Welcome 2016! Part 12

Number Theory Level 4

Find the remainder when the expression below

$\large 9 \times 99 \times 999 \times \ldots \times \underbrace{{999999\ldots 9}}_{999 \ 9's}$

is divided by 1000.

The answer is 109.

1 solution

Aareyan Manzoor
Dec 17, 2015

write it as $(10^1-1)(10^2-1)(10^3-1)(10^4-1)...(10^{999}-1)$ using simple modulo arithmetic this is congruent to $(10-1)(100-1)\underbrace{(-1)(-1)(-1)(-1)....(-1)(-1)}_{997 times}\equiv-891\equiv\boxed{109}\pmod{1000}$

Haha! The same way +1

Department 8 - 5 years, 5 months ago

i don't understand why you wrote 1000-1 as -1.......which concept of simple modulo arithmetic

Mycobacterium Tuberculae - 5 years, 5 months ago

can you elaborate where you require help?

Aareyan Manzoor - 5 years, 5 months ago

in the third step where you write (-1)........upto 997 times .why you write 1000-1 like only-1

Mycobacterium Tuberculae - 5 years, 5 months ago

@Mycobacterium Tuberculae – ok... by modulo arithmetic we know $a+b\pmod{c}\equiv (a\pmod{c}+b\pmod{c})\pmod{c}$ using this $1000+(-1)\pmod{1000}\equiv 1000\pmod{1000}+(-1)\pmod{1000}\equiv 0-1\pmod{1000}$

Aareyan Manzoor - 5 years, 5 months ago

Welcome 2016! Part 12

The answer is 109.

1 solution

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