IIT JEE , AIEEE Forum - Mathematics , Algebra

risin

Let x be the number of base system in which the largest 3 digit perfect square in base 6 can be represented as a 2 digit number.Then x in base 7 is a

a)odd but not prime b)prime c)even perfect square d)even but not perfect square e)none.

mukundmadhav

Largest 3 digit number in base 6 will be 215 (of base 10) (right?)
I don't know its form in base 6 but its not needed
largest 3 digit perfect square less than that is 196.. This is to be represented in a new number system in two digits
So it should be less than the square of the number
Next greatest number is 15. So x = 15
x in base 7 is 20

So i guess it should be even but not perfect square

risin

Could you elaborate on that the calculation of "base" part?Didn't really understand .

mukundmadhav

See Base number systems are basically number systems which use only a particular number of numerals
Like decimal is base 10, it uses 0,1,2...9
Binary is base 2. It uses 0 and 1
Similarly base 6 will use 0,1,2,3,4,5
For a three digit number in base 6, the number of numbers is 216(Because there are six choices each for each of the three digits)
But the first number is 000, so the largest three digit number in base 6 is 215(which is represented in base 6 as 555). Now You need a three digit number which is a perfect square. Greatest number which is a square and less than 215 is 196, which is 14^2.

Now if you want 196 to be expressed as a two digit number, you'll need a number system which has so many numerals that it is possible. Let the number sytem be k. Number of two digit numbers in this system will be k^2. Now 196 should be strictly less than k^2
So k=15..
Now to express 15 in base 7 you proceed like this
0, 1, 2, 3, 4, 5, 6
And instead of 7, you now write 10
So 7 in base 10 = 10 in base 7 :P :P :P
Going on, 10, 11, 12, 13, 14, 15, 16
Now since you can't write 17(Because 7 does not exist in base 7)
you write 20
Hence the answer is 20

risin

Let ABC be the largest 3 digit perfect square in base 6 (A, B, C <= 5) => in base 10 ABC will be 36A + 6B + C and this number is a perfect square. Since 5*(36 + 6 + 1) = 215 => 36A + 6B + C = 14^2 = 196 => ABC = 524.

Now, 524 is written as a 2 digit number CD in base b => b*C + D = 524 where C, D < b and C is a positive integer => b ranges from 23 to 524 => x = 502 = 1315 in base 7.

The properties (odd/even/prime/perfect square) of a number N in base b is same as that of decimal conversion of N => x is even in base 7 also.

=> Choice (4) is the right answer.

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