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If $x^y$ denotes $x$ raised to the power of $y$ find last two digits of

$(1507^{3381})+(1437^{3757})$

MyApproach

($1507^{20 .169 +1})+(1437^{20 . 187+17}$)

=>$07 + (37)^{17}$

=>($37^{17}=_7$ will be too long to calculate(especially its tens digit.)

Is there any shorter way I can calculate the Ans?

Jyrki Lahtonen
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justin takro
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  • $7^4=1$ (mod 100). This will help you with the last two digits of the first term. You cal always use $37^{40}=1$ mod 100 if you do not find something smaller – Shailesh Nov 16 '15 at 10:11
  • $73$ is $37^{-1} \mod 100$, so multiply by $73^3 \cdot 37^3$, and you get that $37^{17} \mod 100 = 73^3 \mod 100$ – D. A. Nov 16 '15 at 10:30
  • How do you know you have to use the exponents modulo $20$? $\varphi(100)=40$. – Bernard Nov 16 '15 at 11:10

2 Answers2

1

To compute $37^{17}$ modulo 100, use exponentiation by squaring:

$37^2 =1369 \equiv 69 \pmod{100}$

$37^4 \equiv 69^2 = 4761 \equiv 61 \pmod{100}$

$37^8 \equiv 61^2 = 3721 \equiv 21 \pmod{100}$

$37^{16} \equiv 21^2 = 441 \equiv 41 \pmod{100}$

$37^{17} = 37^{16}\cdot37 \equiv 41\cdot 37 = 1517 \equiv 17 \pmod{100}$

With larger numbers it would be quicker to compute modulo each prime power in the modulus separately (that is, $37^{17}\equiv 1 \pmod 4$ and $37^{17}\equiv 17\pmod{25}$) and combine them using the Chinese Remainder Theorem at the end, but that is hardly worth it in this particular case.

hmakholm left over Monica
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0

Let's take care of the first term:

$1507\equiv 7\mod 100$. To find the order of $7$ modulo $100$, use the Chinese remainder theorem:

$7$ has order $2\mod4$, and order $4\bmod25$ since $7^2\equiv -1\mod25$, hence it has order $4$ mod $100$ and $$1507^{3381}\equiv 7^{3381\bmod4}\equiv 7\mod 100.$$

Now let's see the second term:

Similarly, $37\equiv 1\bmod4$ and $37\equiv 12\mod 25$. With the Fast exponentiation algorithm, one checks $12^{10}\equiv24 \mod25$, hence it ($37$ as well as $12$) has order $20$. Thus $$1437^{3757}\equiv 37^{3757\bmod 20}\equiv 37^{-3} \mod 100.$$ Now the Extended Euclidean algorithm yields $\;37^{-1}\equiv -27\mod 100$, hence \begin{align*} 37^{-2}&\equiv 27^2=729\equiv 29,\\ 37^{-3}&\equiv -27\cdot 29\equiv -27^2-2\cdot 27\equiv-29-54\equiv -83\equiv 17\mod100, \end{align*} and finally $\enspace1507^{3381}+1437^{3757}\equiv \color{red}{24\mod100}.$

Bernard
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