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

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Joined: 14 Jan 2014
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225, or 15^2, is the first perfect square that begins with  [#permalink]

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Updated on: 28 Jul 2014, 04:51
3
16
00:00

Difficulty:

75% (hard)

Question Stats:

57% (02:06) correct 43% (02:19) wrong based on 263 sessions

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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

Originally posted by Vijayeta on 28 Jul 2014, 03:49.
Last edited by Bunuel on 28 Jul 2014, 04:51, edited 2 times in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 58113
Re: 225, or 15^2, is the first perfect square that begins with  [#permalink]

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28 Jul 2014, 05:07
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Vijayeta wrote:
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

Because 16^2 = 256,y 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.

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Re: 225, or 15^2, is the first perfect square that begins with  [#permalink]

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30 Jul 2014, 00:02
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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

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Re: 225, or 15^2, is the first perfect square that begins with  [#permalink]

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05 Jun 2017, 13:20
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.
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Re: 225, or 15^2, is the first perfect square that begins with  [#permalink]

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30 Dec 2017, 01:17
1
Vijayeta wrote:
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

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)
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Re: 225, or 15^2, is the first perfect square that begins with  [#permalink]

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30 Dec 2017, 01:54
1
Vijayeta wrote:
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

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 these 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
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Re: 225, or 15^2, is the first perfect square that begins with  [#permalink]

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16 Jul 2019, 10:26
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Re: 225, or 15^2, is the first perfect square that begins with   [#permalink] 16 Jul 2019, 10:26
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