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How many three digit even numbers N can be formed that are divisible

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How many three digit even numbers N can be formed that are divisible  [#permalink]

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New post Updated on: 04 Sep 2014, 14:33
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How many three digit even numbers N can be formed that are divisible by 9 such that both N and N^2 have the same units digits?

(A) 20
(B) 21
(C) 22
(D) 23
(E) 24

Originally posted by manish2014 on 04 Sep 2014, 14:10.
Last edited by Bunuel on 04 Sep 2014, 14:33, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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How many three digit even numbers N can be formed that are divisible  [#permalink]

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New post 04 Sep 2014, 14:40
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manish2014 wrote:
How many three digit even numbers N can be formed that are divisible by 9 such that both N and N^2 have the same units digits?

(A) 20
(B) 21
(C) 22
(D) 23
(E) 24


For an even number to have the same units digit as its square, it must have the units digit of 0 (0^2 = 0) or 6 (6^2 = 6). Thus the units digit of N must be 0 or 6.

For a number to be divisible by 9, the sum of its digit must be divisible by 9.

If the units digit of N is 0, then N could be:
900;
810, 180;
720, 270;
630, 360;
540, 450;
990.
10 numbers.

If the units digit of N is 6, then N could be:
306;
216, 126;
936, 396;
846, 486;
756, 576;
666.
10 numbers.

Total = 10 + 10 = 20.

Answer: A.

P.S. Pleas provide OA when posting and do NOT shorten or reword the question in ANY way.


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How many three digit even numbers N can be formed that are divisible  [#permalink]

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New post 07 Dec 2015, 06:15
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manish2014 wrote:
How many three digit even numbers N can be formed that are divisible by 9 such that both N and N^2 have the same units digits?

(A) 20
(B) 21
(C) 22
(D) 23
(E) 24





Unit digit can be "0" or "6"

CASE 1 - When unit digit is "0"

Sum of digits should be divisible by 9

If sum of digits is 9 then first two digits can be -
9,0
1,8 or 8,1
2,7 or 7,2
3,6 or 6,3
4,5 or 5,4

if sum of digits is 18 then -
9,9

CASE 2 - When unit digit is "6"

If sum of digits is 9 then first two digits can be -
3,0
1,2 or 2,1

if sum of digits is 18 then -
9,3 or 3,9
8,4 or 4,8
7,5 or 5,7
6,6

Total = 9 + 1 + 3 + 7 = 20
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Re: How many three digit even numbers N can be formed that are divisible  [#permalink]

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New post 02 Jun 2017, 21:45
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For an even number to have the same units digit as its square, it must have the units digit of 0 (0^2 = 0) or 6 (6^2 = 6). Thus the units digit of N must be 0 or 6.
Now ,
In every 90 numbers(say 1 to 90) , 9 will have 2 factors with 0 or 6 unit digit(i.e 90, 36 )

Therefore 3 digit numbers that area divisible by 9 would be

Total 3 digit numbers = 999-100 +1 = 900

as we know in every 90 numbers we will get two numbers that are divisible by 9( according to the given condition)
Therefore in 900 numbers we will get 20 numbers ((900/10) *2)
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Re: How many three digit even numbers N can be formed that are divisible  [#permalink]

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New post 28 Nov 2019, 09:34
manish2014 wrote:
How many three digit even numbers N can be formed that are divisible by 9 such that both N and N^2 have the same units digits?

(A) 20
(B) 21
(C) 22
(D) 23
(E) 24


Three-digit even numbers must end with {0,2,4,6,8};
Numbers squared with the same units digit are {0,6};
Numbers that are divisible by 9 have multiples of 9 as the sum of its digits;
The UNITS digit, TENS and HUNDREDS must sum to a multiple of 9.

If UNITS = 0; then TENS and HUNDREDS could be:
any of these twice {18,27,36,45} and any of these once {09,99} = 4(2) + 2 = 10

If UNITS = 6; then TENS and HUNDREDS could be:
any of these twice {12,93,48,57} and any of these once {03,66} = 4(2) + 2 = 10

Total = 20

Ans (A)
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Re: How many three digit even numbers N can be formed that are divisible   [#permalink] 28 Nov 2019, 09:34
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