The number x is two-digit positive integer, x^2 is a four-digit intege

Question

The number x is two-digit positive integer,  x^2  is a four-digit integer, and the thousands digit of  x^2  is 9. Which of the following must be true?

I. The tens digit of x is 9.
II. The sum of all digits of x is smaller than the sum of all digits of  x^2 .
III. The units digit of x is at least 5.

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III

chetan2u · Accepted Answer

So, the info we have 1) x = ab , where a and b are digits. 2) 1000\leq{x^2}<{10000}........32\leq{x}<{100} 3) x^2=9pqr=(ab)^2......so ab\geq{\sqrt{9000}} Now \sqrt{9000}<95^2 So, x = 95, 96, 97, 98 or 99 I. The tens digit of x is 9. Always true II. The sum of all digits of x is smaller than the sum of all digits of x^2 . 95^2 = 9025....9+0+2+5>9+5 ..Yes 96^2=9216....9+2+1+6>9+6 ..Yes 97^2=9409...9+4+0+9>9+7 ..Yes But 99^2=9801...9+8+0+1=9+9 ...Not true Not always true III. The units digit of x is at least 5. Always true I and III....D You can skip the work for statement II when you see that I and III can be straight way found to be true. However, the question requires us to do some intensive calculations, such as \sqrt{9000} , which may not be true in actuals.

jhaamod10 · Answer

X is two digit integer x2 is four digit integer and thousands digit of x2 is 9 
All above condition satisfy for the numbers 94,95,96,97,98 & 99.
So tens digit of all possible values of x is 9
Now 95^2=9025
 96^2=9216
 97^2=9409
 98^2=9604
 .99^2=9801
So sub options ll, lll, and llll not necessarily true. 
Hence option A is correct answer.

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jhaamod10 · Answer

Correction 
Conditions not satisfy for 94.

Posted from my mobile device

dracarys007 · Answer

10^2 = 100
90^2 = 8100
95^2 = 9025
100^2 = 10000
Hence option D is right answer.

singhamit06 · Answer

The condition in the question is true for any integer greater than square root of 9000 and less than 100. This condition is satisfied for 95, 96, 97 98 and 99. Hence D is the right answer,

jhaamod10 · Answer

Correction
I thought sub option lll is x^2 has unit digit atleast 5
But it is asked x has unit digit at least 5
Hence answer is option D.

Posted from my mobile device

CrackverbalGMAT · Answer

When we have such questions, we need to check if the question asks which statements   can be   true or which   must be   true.

If it asks for must be true as in the case above, then under no condition can the statement be false.

If it asks for can be true, we need to test for a single condition where it would be true (even if other conditions do not satisfy the statements)

We have the following data

(a) A 2 digit number (say x), who's square is a 4 digit number. So 32  \leq  x < 100

(b) The thousands place of  x^2  is 9. So, x can be either 95, 96, 97, 98 or 99

Statement I:   Tens digit of x is 9. This is always true for the numbers listed.

Statement II:   This requires calculation and an area where time can be lost.

If we glance through the options, since statement I is true, Options A, B and E are eliminated.

If statement III is true, then the answer is D, else Option C.

Statement III:   The units digit of x is at least 5 (i.e the last digit can be 5, 6, 7, 8 or 9).

This is always true.

Option D

Arun Kumar

KarishmaB · Answer

x is a two digit number and x^2 is a 4 digit number something like 9abc i.e. x^2 is 9000 something. 
Now note that 100^2 = 10,000 and 90^2 = 8100. So x will lie between 90 and 100. 
Let's narrow down the range further. We can easily calculate the squares of numbers ending in 5. 
95^2 = 9025 (We write 25 at the end and multiply the previous leftover number with its successor i.e. 9*10 = 90. So we get 9025)

Certainly, this would be the smallest perfect square greater than 9000. So x has 5 possible values from 95 to 99.

So we see that statements I and III are certainly true.

Answer (D)

Statement II is a bit unwieldy. 
It is possible that the sum of digits of the square may not be greater than the sum of digits of x. We will need to check the squares of all numbers from 95 to 99. But since the options do not have I, II and III as one of the options, we don't need to do this extra work. 
Though there are some tips that we can use to make it easier if we were required to check that too. 
96^2 would have 9 in the beginning and 6 at the end. In the middle it can't be two 0s (because 9006 will not be the square of 96 since 95^2 is 9025) so sum of digits of the square will be greater. 
97^2 will have 9 in the beginning and 9 at the end (because 7^2 is 49) so sum of square will certainly be greater.
98^2 will have 9 in the beginning and 4 at the end but will be above 9500 so it will have at least another 5 in the middle so sum of digits of the square will be greater. 
99^2 will have 9 in the beginning and 1 at the end. It will need to be found out but we can use (100 - 1)^2 = 10,000 + 1 - 200 = 9801 to get its value.

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The number x is two-digit positive integer, x^2 is a four-digit intege

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