225, or 15^2, is the first perfect square that begins with

Question

225, or 15^2, is the first perfect square that begins with two 2s. What is the sum of the digits of the next perfect square to begin with two 2s?

A. 9
B. 12
C. 13
D. 14
E. 17

Bunuel · Accepted Answer

Because 16^2 = 256, the next perfect square to begin with two 2s cannot be 3-digit.

Let's try 4-digit. 40^2 = 1,600 and 50^2 = 2,500. So, if there is a 4-digit perfect square to begin with two 2s it must be between 41^2 and 49^2. Test the middle number 45^2 = 2,025. Test larger number: 46^2 = 2,116. Test next number: 47^2 = 2,209. Bingo! The sum of the digits = 2 + 2 + 0 + 9 = 13.

Answer: C.

PareshGmat · Answer

The next perfect square would be 4 digit number "22xx"; we require to find the addition

45^2 = 2025

The required should be somewhat greater than 45; but would be way less than 50

46^2 = 1600 + 36 + 480 = 2116  (Not correct)

47^2 = 1600 + 49 + 560 = 2209

Sum of digits = 2 + 2 + 0 + 9 = 13

Answer = C

sisirkant · Answer

The next perfect square starting with 22 is, say 22xx. Let's say it's 2200. The square root of it 46.x. Now try with next no which is 47 and it's 2209. So the sum of digits is 13.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

First, notice that consecutive squares get further and further apart as they grow larger: 14^2 = 196, 15^2 = 225, 16^2 = 256, etc. So the next perfect square will have to have at least four digits. We know that 40^2 = 1600 and 50^2=2500, so a four-digit perfect square beginning with 22 must be somewhere in this range. 45^2 = 2025, which is closer, but not close enough, so 46^2 won't be close enough either. 47^2 = 2209 works, however, so that's our answer! (C)

EgmatQuantExpert · Answer

I used a slightly unconventional approach.

First we need to eliminate options B,D and E immediately  .  (I will explain why at the end of the solution).
 
We will be left with option A and C.

Let's take option A.

If the first two digits are 22, then the sum of digits of the first two digits is already 4 and we are left with (9-4) = 5 for the last two digits.

The possible cases would be 2205, 2250, 2214, 2241, 2223 and 2232

Again t hese can be discarded easily because number ending with 5, will have 2 as the tens digits. Number ending with 0 must have 0 as the tens digit. Number ending with 4 number have an even number as the tens digit. Number ending with 3 and 2 cannot be perfect square . A slight check might be needed for 2241, but then this can also be easily eliminated because 40^2 is 1600 and 41^2 cannot be 2241.

Since none of the numbers can be squares and have sum of digits as 9, thus    the answer has to be Option C   .

Note  : I am not trying to find the actual number. All I am doing is using the properties of a perfect square to eliminate the options.

Now if you are wondering why I eliminated options B, D and E. then here is the reason.

Take a few perfect squares and try to find the digital sum of the number. By digital sum, I mean, keep on adding the digits, till you get a single number.

1^2 = 1 ( Sum is 1)
2^2 = 4 ( Sum is 4) 
3^2 = 9 ( Sum is 9)
4^2 = 16 ( Sum is 7)
5^2 = 25 ( Sum is 7)
6^2 = 36 ( Sum is 9)
10^2 = 100 ( Sum is 1)
11^2 = 121 ( Sum is 4)
13^2 = 196 ( Sum is 16 = 1+ 6 = 7)
.....

You can keep on trying with as many perfect squares as possible, there will be one pattern that you will notice -   The digital sum is either 1 or 4 or 7 or 9 and nothing else  . You will not get any other number.

I used to this property to eliminate the options.

The digital sum of all the 5 options are -

A. 9 (digital sum is 9, this could be the answer)
B. 12 ( digital sum is 3, this cannot be the answer)
C. 13 ( digital sum is 4, this can be the answer) 
D. 14 ( digital sum is 5, this cannot be the answer)
E. 17 ( digital sum is 8, this cannot be the answer)

I was kind of unlucky and had to do some calculation to eliminate 9. If 9 would not have been in the answer options, I would have straight away marked 13 as the answer, without thinking twice. :D

Regards,
Saquib
e-GMAT
Quant Expert

CrackverbalGMAT · Answer

There are only 3 perfect squares between 200 and 300 i.e. 225, 256 and 289. Therefore, the next perfect square to begin with 22 HAS to necessarily be a 4-digit perfect square.

This means it has to be a number in the range of 2200 to 2300 (not equal to 2300). 
Since both these numbers are closer to  50^2  = 2500 than  40^2  = 1600, we can say that the perfect square we are looking for is bigger than  45^2 .

45^2  = {4(4+1}25 = 2025.

If a number (x5) is squared, the resultant perfect square will always have 25 as the last 2 digits and the remaining digits can be represented as x(x+1).

Therefore, the square we are looking for is bigger than 2025.

46^2 = 45^2  + 45 + 46 = 2025 + 91 = 2116

Similarly,  47^2= 46^2  + 46 + 47 = 2116 + 93 = 2209. 
The perfect square we are looking for is 2209. The sum of the digits of 2209 = 13.

The correct answer option is C.

Hope that helps!
Aravind B T

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225, or 15^2, is the first perfect square that begins with

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