Just ones and zeroes!

Number Theory Level 4

How many ones are there in the expansion of

( 101 ) × ( 10001 ) × ( 100000001 ) . . . . . × ( 100.... ( 2 7 − 1 ) z e r o e s . . . 0001 ) ? (101) \times (10001) \times (100000001) \ .....\times (100.... ( 2^7-1) zeroes ...0001) ?


The answer is 128.

Jaiveer Shekhawat by jaiveer shekhawat , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Math Man
Aug 19, 2015

Let x = 10

Then the expression is ( x 2 + 1 ) ( x 4 + 1 ) . . . . ( x 128 + 1 ) (x^2+1)(x^4+1)....(x^{128}+1) We know there are 128 terms, and each term coefficients is 1. So there are 128 1's//

1 Helpful 0 Interesting 1 Brilliant 0 Confused
Jaiveer Shekhawat
Aug 13, 2015

101 has 2 2 − 1 2^{2}-1 digits

(101)(10001) = 1010101 2 3 − 1 2^{3}-1 digits

(101)(10001)(100000001) = 101.....101 2 4 − 1 2^{4}-1 digits

.

.

(101)(10001)............ (100.......127 zeroes ....001) = 2 8 − 1 2^{8}-1 digits

You will observe that ( 10 0 1 + 1 ) × ( 10 0 2 + 1 ) (100^{1}+1) \times(100^{2}+1) ... X... ( 10 0 2 n + 1 ) (100^{2^{n}}+1) will have 2 n + 1 − 1 2^{n+1}-1 digits with 2 n 2^{n} ones.

Thus, the expansion has 2 7 2^{7} = 128 o n e s \boxed{128 ones }

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

How do you know it works for all such integers?

Ok!, see the other way round!!

101 = 1 0 2 + 1 10^{2}+1

10001 = 1 0 4 + 1 10^{4}+1

If we multiply just these two, we get ( 1 0 2 + 1 ) (10^{2}+1) X ( 1 0 4 + 1 (10^{4}+1 ) = ( 1 0 6 ) (10^{6}) + ( 1 0 4 ) (10^{4}) + ( 1 0 2 ) (10^{2}) +1 .

Hence, it will have ( 2 2 + 2 1 + 1 ) (2^{2}+2^{1}+1) digits = 2 3 − 1 2^{3}-1 = 7 digits with 4 ones.

Even if you increase the no. of factors, the product will be of the form

( 1 0 2 n + 2 n − 1 . . . + . . . . 2 2 + 2 ) (10^{2^{n}+2^{n-1} ...+.... 2^{2}+2}) + ( 1 0 ( 2 n + 2 n − 1 . . . + . . . . 2 2 + 2 ) − 2 ) (10^{(2^{n}+2^{n-1} ...+.... 2^{2}+2)-2}) + ( 1 0 ( 2 n + 2 n − 1 . . . + . . . . 2 2 + 2 ) − 4 ) (10^{(2^{n}+2^{n-1} ...+.... 2^{2}+2)-4}) + ( 1 0 ( 2 n + 2 n − 1 . . . + . . . . 2 2 + 2 ) − 6 ) (10^{(2^{n}+2^{n-1} ...+.... 2^{2}+2)-6}) ..... + ( 1 0 6 ) (10^{6}) + ( 1 0 4 ) (10^{4}) + ( 1 0 2 ) (10^{2}) +1

Now, it is clear that, if there are n factors then it will have a total of ( 2 n + 2 n − 1 . . . + . . . . 2 2 + 2 + 1 ) (2^{n}+2^{n-1} ...+.... 2^{2}+2+1) digits = 2 n + 1 − 1 2^{n+1}-1 .

...and thus, it will have 2 n 2^{n} ones.

jaiveer shekhawat - 5 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...