# How do you find the last three digits of 999999999999999999 ÷ 7 without using a calculator?

The last [math3 [/math digits of [math (10 ^ {18} – 1)/7 [/math.

I will show that finding all the figures is also possible.

Splitting [Math10 ^ {18} – 1 [/math:

[Mathx ^ 3 – 1 = (X-1) (x ^ 2 + x + 1) [/math

Fill in [mathx = 10 ^ 6 [/math:

[Math10 ^ {18} – 1 = (10 ^ 6 – 1) (10 ^ {12} + 10 ^ 6 + 1) = (10 ^ 6 – 1) \\cdot 1000001000001 [/math

Share by 7:

According to Fermat’s ‘ small theorem ‘, [Math7 [/math is a divisor of [math10 ^ 6 – 1 [/math].

There is:

[Math (10 ^ 6 – 1)/7 = 142857 [/math

Above you need to share your hand.

Perhaps nice to mention that these [Math6 [/math figures are also the first [Math6 [/math digits behind the comma of [math1/7 [/math.I know this from my head. But that does not apply to everyone.

[Math (10 ^ {18}-1)/7 = 142857142857142857 [/math

The answer to the question is therefore [math857 [/math

The most beautiful answer is from Jan van Delden, but a straight-grounded mathematician is lazy, so that doesn’t calculate 18 digits when 3 are asked.Be asked the last 3 digits of [math (10 ^ {18}-1)/7 [/math and indeed this division comes out exactly as a result of the small theorem of Fermat.So basically we need to have the last digits of [Math (x999)/7 [/math where [MATHX [/math somewhere between [Math1 [/math and [MATH6 [/math so that the division comes out exactly.We can of course try all 6 possibilities, but that’s not fun, we do modulo arrhythmetics. [math999 \\equiv 5 \\mod 7, 10 \ \ equiv 3 \\mod 7 [/math and [math1000 = 10 ^ 3 \ equiv 3 ^ 3 = 27 \ \ equiv-1 \\mod 7. [ /math So if we bite [math5000 [/math It should be correct and yes, [math5999/7 = 857, [/math exactly the [ math9-[/math complement of [math999/7 = 142 [/mathrest [Math5 [/math.

This way you can see that you have to do very little work with a lot of effort.:)

The last three digits are the remainder at division by 1000.So basically you search the result of [math999999999999999999/7 (\\mod 1000) [/math.

Using the extended algorithm of Euclid you can calculate that 143 is the multiplicative inverse of 7 modulo 1000, ie that 143 x 7 = 1 (mod 1000) (I do not know how to make a table here but if you Are on Fast mod inverse calculator n = 7, p = 1000 fills up you will get what you would also find by hand).

So we can replace the division with a multiplication:

[math999999999999999999/7 (\\mod 1000) = 999999999999999999 \\cdot 143 (\\mod 1000) [/math

And since AB (mod n) = (a mod n) (b mod n) (mod n),

[math999999999999999999 \\cdot 143 (\\mod 1000) = 999 \\cdot 143 (\\mod 1000) [/math

The latter is easy to do: 143 x 999 = 143000-143 = 142857 and that is 857 mod 1000.

Reasoning from the assumption that it yields an integer (which is not guaranteed or given, but question is in other case nonsense).

The last division is P9/7, p entirely, that can only be with P = 4, because 7×7 = 49 (no other P). So last is 7.

The one after last is P5/7 (5 stays over from previous reasoning, 9-4 = 5). So P = 3, 35/7 = 5. One After last = 5.

Another P6 (9-3 is 6), p = 5, 56/7 = 8.

So 857 is only possibility.

The rest is left as an excersise to the reader.

Stated that the counter contains only 9ens, which counters can then I.H.A.?

999999999999999999 ÷ 7

= 1.42857 142857 142857….. 142857

Nein find three digits.

I would make a tail division of it since I was very bad in mathematics in high school and did not have an exam in it either.

Already after 6 Nines I come out on 0 and start the tail division 2 times again as there are 18 nines.I’m coming out on 142857142857142857. So the answer is 857

9:7 = 1 and 2 remember.29:7 = 4 and 1 remember. 19:7 = 2 and 5 remember. 59:7 = 8 and 3 remember. 39:7 = 5 and 4 remember. 49:7 = 7

Makes 142857 together

Grab the first 9. 9 ÷ 7 = 1 of which 2 remains.

So the number is now 299,999,999,999,999,999.

Now grab the first 2 digits, 29. 29 ÷ 7 = 4 where 1 remains.Now the number is 19,999,999,999,999,999.

Grab the first two digits again, 19. 19 ÷ 7 = 2 where 5 overblijdt.Now the number is 5,999,999,999,999,999. Continue to continue until there is no more of the number left to divide by 7. If it is good you have finally produced this number 142,857,142,857,142,857. The last 3 digits are 857. I accidentally saw that you could count it out.