CAT 1999 Question 2

Question 2

Let a, b, c be distinct digits. Consider a two digit number $$'ab'$$ and a three digit number $$'ccb'$$, both defined under the usual decimal number system. If ($$ab^{2} = ccb$$) and $$ccb > 300$$ then the value of b is

Solution

$$(ab)^2$$ = ccb

ccb > 300

The last digit of the number ab must be same as that of the square of ab.

So, b can be 0, 1, 5 or 6.

$$20^2$$=400 and $$30^2$$=900 are three digit numbers and greater than 300. But the first 2 digits are not same. Hence, b is not 0.

If b is 5, then the ten's digit of ab's square will be 2 => c = 2. But if c is 2, then ccb is not greater than 300. Hence, b is not 5.

If b is 6, then $$26^2$$ = 576 is the only three digit number that is greater than 300. But, it is not in the form of ccb => b is not 6.

If b is 1, then $$21^2$$=441 satisfies all the given conditions => b is 1.



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