# Calculate $1\times 3\times 5\times \cdots \times 2013$ last three digits.

Calculate $1\times 3\times 5\times \cdots \times 2013$ last three digits.

If $N=1\times 3\times 5\times \cdots \times 2013$

we need to find $N\pmod{1000}$

Now $1000=8\cdot125$

Now, $1\cdot3\cdot5\cdot7\equiv1\pmod 8$ (Find one generalization here )

So, $(1\cdot3\cdot5\cdot7)\cdots (2001\cdot2003\cdot2005\cdot2007)\equiv1\cdot1\cdots\cdot1\cdot1\equiv1\pmod 8$

So, $N\equiv 1\cdot2009\cdot2011\cdot2013\pmod 8\equiv9\cdot11\cdot13\equiv1\cdot3\cdot5\equiv 7\pmod 8$

Clearly, $125|N\implies N\equiv0\pmod{125}$ as $125$ is a factor of $N$

Now applying the Chinese Remainder Theorem,

$$N\equiv 0\cdot b_1\cdot\frac{1000}{125}+7\cdot b_2\cdot\frac{1000}8\pmod{1000}$$ where $b_1\cdot\frac{1000}{125}\equiv1\pmod{125}\text{ and } b_2\cdot\frac{1000}8\equiv1\pmod 8$

We don't need to calculate $b_1$ as it's multiplier is $0$

$125b_2\equiv1\pmod8\iff 5b_2\equiv1\pmod8$

Trying multiplication of $5$ with numbers coprime to and $<8,$ we get $b_2\equiv5\pmod 8$

So, $$N\equiv7\cdot5\cdot125\pmod{1000}\equiv(8\cdot4+3)125\equiv3\cdot125\equiv375\pmod{1000}$$

You don't have to calculate all that. What are the possible values of the last three digits given that N is odd and divisible by 125? Now use lab bhattacharjee's result that N is congruent to 7 mod 8.

you need to calculate $1\times3\times5...\times55$,then you see the pattern,you only need to calculate last 3 digits.

for last 2 digits, the circle is from 15, the pattern is 25,25,75,75. for last 3 digits, the circle is from 25, the pattern is,625,875,375,625 for last 4 digits, the circle is from 25, the pattern is, 625,6875,9375,625,625,1875,9375,5625 for the last 5 digits, it also repeat at 16 numbers.

when you calculate last 3 digits, you must care if it is over 101, ie:$1\times 3\times5...\times101$,then for the 3rd digit, if it is odd number ,you have to add 500 to last 3 digits, it it is even, then it is same.

for example, 2013, the 3rd digit is 0, so you can direct calculate: 2013 same as 1013(8 times),1013 same as 1013-800=213,213 same as 213-160=53,53 is same as 53-24=27, which is 375.

if we ask 2113, then 3rd digit is 1,so for the final number ,you need to add 500. 2113 same as 113,113 same as = 113-88=25, the original number is 625, add 500 ,it is 1125, so the last number is 125.

for last 4 or 5 digits, since more number will effect the result, so you have to more complex calculation which may not so straight forward.

this method is not so "mathematical", but it can directly show the period of last some digits which may cause some interesting to the students.

And my question is :why the last 2 or 3 is repeat in such way, can anyone predict what is the circle numbers(16 numbers) for last 5 digits and from where?

• could you please share how you have identified $55?$ Commented Mar 2, 2013 at 4:39
• I think $55$ is the smallest positive integer$\equiv7\pmod 8$ when $N$ is divisible by $5^3.$ But how you are going to calculate $1\cdot3\cdot5\cdots53\cdot55?$ Also $2013\equiv5\pmod 8\not\equiv7$ Commented Mar 2, 2013 at 5:12
• your method is right one. the 55 is simply the pattern repeat 3 times. the pattern is 625,625,875,375 from 31 and then repeat. when I post my result, I haven't see your answer. Commented Mar 2, 2013 at 6:11
• yes, I calculate for last 3 digits in excel table, it is a small task. it is a "short cut".;-P.+1 for your answer. Commented Mar 2, 2013 at 6:21