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

Question

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

Bunuel · Accepted Answer

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.

subhamgarg91 · Answer

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

khrystal · Answer

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)

exc4libur · Answer

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)

Anirban_iimisb · Answer

why not cases for units digit as 1 or 5 ? Squaring them will also give same unit digit

Bunuel · Answer

Checek the highligted text in the stem:

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?

Kinshook · Answer

@manish2014Asked: 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?
Unit digit = {0,1,5,6}
N is a multiple of 9 & 2 = 18 and is of 3 digits

N = {126, 180, 216, 270, 306, 360, 396, 450, 486, 540, 576, 630, 666, 720, 756, 810, 846, 900, 936, 990} : 20 possibilities of N

IMO A

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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

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