Convert the base 10 numeral to a numeral in the bse indicate

Convert the base 10 numeral to a numeral in the bse indicated.

5+9= ?(mod 6)

4-1=2 (mod 8)

4 * ?= 3 (mod 2)

?+3=2(mod 6)

4*9=? (mod 5)

72= ?(mod 7)

Answer: Is this correct?

14/6 = 2 R 2 so 2 mod 6
?
?
8 ÷ 6 = 1 R 2 so ? = 5
36 ÷ 5 = 7 R 1 so 1
72 ÷ 7 = 10 R 2 so 2

Re: Convert the base 10 numeral to a numeral in the bse indi

Are we converting bases or discussing modular arithmetic?

#2 makes no sense. Are you certain youcopied it correctly? Shouldn't there be a question mark inthere, somewhere?

"3 (mod 2)" makes little sense.

You do seem to have the idea of the remainder, but much of this is a little odd-looking.

Re: Convert the base 10 numeral to a numeral in the bse indi

Can you please see if the following are correct? I am lot on the last 2, please help me solve. Thank you!



5+9= ? (mod 6)
14=? (mod 6)
14/6=2, remainder 2
Therefore 9+5=2 (mod 6)


4*9=? (mod 5)
36=?(mod 5)
36/5=7, remainder 1
Thus 4*9=1 (mod 5)

4-?=2 (mod 8)
(8+4)-?=2 (mod 8)
12-?=2 (mod 8)
12-10=2 (mod 8)
Therefore, 12-10=2 and 4-10=2 (mod 8)


72=? (mod 7)
7/72= 10 remainder 2
Thus 72=2(mod 7)


These last 2 I really need help please...

1) 4*?=3 (mod 2)

2) ?+3=2 (mod6)

Re: Convert the base 10 numeral to a numeral in the bse indi

1) 4*?=3 (mod 2)

Click to expand...

Any multiple of 4 is going to be a multiple of 2.

If you subtract 3, you will get an odd number.

\(\displaystyle \frac{4k-3}{2}=p\)

Where p is an integer. Is there a value of k that will result in an integer?.

2) ?+3=2 (mod6)

Click to expand...

How about 5?.

5+3=8

\(\displaystyle 8\equiv 2(mod \;\ 6)\)