How many 8 digit numbers that are divisible by 9 can be formed so that

Question

How many 8 digit numbers that are divisible by 9 can be formed so that no digit is repeated?

A. 3(7!)
B. 4(7!)
C. 8!
D. 36(7!)
E. 9!

GMATinsight · Accepted Answer

Divisibility test of 9 = Sum of the digits of number must be divisible by 9 for number to be divisible by 9

Sum of all digits from 0 to 9 = 45 (DIvisible by 9)

Two digits to be excluded out of 10 digits can be {0,9} or {1,8} or {2,7} or {3,6} or {4,5}

if {0,9} are not used, Total Numbers = 8!
if {1,8} are not used, Total Numbers = 7*7!
if {2,7} are not used, Total Numbers = 7*7!
if {3,6} are not used, Total Numbers = 7*7!
if {4,5} are not used, Total Numbers = 7*7!

Total Numbers = 7*7!*4 + 8! = 7!*(28+8) = 36*7!

Answer: Option D

nick1816 · Answer

Number of 8 digits in which no digit is repeated= 9*9*8*7*6*5*4*3

As sum of all 10 digits is multiple of 9 and sum of digits of required 8-digit numbers is also divisible by 9, sum of 2 excluded digits must be equal to 9.

Total Number of ways to select 2 distinct digits out of 10 = 10C2=45
 
Total Number of ways to select 2 distinct digits out of 10 such that their sum is 9= 5 
{(0,9), (1,8), (2,7), (3,6) and (4,5)}

Number of 8 digit numbers that are divisible by 9 can be formed so that no digit is repeated
=  \frac{5}{45} (9*9*8*7*6*5*4*3) 
= 9*4*2 *7*6*5*4*3
= 36*7!

preetamsaha · Answer

As ( 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 )= 45 & 45 mod 9 = 0,
sum of 2 digits to be dropped  (8 digit numbers to be formed )   must be a multiple of 9 (obviously 9).
we get 9 = 0+9 =1+8 = 2+7 = 3+6 = 4+5.

So, there are 5 cases 
Now, number formed by removing (0,9) can be arranged in 8! ways.

But for other 4 cases, we've to exclude numbers having 0 as 8th digit (leading digit ).
So, there are 8!-7! = 7(7!) ways for each case.

We can form 8! + 28(7!)= 36(7!) numbers in this way.

CORRECT ANSWER IS D

GMATinsight · Answer

Hey nick1816 I really like this method of solving... That's great buddy... +1 Kudos to you.

Aravind04 · Answer

S={0,1,2,3,4,5,6,7,8,9}→10digits

sum of 10 digits is 45⇒ divisible by 9

∴ For 8 digit number we need to remove two digits from S

After removing ∑⇒ divisible by 9

∴ We can only remove the pairs (0,9),(1,8),(2,7),(3,6),(4,5)

Since 0+9=9,45−9=36⇒ divisible by 9

1+8=9,45−9=36⇒ divisible by 9

4+5=9

3+6=9

∴ If (0,9) are removed then no. of 8 digits nos possible =8!

if (1,8) are removed the no. of 8 digit nos =8!−7! (subtracting the number of cases where '0' is at the left most place)

Similarly, when we remove (2,7), (3,6) and (4,5) we get 8!−7! in each case.

∴ Total 8 digit nos =8!+4(8!−7!)=5⋅8!−4⋅7!

=40⋅7!−4⋅7!

=36(7!)

jnmudd15 · Answer

Could you explain why the total numbers are 7*7! when {1,8}, {2,7}, {3,6} and {4,5} are not used, respectively? I'm not following that logic.

Bunuel · Answer

When constructing an 8-digit number using the digit 0, the first digit cannot be 0. If 0 were the first digit, the number would effectively become a 7-digit number. Therefore, the first digit can be any of the other 7 digits (any digit but 0). The second digit can then take any of the remaining 7 values (excluding the digit already used for the first digit), and so on. Hence, when 0 is included among the 8 digits, the total number of possible combinations is 7 * 7 * 6 * ... * 1 = 7 * 7!.

Hope it's clear.

How many 8 digit numbers that are divisible by 9 can be formed so that

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How many 8 digit numbers that are divisible by 9 can be formed so that

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