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# Question of the week - 34 (A, B and C are three distinct ............)

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Joined: 04 Jan 2015
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Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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Updated on: 28 Feb 2019, 00:28
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95% (hard)

Question Stats:

21% (02:43) correct 79% (02:16) wrong based on 89 sessions

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Question of the Week #34

A, B and C are three distinct single-digit positive numbers. If the units digit of $$A^2$$, $$B^2$$ and $$C^2$$ are distinct perfect squares, then what is the number of possible values of (A, B, C)?

A. 4
B. 8
C. 20
D. 48
E. 120

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Originally posted by EgmatQuantExpert on 01 Feb 2019, 04:32.
Last edited by EgmatQuantExpert on 28 Feb 2019, 00:28, edited 2 times in total.
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Posts: 7957
Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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01 Feb 2019, 10:10
EgmatQuantExpert wrote:
A, B and C are three distinct single-digit positive numbers. If the units digit of $$A^2$$, $$B^2$$ and $$C^2$$ are distinct perfect squares, then what is the number of possible pairs of (A, B, C)?

A. 4
B. 8
C. 20
D. 48
E. 120

The single digit that are perfect square are 1, 4 and 9, so the units digit of $$A^2$$, $$B^2$$ and $$C^2$$ are 1, 4 and 9.
Which all numbers have their square ending with 1-- 1 and 9
Which all numbers have their square ending with 4-- 2 and 8
Which all numbers have their square ending with 9-- 3 and 7.

(I) Thus A, B and C can be any of 6 but with restriction.
Without restriction we can fill up A,B and C in 6*5*4 = 120 ways.
Now (1 and 9), (2 and 4) and (3 and 7) cannot be together. Let us take them together.
If 1 and 9 are together, the third can be any of the four, so 4*3!=24. Similarly if 3 and 7 are together or 2 and 8 are together, the ways are 4*3!=24, thus 24+24+24=72.

(II) Other way
(1,9), (2,8), (3,7)..
ways (1,2,3); (1,2,7); (9,2,3); (9,2,7) and another 4 with 8 instead of 2, that is (1,8,3); (1,8,7); (9,8,3); (9,8,7)
All EIGHT can be arranged in 3! ways so answer is 8*3!=8*6=48..

D
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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01 Feb 2019, 10:22
It should be 8, as distinct square ending number will be 1, 4, 9.

Possible distinct set can be (312, 318, 392, 398, 712,718, 792, 798).

If repetition is allowed than answer can be different this is for distinct only.
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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02 Feb 2019, 03:27
chetan,

pls elaborate on
If 1 and 9 are together, the third can be any of the four, so 4*3!=24. Similarly if 3 and 7 are together or 2 and 8 are together, the ways are 4*3!=24, thus 24+24+24=72.
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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02 Feb 2019, 04:19
A,B,C can be 1,4,9 as they are the only single digit perfect square

so A= 1; 1,9
B=4; 2,8
C=9; 3,7

(1,9), ( 2,8), ( 3,7)
total pairs of these digits can be
(1,2,3), ( 1,8,7), ( 1,2,7), ( 1,8,3) ....

2c1*2c1*2c1 * 6 = 8*6 = 48
IMO D

EgmatQuantExpert wrote:
A, B and C are three distinct single-digit positive numbers. If the units digit of $$A^2$$, $$B^2$$ and $$C^2$$ are distinct perfect squares, then what is the number of possible pairs of (A, B, C)?

A. 4
B. 8
C. 20
D. 48
E. 120

_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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02 Feb 2019, 08:01
1
Ana4001 wrote:
chetan,

pls elaborate on
If 1 and 9 are together, the third can be any of the four, so 4*3!=24. Similarly, if 3 and 7 are together or 2 and 8 are together, the ways are 4*3!=24, thus 24+24+24=72.

if 1 and 9 are together, the third number can be selected in 4 ways..( 1 out of 2,8,3,7), and these 3 numbers i.e. 1,9, and the selected number can be arranged in 3! ways. thus total ways = 4*3!.
Similarly for selection when 2 & 8 and 3 & 7 are taken together.
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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02 Feb 2019, 08:51
2
EgmatQuantExpert wrote:
A, B and C are three distinct single-digit positive numbers. If the units digit of $$A^2$$, $$B^2$$ and $$C^2$$ are distinct perfect squares, then what is the number of possible pairs of (A, B, C)?

A. 4
B. 8
C. 20
D. 48
E. 120

IMO D

A, B and C are three distinct single-digit positive numbers, so they can hold values from 1-9.
Also the unit digit of their squared values is a perfect square. This is true only for the square of the numbers 1, 2, 3, 7, 8 and 9.

Therefore, after grouping we have -
1, 9 --> 1
2, 8 --> 4
3, 7 --> 9

LHS are values of A, B and C. And RHS is the unit digit of squared values of A, B and C.
A, B and C can take only one value from each of the LHS. Therefore, A, B and C each can take $$C_1^2*C_1^2*C_1^2=8$$.
And they can be rearranged into $$8*3!=48$$

-----------------------------------------------------------------------------------
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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02 Feb 2019, 09:40
chetan2u wrote:
EgmatQuantExpert wrote:

Now (1 and 9), (2 and 4) and (3 and 7) cannot be together. Let us take them together.
If 1 and 9 are together, the third can be any of the four, so 4*3!=24. Similarly if 3 and 7 are together or 2 and 8 are together, the ways are 4*3!=24, thus 24+24+24=72.

I am sorry, but I don't get why (1 and 9), (2 and 4) and (3 and 7) cannot be together, could you please help me clarify this doubt?
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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02 Feb 2019, 19:20
Hi Friend,

I think question is asking for pairs, not for arrangements. Thus, multiplication with 3! is not required here.
It is combinations problem,not a permutation problem.

B.

GMATMBA5 wrote:
EgmatQuantExpert wrote:
A, B and C are three distinct single-digit positive numbers. If the units digit of $$A^2$$, $$B^2$$ and $$C^2$$ are distinct perfect squares, then what is the number of possible pairs of (A, B, C)?

A. 4
B. 8
C. 20
D. 48
E. 120

IMO D

A, B and C are three distinct single-digit positive numbers, so they can hold values from 1-9.
Also the unit digit of their squared values is a perfect square. This is true only for the square of the numbers 1, 2, 3, 7, 8 and 9.

Therefore, after grouping we have -
1, 9 --> 1
2, 8 --> 4
3, 7 --> 9

LHS are values of A, B and C. And RHS is the unit digit of squared values of A, B and C.
A, B and C can take only one value from each of the LHS. Therefore, A, B and C each can take $$C_1^2*C_1^2*C_1^2=8$$.
And they can be rearranged into $$8*3!=48$$

-----------------------------------------------------------------------------------
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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02 Feb 2019, 21:19
gvij2017 wrote:
Hi Friend,

I think question is asking for pairs, not for arrangements. Thus, multiplication with 3! is not required here.
It is combinations problem,not a permutation problem.

B.

GMATMBA5 wrote:
EgmatQuantExpert wrote:
A, B and C are three distinct single-digit positive numbers. If the units digit of $$A^2$$, $$B^2$$ and $$C^2$$ are distinct perfect squares, then what is the number of possible pairs of (A, B, C)?

A. 4
B. 8
C. 20
D. 48
E. 120

IMO D

A, B and C are three distinct single-digit positive numbers, so they can hold values from 1-9.
Also the unit digit of their squared values is a perfect square. This is true only for the square of the numbers 1, 2, 3, 7, 8 and 9.

Therefore, after grouping we have -
1, 9 --> 1
2, 8 --> 4
3, 7 --> 9

LHS are values of A, B and C. And RHS is the unit digit of squared values of A, B and C.
A, B and C can take only one value from each of the LHS. Therefore, A, B and C each can take $$C_1^2*C_1^2*C_1^2=8$$.
And they can be rearranged into $$8*3!=48$$

-----------------------------------------------------------------------------------

Hello,

Combination will give you only 8 possible selections of A, B and Cs to form the pair (A, B, C). For example, (1,2,3), (1,2,7), (1,8,3), (1,8,7).....
But these 8 possible selections can be rearranged among A, B and Cs --> (2,1,3), (2,1,7), (8,1,3), (7,1,8)..... This rearrangement will give 3! times the original 8 selections.
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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05 Feb 2019, 14:49
Helo chetan2u !

Which is the concept behind considering just 1,4 and 9 as perfect squares digits?

Because 1*1 =1, 2*2 = 4 and 3*3 = 9?

So, for ex 4, 4*4 = 16 and because 16 is not a single digit then 4 is not considered as a perfect square single digit?

Kind regards!
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Re: Question of the week - 34 (A, B and C are three distinct ............)  [#permalink]

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06 Feb 2019, 20:18

Solution

Given:
• A, B, and C are three single-digit positive numbers
• The units digit of $$A^2$$, $$B^2$$, and $$C^2$$ are distinct perfect squares

To find:
• The number of possible values of (A, B, C)

Approach and Working:
• The list of all single digit positive numbers are, {1, 2, 3, 4, 5, 6, 7, 8, 9}
• The units digit of squares of these numbers are, {1, 4, 9, 6, 5, 6, 9, 4, 1}
• Among these the perfect squares are {1, 4, 9}
o The numbers whose perfect squares end with 1 are {1, 9}
o The numbers whose perfect squares end with 4 are {2, 8}
o The numbers whose perfect squares end with 9 are {3, 7}

• Thus, the total number of ways of selecting one number from each pair = 2 * 2 * 2 = 8 ways.
• These three numbers can be arranged in 3! = 6 ways

Therefore, total number of values of (A, B, C) = 8 * 3! = 48

Hence the correct answer is Option D.

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Re: Question of the week - 34 (A, B and C are three distinct ............)   [#permalink] 06 Feb 2019, 20:18
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