Fun with Numbers

Number Theory Level 3

$\large 19^8+18^7+17^6+16^5+15^4+14^3+13^2+12^1+11^0$

Find the remainder when the number above is divided by 11.

This is a part of the set 11≡ awesome (mod remainders) .

The answer is 9.

3 solutions

Aditya Raut
Jun 18, 2014

$19^8 \equiv 3^8 \pmod{11}$ , $3^5 \equiv 1 \pmod{11}$ so $19^8 \equiv 3^3 \equiv 5 \pmod{11}$

$18^7 \equiv 7^7 \equiv 5^3 \times 7 \equiv 28 \equiv 6 \pmod{11}$

$17^6 \equiv 6^6 \equiv (6^2)^3\equiv 3^3 \equiv 5 \pmod{11}$

$16^5 \equiv 5^5 \equiv 25 \times 25 \times 5\equiv 45 \equiv 1 \pmod{11}$

$15^4 \equiv 4^4 \equiv 2^8 \equiv -2^3 \equiv -8\equiv 3 \pmod{11}$

$14^3 \equiv 3^3 \equiv 5 \pmod{11}$

$13^2 \equiv 2^2 \equiv 4 \pmod{11}$

$12^1 \equiv 1 \pmod{11}$

$11^0 \equiv 1 \pmod{11}$

Hence answer is $5+6+5+1+3+5+4+1+1 \equiv \boxed{9} \pmod{11}$

The last one 11^0 is tricky, and I forgot it is 1 and thought it is 0 at first to give me the answer of 8 instead of 9.

Kobe Cheung - 6 years, 8 months ago

Oh yes, I missed that too. I was getting 8.

Roman Frago - 6 years, 4 months ago

Can you explain me the first two lines of his solution?

Vishal Yadav - 5 years, 8 months ago

Actually it should be 19^8=-3(mod11). But since you have even exponent it has been simplified to a positive 3.

Satyajit Ghosh - 5 years, 8 months ago

@Satyajit Ghosh – Thanks. But what about the next two conversions?

Vishal Yadav - 5 years, 8 months ago

@Vishal Yadav – ${ 18 }^{ 7 }\equiv { (7) }^{ 7 }\quad (mod11)$

This is a property of modulo, when you divide 18 by 11, you have 7 as remainder, so it is again simplified to ${ (7) }^{ 7 }$

similarly, ${ 17 }^{ 6 }\equiv { (6) }^{ 6 }\quad (mod11)$

You can check out this wiki for more information

Satyajit Ghosh - 5 years, 7 months ago

I have done the same silly mistake.

Gopal Santra - 5 years, 3 months ago

me too . kept thinking that 8 is correct but had to see solution at last when it showed me wrong

A Former Brilliant Member - 4 years, 9 months ago

Can you explain me the first two lines of your solution ?

Vishal Yadav - 5 years, 8 months ago

So taken from $\text{Gamal Sultan}$ 's solution:

$19^8\equiv (-3)^8$ because ignoring the exponents $22\equiv 0\pmod{11}$ , $22-3=19$ , Therefore $19^8\equiv (-3)^8 \pmod{11}$ , and $(-3)^8$ becomes $3^8$ because of positive exponent. Then $3^8\equiv 9^4 \equiv (-2)^4 \pmod{11}$ because $11-2=9$ , $(-2)^4 \equiv 16 \equiv 5 \pmod{11}$

$18^7\equiv (-4)^7 \equiv (-4)(-4)^6 \equiv (-4)(64)^2 \equiv (-4)(-2)^2 \equiv -16 \equiv 6 \pmod{11}$

$WLOG$ this applies to the rest of the numbers and you should get $\boxed{9} \pmod{11}$

Bloons Qoth - 4 years, 11 months ago

Yay.. Got it for the first try itself.... 👍

Sparsh Sarode - 5 years ago

Hey you are in Safe hands how did you cleared RMO

Aparna Kalbande - 4 years, 10 months ago

what does this "mod" mean ? i solved it using bionomial? plz tell

A Former Brilliant Member - 4 years, 9 months ago

See mod <--hopefully this helps

Bloons Qoth - 4 years, 9 months ago

Gamal Sultan
Jan 30, 2015

19^8 = (22 - 3)^8 = (-3)^8 = 9^4 = (-2)^4 = 16 = 5 (mod 11)

18^7 = (22 - 4)^7 = (-4)^7 = (-4)(4)^6 = (-4)(64)^2 = (-4)(-2)^2 = -16 = 6 (mod 11)

17^6 = (11 + 6)^6 = 6^6 = 36^3 = (33 + 3)^3 = 27 = 5 (mod 11)

16^5 = 5^5 = (5)(25)^2 = (5)(3)^2 = 45 = 1 (mod 11)

15^4 = 4^4 = 16^2 = 5^2 = 25 = 3 (mod 11)

14^3 = 3^3 = 27 = 5 (mod 11)

13^2 = 2^2 = 4 (mod 11)

12^1 = 1 (mod 11)

11^0 = 1 (mod 11)

Then

5 + 6 + 5 + 3 + 5 + 1 + 4 + 1 + 1 = 9 (mod 11)

The answer is 9

Looks pretty clean even without LaTeX :)

Viki Zeta - 4 years, 10 months ago

Dillon Chew
Aug 24, 2015

> mod(19^8+18^7+17^6+16^5+15^4+14^3+13^2+12^1+11^0,11)

ans =

Fun with Numbers

This is a part of the set 11≡ awesome (mod remainders) .

The answer is 9.

3 solutions

0 pending reports