4 Digit Integers

Probability Level 3

The number of 4 digit positive integers such that the product of their digits is divisible by 3 is:

6628 2700 7704 5464

Arijit Banerjee by Arijit Banerjee , 25, India

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2 solutions

Arijit Banerjee
Apr 5, 2014

Product will be divisible by 3 if at least one is 0, 3 , 6, 9 . Hence the total 4 digit numbers = 9 * 10^3 = 9000 no. of 4 digit numbers without 0 , 3, 6, 9 = 6^4 = 1296. Hence the required no. numbers are = 9000 - 1296 = 7704

27 Helpful 0 Interesting 0 Brilliant 0 Confused

Oh you are asking for how many 4 digit positive integers that the product of their digits is divisible by 3? I think you should rephrase your question. Many will attempt to answer that by multiplying the digit of your multiple choices. For example, 5 x 4 x 6 x 4 = 480 which is divisible by 3 etc...

Lexter Natividad - 7 years, 2 months ago

I think the question is misleading

lex montano - 7 years, 2 months ago

oh. Rephrase your question please.

Muhammad Hamza - 7 years, 2 months ago

damn man i thought u were asking for one of the no among the givens..... :P

Vivek Routh - 7 years, 2 months ago

i m not satisfied with this answer...please solve with some details.

Inzamam Ali Khawaja - 7 years, 2 months ago

Well I didn't create this question I took it down from a magazine "Mathematics Today " of MTG group ...

Arijit Banerjee - 7 years, 2 months ago

why is the 0 case considered?...excluding 0 case 5225 is the answer...u cud do it either way ....either above or 2. _ _ _ _ a 4 digit no with 3,6 or9 at one space and remaining ( 1,2,4,5,7,8) at the other places +3,6or9 at 2 places and remaining at rest of 2 places and so on and add all cases

Prakhar Kapoor - 7 years, 2 months ago

Log in to reply

0 is divisible by all the numbers.

A hint is also provided in the image with the question as it contains the digit "0"

vipul soni - 7 years, 1 month ago

"the product of their digits is divisible by 3." 7 * 7 * 0 * 4 = 0

John Bicare - 7 years, 2 months ago

Common............... Clear the question.

Mahbubur Rahman - 7 years, 2 months ago

You should rewrite your question .. It's really misleading

Muhammad Abdeen - 7 years, 1 month ago

VBA COde: k = 0 For a = 1 To 9 For b = 0 To 9 For c = 0 To 9 For d = 0 To 9 If (a * b * c * d) Mod 3 = 0 Then

k = k + 1 End If Next d, c, b, a MsgBox (k)

Masba Islam - 7 years, 1 month ago
Ayush Kumar
Apr 13, 2014

simply first calculate the number containning atleast one of four digit is 0,3,6,9 by applying permutation then number containning not 0,3,6,9.then subtract the second from first you will get the answer 7704

1 Helpful 0 Interesting 0 Brilliant 0 Confused

The number can't contain the digits 0, 3, 6, or 9. Thus we have 6 digits to work with, giving us 6^4 = 1296 numbers that don't work. Since there's 9 10 10*10 = 9000 total 4-digit numbers, our answer is 9000 - 1296 = 7704 numbers.

Shawn Ong - 7 years, 1 month ago

Rewrite your question with clear understanding!!!!!

Abhishek Kumar - 7 years ago

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