A four digit numeric code is to be formed using the digits 0, 1, 2, 3,

Question

A four digit numeric code is to be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.The sum of the first and second digit is 3 and the fourth digit is a multiple of 4.How many different codes are possible?

Answer:

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120

(Sorry I don't have the answer choices for the question above.)

please explain

Answer:  120

(Sorry I don't have the answer choices for the question above.)

please explain

Anurag@Gurome · Accepted Answer

As the sum of first two digits is 3, the first two digits are either {0 and 3} or {1 and 2}. Hence first two places of the code can be filled in 4 ways : 03, 30, 12, 21

Last digit should be multiple of 4. Hence last place of the code can be filled in 3 ways : 0, 4, 8

The remaining place, i.e. the third place has no restriction. Hence it can be filled by any of the 10 digits, i.e. in 10 ways.

Therefore, number of different codes possible = 4*3*10 = 120

MichelleSavina · Answer

thanks

... i did it the same way....but did not take 0 into consideration.... Is 0 also considered a multiple of any positive integer ???

Anurag@Gurome · Answer

Zero is considered as a multiple of any number.

Bunuel · Answer

Note that an integer  a  is a multiple of an integer  b  (integer  a  is a divisible by an integer  b ) means that  \frac{a}{b}=integer : so, as 0 divided by any integer (except zero itself) yields an integer then yes,  zero is a multiple of every integer .

Also on GMAT when we are told that  a  is divisible by  b  (or which is the same: " a  is multiple of  b ", or " b  is a factor of  a "), we can say that: 
1.  a  is an integer;
2.  b  is an integer;
3.  \frac{a}{b}=integer .

Hope it helps.

akkifreaky · Answer

First 2 digits cant be 03 because if the first digit is 0, then the no. becomes a 3 digit no. Hence I think the answer is 90 and not 120. Can someone correct me if I am wrong ?

sudhir18n · Answer

yes even i think it cant be 03.

taking numbers 0 and 3
first place in 1 way( 0 cant be taken)
2nd place in 1 way 
3rd place in 10 ways( any number from 0-9)
fourth place in 3ways ( 0,4,8)

total 1*1*10*3 = 30 ways

taking number 1 and 2
first place in 2 ways
2nd place in 1 way
3rd place in 10 ways
4th place in 3 ways

total 2*1*10*3= 60

hence is 30+60 = 90 ways

scbguy · Answer

I thought it should be 96 using the slot method

It should be 4x3x2x4
4 because of the four options for summing the two digits to sum and make 3
3 because one of the combos has already been used
Two multiples of 4 so 2 there
And finally those numbers that are left out.

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AnkitK · Answer

0 cannot come in first place .OA shd be 90.

jamifahad · Answer

Bunuel, so a and b can never be 4.4 and 2.2 respectively on GMAT when question says a is divisible by b?

Important point to note! .

Thanks.

vitaliyGMAT · Answer

Sum of the first and second digit =  (3+0) ,  (2+1)  and  (1+2)  --->  3  ways

We can't choose 0+3 because in this case number won't have 4 digits.

4th digit is a multiple of 4 =  0, 4, 8  ------>  3  ways

for the 3d digit all options are awailable -  10  ways

Total:  3*3*10 = 90  ways

ilaukikt · Answer

vitaliyGMAT

It says a numeric code. A code can start with 0. example a briefcase code can be 0000 and still be a 4 digit code.

ilaukikt · Answer

akkifreaky  
It says a numeric code. A code can start with 0. example a briefcase code can be 0000 and still be a 4 digit code.

vitaliyGMAT · Answer

Yes you are right, by some reason I was thinking about 4 digit number. Codes can have 0 in front.
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A four digit numeric code is to be formed using the digits 0, 1, 2, 3,

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