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CAT-2007-Previous Years Paper

Question
9 out of 25
 

Consider four digit numbers for which the first two digits are equal and also equal. How many such numbers are perfect squares?



A 4

B 0

C 1

D 3

E 2

Ans. C

Let the four-digit number be

1000a + 100a + 10b + b = 1100a + 11b

This number will be a perfect square if

1100a + 11b = k2; where k is an integer

Therefore, k should be a multiple of 11 such that 100a + b is a 3-digit number k = 44, 55, 66, …99. Corresponding values of 100a + b will be 176, 275,396,539,704,891.

Therefore, a, b < 9 only ‘704’ satisfies this.

Therefore, 1100a + 11b = 7700 + 44 = 7744 = 882

Alternatively, we can solve this question by using elimination and fundas of number system.

Unit digit of a perfect square can be – 0, 1, 4, 5, 6, 9

Only these digits will get repeated at the last two places of the said square in this question.

So, the last two digits can only be – 00, 11, 44, 55, 66, 99

A perfect square cannot have 11, 55, 66, 99 as its last two digits.

(Reason – A perfect square, when divided by 4, should give either of 0 or 1 as the remainder).

So, now we are left with = 00 and 44.

Perfect square lies in between 1000 and 9999 (both including). If the last two digits are 00, then the unit digit should be ‘0’ in the number. Then it is not possible to have other two digits (at the thousands and hundreds place) same.

So the only option left, if possible = 44

Taking a cue from 122=144, 882= 7744

This is the only number satisfying the condition given in the question.

Note:

1. The same question with a different format was asked in CAT 99. The question was – Given (aa)2= bbcc, where a, b and c are non-zero digits. Find the value of b.

2. 382= 1444 is another such interesting number. “444” occurring at the end of this square is the maximum number of times a non-zero digit is occurring at the end of any perfect square.

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