[superprismatic]

If 182 + 235 were equal to 427, what would 21 times 17 be?

Why?

[dyalDragon]

357

527

567

[Guest]

I'm not so good at this sort of thing, so I've probably messed up something.

367

Edit:

Oh. Forgot the "Why?"

Were this a base 9 system, then 7*2 would be 15. The rest of the numbers would stay the same, I think. My problem is knowing how to shift the second row of multiplication.

Edited July 1, 2009 by AnapesticTetrameter

[Guest]

367. It's base 9

[dyalDragon]

I'm not so good at this sort of thing, so I've probably messed up something.

367

oops...thx realized i had a typo because of your answer... my first should match yours

[Guest]

The calculations are done in base 9: so 21 x 17 in base 9 is the equivalent of 19 x 16 in base 10 = 304 which is 367 in base 9.

[Guest]

If 182 + 235 were equal to 427, what would 21 times 17 be?

Why?

It is 367. The numbers are base 9.

[Guest]

i think It's in base 9 so 246?

[Guest]

I'm not so good at this sort of thing, so I've probably messed up something.

367

Edit:

Oh. Forgot the "Why?"

Were this a base 9 system, then 7*2 would be 15. The rest of the numbers would stay the same, I think. My problem is knowing how to shift the second row of multiplication.

182 + 235 = 427

Find the base:

b² + 8b + 2 + 2b² + 3b + 5 = 4b² + 2b + 7

3b² + 11b + 7 = 4b² + 2b + 7

b² = 9b

b=9

transform 21*17

(2b + 1)(b+7)

Combine:

2b² + 15b + 7

go from right to left - for every coeeficient greater than b, subtract b from the coeffecient and add 1 to the next higher power until it is less than b.

for b=9

2b² + 15b + 7 = 3b² + 6b + 7

Now you have your digits: 367_b

[Guest]

i think It's in base 9 so 246?

[/qoops i multiplied 19 * 16 and got 204 not 304 it is 367...

[Guest]

169

21

17

---------

147

+ 21

--------

169

[Guest]

17x21={(17x6)+(17x15)}=182+235=427

[Guest]

182 + 235 = 427

Find the base:

b² + 8b + 2 + 2b² + 3b + 5 = 4b² + 2b + 7

3b² + 11b + 7 = 4b² + 2b + 7

b² = 9b

b=9

transform 21*17

(2b + 1)(b+7)

Combine:

2b² + 15b + 7

go from right to left - for every coeeficient greater than b, subtract b from the coeffecient and add 1 to the next higher power until it is less than b.

for b=9

2b² + 15b + 7 = 3b² + 6b + 7

Now you have your digits: 367_b

Oh my. I understood everything except for "add 1 to the next higher power until it is less than b." For some reason, that sounds like "I cut off an inch and it was still too short."

I saw that the biggest number in the equation was 8. Therefore, it had to be greater than base 8. It also probably couldn't be base 10 or greater, since the initial equation then wouldn't have been any different than it would had it been base 10. So that leaves base 9.

Then I multiplied the numbers assuming they were base 9. That's pretty simple; any number greater than 8 is represented by 1 more than the number (e.g. "9" in base 10 = "10" in base 9). Consequently it came out to be...well, whatever I'd said it was, assuming my multiplication was right! I was just unsure if, when 'long multiplying', you still shifted one place over for the second set of multiplying.

That probably made no sense, but I understand!!

[Guest]

Oh my. I understood everything except for "add 1 to the next higher power until it is less than b." For some reason, that sounds like "I cut off an inch and it was still too short."

from this:

2b² + 15b + 7

go from right to left - for every coeeficient greater than b, subtract b from the coeffecient and add 1 to the next higher power until it is less than b.

OK. You have 3 coefficients in the equation: 2, 15, and 7. Well, you know that none of the coeeficients can be greater than 9, since you know that b is 9. So, go from right to left:

7 < 9. So you don't touch it.

15 >= 9. Subtract 9 from 15, you get 6. you now need to add 1 to the next coeeficient (you're at 1, the b¹, so add 1 to b²'s coefficient). So now you have 3, 6, 7.

3 < 9. So you don't touch it. That gives you

3b² + 6b + 7 = 367 ...

The good thing about the method that i showed was that it works for any base b (as long as the original equation is actually correct ). You don't need to hunt-and-peck for the base. For example:

Change the original equation

182 + 235 = 427

to

192 + 295 = 417

b² + 9b + 2 + 2b² + 9b + 5 = 4b² + 1b + 7

3b² + 18b + 7 = 4b² + 1b + 7

17b = b²

b = 17.

now transform 21*17

(2b+1)(b+7)

2b² + b + 14b + 7

2b² + 15b + 7

A=10, B=11, C=12, D=13, E=14, etc.

so now you have 2F7 in base 17. Or, plugging in 17 in the above polynomial, 840 base 10.