+0

0
1317
9

Vinny wrote down all the single-digit base-$b$ numbers and added them in base $b$, getting $34_b$.What is $b$?

Dec 22, 2018

#7
+118459
+2

Good work guest!

I'll just show you how to do it formally.

Vinny wrote down all the single-digit base-$b$ numbers and added them in base $b$, getting $34_b$.What is $b$?

let the base be b then

1+2+3+.....(b-1)= 3b+4

This as an AP

S= (b/2)(1+b)

so

$$T_1=0,\qquad T_b=b-1\\ \frac{b}{2}(0+b-1)=3b+4\\b(b-1)=6b+8\\ b^2-b-6b-8=0\\ b^2-7b-8=0\\ (b-8)(b+1)\\ b=-1\;\;or\;\;8$$

But b must be a possitive integer so the base is 8

Dec 23, 2018

#1
+4610
0

Hmm, I give a take at this.

Dec 22, 2018
edited by tertre  Dec 22, 2018
#2
0

No, that's incorrect.

Guest Dec 22, 2018
#3
+883
0

Incorrect, tertre. we know it has to be 5 and up.

ant101  Dec 22, 2018
#4
0

ant101 what is the answer then?

Guest Dec 22, 2018
#5
+1

0+1+2+3+4+5+6+7 =28 base 10

28/8 =3 + 4 as a remainder

so, the answer is 34 base 8.

Dec 22, 2018
edited by Guest  Dec 22, 2018
#6
0

yup that is correct.

Guest Dec 22, 2018
#7
+118459
+2

Good work guest!

I'll just show you how to do it formally.

Vinny wrote down all the single-digit base-$b$ numbers and added them in base $b$, getting $34_b$.What is $b$?

let the base be b then

1+2+3+.....(b-1)= 3b+4

This as an AP

S= (b/2)(1+b)

so

$$T_1=0,\qquad T_b=b-1\\ \frac{b}{2}(0+b-1)=3b+4\\b(b-1)=6b+8\\ b^2-b-6b-8=0\\ b^2-7b-8=0\\ (b-8)(b+1)\\ b=-1\;\;or\;\;8$$

But b must be a possitive integer so the base is 8

Melody Dec 23, 2018
#8
+124707
+1

Nice, Melody......not as hard as I was making it out to be......!!!

CPhill  Dec 23, 2018
#9
+118459
+1

Thanks Chris,

Sometimes we just get on the wrong track.

It was a nice little question actually :)

Melody  Dec 23, 2018