# PROOF-mod arith

• Oct 30th 2007, 05:44 AM
anncar
PROOF-mod arith
Let N=22*31 + 11*17 + 13*19. Without actually computing N, determine the parity of N, the units digits of N, and the remainder when N is divided by 7.
• Oct 30th 2007, 06:26 AM
kalagota
Quote:

Originally Posted by anncar
Let N=22*31 + 11*17 + 13*19. Without actually computing N, determine the parity of N, the units digits of N, and the remainder when N is divided by 7.

parity of N is even.. since, odd*odd=odd, even*odd=even, odd+odd=even and odd+even=odd..
the unit digit is 6.. since 2*1 + 1*7 + 3*9, and 3*9=27, hence, you just need to add 2+7+7=16.. Ü
for the remainder, i had a long method. you just need to express N=7p+r and by Division Algo., r is the remainder which is in that case, r=3. Ü
• Oct 30th 2007, 11:28 AM
DivideBy0
Quote:

Originally Posted by anncar
Let N=22*31 + 11*17 + 13*19. Without actually computing N, determine the parity of N, the units digits of N, and the remainder when N is divided by 7.

To find the parity set N mod 2:

$\displaystyle 22 \times 31 + 11 \times 17 + 13 \times 19 \equiv 0 \times 1 + 1 \times 1 + 1 \times 1 \equiv 2 \equiv 0 \pmod 2$

Therefore it is even.

To find the remainder when divided by 7:

$\displaystyle 22 \times 31 + 11 \times 17 + 13 \times 19 \equiv 1 \times 3 + 4 \times 3 + 6 \times 5 \equiv 1 \times 3 + 4 \times 3 + (-1) \times (-2) \equiv 3+12+2$

$\displaystyle \equiv 17 \equiv 3 \pmod 7$

Therefore it leaves remainder 3.