Dtjytyk

Can I know what are the last two digits of $35^{73} times 53^{25}$.

Please explain it in detail to me. Thanks!

The last two digits of a number $a$ are is the non-negative remainder of that number over $100$ as $amod 100$, for example for $a=1256$ the last two digits are $56=1256mod 100$. Therefore we need to find $35^{73}mod 100$ and $53^{25}mod 100$. Note that $$5^{2}equiv 25mod 100\5^{3}equiv 25mod 100\5^{4}equiv 25mod 100\.\.\.\5^{73}equiv 25mod 100$$also $$7^2equiv -1mod 25,4$$therefore $$7^{73}equiv 7cdot 7^{72}equiv 7mod 25,4to\7^{73}=25k+7=4k'-1to 7^{73}=100k''-1to \7^{73}equiv -1mod 100$$finally $$35^{73}equiv 7^{73}times 5^{73}equiv -25mod 100$$also $$53^{25}=(50+3)^{25}=3^{25}+25cdot 3^{24}cdot 50+50^2cdot(text{the rest of the terms})\equiv 3^{25}+25cdot 3^{24}cdot 50mod 100$$with a similar manner, we can conclude that $$3^{24}equiv 81mod 100$$ therefore $$53^{25}equiv 3^{25}+25cdot 3^{24}cdot 50equiv 43+25cdot 50cdot 81equiv 43+50cdot 25equiv 93equiv -7mod 100$$integrating the parts together, we attain to the answer:$$35^{73}times 53^{25}equiv75mod100$$therefore the last two digits are $75$.

$ 35^{large 73} 53^{large 25}bmod 100, =, 25left[dfrac{35^{large 73} 53^{large 25}}{25}bmod 4right] = 25overbrace{left[dfrac{(-1)^{large 73} 1^{large 25}}{1}bmod 4right]}^{ large equiv -1 equiv color{#c00}{ 3} pmod{!4}} = 25[color{#c00}3] $

We used $, abbmod ac, =, a,(bbmod c), =, $ mod Distributive Law

A slightly shorter variant:

We need to determine the congruence class of $;35^{33}cdot 53^{25}=5^{33}cdot5^{33}cdot 53^{25}$.

• First, it easy to check the $5^kequiv 25mod 10,$ for all $kge 2$. So $;5^{33}equiv 25mod 100$.


• As $7$ is coprime to $100$ and $varphi(100)=40$, lil' Fermat asserts that $7^{40}equiv 1mod 100$, so the order of $7$ modulo $100$ is a divisor of $40$. Indeed
  $$7^2=49,quadtext{ so } 7^4=(50-1)^2=2500-100+1equiv 1mod 100.$$
  Thus $7$ has order $4$, and $;7^{33}equiv 7^{33bmod 4}equiv 7mod 100$.


• To compute $P=53^{25}bmod 100$, we use the fast exponentiation algorithm:
  begin{array}{r|lc|l}
  hline
  n&&&P\hline
  25&53 &53&53 \
  12& 53^2& 9&53 \
  6&53^4&81&53 \
  3&53^8&61&53cdot61equiv 33 \
  1&53^{16}&21&33cdot 21equiv 93\
  hline
  end{array}
  We'll take $53^{25}equiv -7mod 100$. We obtain
  $$53^{25}equiv 25cdot 7cdot-7=25cdot(-49)equiv 25cdot 51=1275equiv color{red}{75}mod 100. $$
  

$35^{73}*53^{25} = 5^{73}*7^{73}*53^{25}=25*(5^{71}*7^{73}*53^{25})=25 times$ an odd number.

So the last two digits are either $25$ or $75$.

$5 = 4 + 1$, $7=2*4-1$ and $53 = 4*13+1$.

So $5^{71} = (4 + 1)^{71} = $some multiple of $4$ $+1$.

$7^{73} = (2*4-1)^{73}= $some multiple of $4$ $+(-1)^{73}=$ some multiple of $4$ $-1$.

$53^{25} = (4*13+1)^{25} = $ some multiple of $4$ $+1$.

So $5^{71}*7^{73}*53^{25} =$ (some multiple of $4$ $+1$)(some mulitple of $4 -1$)(some multiple of $4 +1$) $=$ some multiple of $4-1$.

So $35^{73}53^{25} = 25*($some multiple of $4 - 1) = ($some multiple of $100 - 25)$. So the last two didigits are $75$.

=== old answer ===

In detail?

In figuring out the last two digits of a bunch of arithmetic we never have to worry about anything along the way except the last two digits because all the later digits represent multiples of $100$ and won't affect the last two digits at all.

$35^{73} = (30 + 5)^{73} = sumlimits_{k=0}^{73} {73 choose k}30^k*5^{73-k}$. That's the binomial theorem.

So $30^k$ will be a multiple of $100$ if $k ge 2$ and we only have to worry about the last two digits we don't have to worry about $kge 2$.

$35^{73} = (5 + 30)^{73} = 5^{73} + 73*5^{72}* 30 + ....$ a bunch of multiples of $100$ that we don't have to worry about.

$5^2 = 25; 5^3 = 125$ but we only care about the last to digits. If the last two digits of $5^k$ are $25$ then the last to digits of $5^{k+1} = 5^k*5 = ($ some multiple of $100$ that we don't have to worry about$ + 25)*5 = ($some multiple of $500$ that we don't have to worry about$) + 125$ and the last two digits are $25$.

so the last two digits of $5^{73}$ are $25$ and then last two digits of $5^{72}$ are $25$. So the last two digits of $73*5^{72}*30$ are the same as the last to digits of $73*25*30 = 73*15*50=$some odd number $*50$. And odd number is $2k+1$ so $(2k+1)*50 = 100k + 50$ so the last two digits of $73*25*30$ is $50$.

so... the last two digits of $35^{73}$ are the same as the last two $5^{73} + 73*5^{72}*30$ which are the same as the last two digits of $25 + 50$ which are $75$.

And we do the same thing for $53^{25}$.

$53^{25} = (3+ 50)^{25} = 3^{25} + 25*3^{25}*50 + ...$ a whole bunch of multiples of $100$ that we don't have to worry about.

$3^2 = 9; 3^3= 27; 3^4 = 81; 3^5 = 243$ .... okay, we need to be a little more clever.

$3^2 = 9 = 10 - 1$. So $3^{2k} = (10-1)^k = (-1)^k + k*(-1)^{k-1}*10 + $ ... a bunch of multiples of $100$ we don't have to worry about.

So $3^{24} = (10-1)^{12} = (-1)^{12} + 12*(-1)^{11}*10 +$ ... stuff and will have the same last two digits as $1 -120 + $ ... multiples of $100$. Which will have the same last two digits as $-119 + $... multiples of $100$ which is the same last two digitas as $w00 -119 =81$.

So $3^{25}$ will have the same last to digits as $81*3 = 243$ so the last two digits of $3^{25} $ are 43$.

The last two digits of $25*3^{25}*50$ are the same as (some odd number)$*50$ so are $50$.

And the last two digits of $53^{25}$ are the same as the last two digits of $43+50=93$.

...

So that last two digits of $35^{73}*53^{25}$ are the same as the last two digits of $75*93 = (50 + 25)(92 + 1) = ($some even number$)times 50 + ($some multiple of $4)times 25 + 50 + 25=$ some multiples of $100$ that we don't care about $+75$. So the last two digits are $75$.

$35^{75}cdot53^{25}$ is an odd number divisible by $25$, so the last two digits are either $25$ or $75$. Taken mod $4$, it is congruent to $(-1)cdot1=-1$, so the answer must be $75$.