiv\;\; 16^{[5(59) + 1] + 2} \;\;\equiv\;\;16^{5(59) + 1}\cdot16^2 \;\;\equiv\;\; 16\cdot16^2 \;\;\equiv\;\;16^3 \;\;\equiv\;\;96\text{ (mod 100)}$

Therefore, the tens digit of $16^{298}$ is ${\color{red}9}.$

Edit: corrected typo . . . Thanks, Galactus!

**jiboom** · December 24th 2009, 09:25 AM

workin mod 100 noting 16^3=4^6=-4

16^298= 16(16)^297
=16(-4)^99
=16(-4)^3.(-4)^96
=-24(4)^16
=-24(4^4)((-4)^2)
=-24.-4=96

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