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numbers gmat prob

by quantskillsgmat » Tue Dec 27, 2011 9:14 pm
what can be best approach for this question.I dont want to use pluginn method.
Alphabets a,b and c integers forming a two digit number ab and three digit number ccb.if square of ab =ccb.what is value of b.
a)1
b)0
c)5
d)6
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by user123321 » Tue Dec 27, 2011 9:41 pm
quantskillsgmat wrote:what can be best approach for this question.I dont want to use pluginn method.
Alphabets a,b and c integers forming a two digit number ab and three digit number ccb.if square of ab =ccb.what is value of b.
a)1
b)0
c)5
d)6
if this has to be solved in 2 min.
my approach will be to find the ones that end with 0,1,5,6
since all are present. I quickly find any square is present with 11_ or not.
then in 22_ yes we have 225 here so 15*15 = 225 => b = 5
//ly in 33_ nothing
//ly 44_ yes we have 441 here so 21*21 = 441 => b = 1

wait it seems the question has multiple answers. is it right?

user123321
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by quantskillsgmat » Tue Dec 27, 2011 9:54 pm
u r ryt i missed one condition ccb>300so ans is 1
thanx
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by neelgandham » Wed Dec 28, 2011 11:31 am
Solution:

(10a+b)^2 = 100c + 10c + b
100a^2 + b^2 + 20ab = 100c + 10c + b

The unit's digit of 100a^2,20ab,100c, and 10c is 0 as these are multiples of 10. So, the unit's digit of b^2 is equal to the unit's digit of b. The value of b can be 0,1,5 or 6. Since the value of (10a+b)^2 (= 100c + 10c + b) is
1) a three digit number,
2) a number greater than 300 and
3) a number with unit's digit = 0,1,5,or 6, the value of 10a+b should be one among 20,21,25,26,30,31.

Squaring of the numbers
20^2 = 400
21^2 = 441
25^2 = 625
26^2 = 676
30^2 = 900
31^2 = 961. and the only number of the form CCB is 441 (10's digit = 100's digit). So the answer is A
Anil Gandham
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