### Where's the Base At?

Over lunch the other day, I was posed the following question. Given,

24^{2} = 554

In between stuffing my face and scratching my head, I did manage to solve the problem, albeit not half as quickly as I would have hoped. I offered the lack of a readily available pen as my excuse. One can't really solve quadratic equations without a pen now, can one? As it turns out, there is a faster, more intuitive method that doesn’t really require a quadratic equation at all. It seemed so obvious once revealed that I had to kick myself for not thinking it.

My question to you, dear reader, is can you come up with the quick solution for this problem?

Bonus question: Now that I have you reeled in, pray tell me what the base of the following equation should be

2^{13} = 80000

## 5 Comments:

base at 'dozen'

I expanded 554 to decimal from base 5 onwards till I hit a square num in decimal(784).

What's the easier method?

ans to 2nd q: 16

I resorted to a quickfire prog for the soln:

#include "stdio.h" //tags not allowed in blogger.com comments

int

powof2(int d)

{

int i,j;

for (i =1,j=1;i<=25;i++)

{

if ( d == (j *= 2) )

{

return i;

}

}

return 0;

}

main()

{

int i,j,k;

for (i =1,j=1; i<=25;i++)

{

j = 8 * (i*i*i*i);

printf("i=%d\tsq=%d\n",i,j);

if ( (k = powof2(j)))

printf("%d, %d\n",k,j);

}

}

Sujith, Full marks to you (and JK steel radials).

Here are the solutions for anyone who for anyone who cares:

Problem 1, standard approach (the one I used):

Assume that the given equation is true in some base x.

Converting the equation to base 10, we get

(2*x^1 + 4*x^0)^2 = 5*x^2 + 5*x^1 + 4

or

4x^2 + 16x + 16 = 5x^2 + 5x + 4

or

x^2 - 11x - 12 = 0

Solving, we get x=-1, and x=12.

Discarding x=-1, we get x=12.

Problem 1, intuitive solution:

Given that 24^2 = 554 in some base x, we can deduce that x>10 (since 24^2 = 576 in base 10).

Also, we can 'intuitively' deduce that x must be not much greater than 10.

Now for the clever bit.

Look at the last digit of the number on the LHS. It is a 4. The last digit on the RHS is also a 4.

Thus, the base x must be such in which 4^2 ends in the digit 4.

We know that in base 10, 4^2 = 16 = ends in 6

Thus in base 11, 4^2 = 15 = ends in 5

And in base 12, 4^2 = 14 = ends in 4

Substituting for x = 12, we find it is the correct answer.

In fact, the only other base in which 4^2 ends in 4 is 6, which we had already ruled out (since we'd established that x>10)

Problem 2:

This problem was a little bit trickier. We start out assuming the base to be some x as before.

Thus, we get

2^(x+3) = 8 * x^4

or

2^x * 2^3 = 8 * x^4

or

2^x = x^4

Taking log (to the base 2), we get

x = 4*lg(x)

At this point I was stymied, since I couldn't figure

out how to solve the eqn. Friend of mine suggested

that the only way to solve an eqn. of this sort is to

plot it out, and find the point of intersection.

I used Matlab to plot the graphs and found them

intersecting at x=1.3333 and x=16. Rejecting 1.3333, I

substituted for x=16, and voila!

Another friend took the log to the base x, and got

x.lg 2 = 4.

Substituting for x in powers of 2, he reached the solution at x=16.

Or you could solve programmatically as Sujith did.

I got the easy solution to problem 1 immediately because, many years ago, in an IIT-JEE training class, someone used the last-digit trick to compute a solution to a problem (see below).

Wow, problem 2 was interesting. I wasn't able to solve it mentally. I realised that the base had to be larger than 10 and that it had to be a power of 2, but I screwed up when I tried 16, because I didn't apply the base to 13 correctly. I forgot that the base of the number 13 could change (dang!).

Here's the type of problem that the IIT-JEE training program had posed. 'What is the last digit of the number 287 raised to the power of 100?'

C--

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