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AB4AC If teh 5 digit number above is divisible by 45, and A, [#permalink]
05 Oct 2004, 17:57

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AB4AC
If teh 5 digit number above is divisible by 45, and A, B, and C are distinct digits, what is the value of the number?
I) A + C = 3
II) A * C = 0 and A * B = 24

AB4AC is divisible by 45 if it can be divisible by both 9 and 5.

Divisibility of 9:sum of digits is 9.
Divisibility of 5:last digit is 0 or 5.

1. We have A+C =0.Possibilities are A=0 or 1 or 2 or 3, C=3 or 2 or 1 or 0. Since the last digit is C, the only option we have is C=0 (for the number to be divisible by 5). So, A=3

We then have the number as 3B430. For the number to be divisible by 9, B has to be 8. Only number. hence sufficient.

2.A*C=0, AB=24. combining these two we can say that A cannot be 0. So, C should be 0. Now, forA and B the possibilities are A=3,6,4 or 8, B =8,4,6 or 3. but 4 is not a possibility because there is already a 4 in the 5 digit number and we have the numbers are distinct. this also means 6 is not possible, because if 6 there has to be a 4.

Only possibility is 3and 8 for either A or B. However, if A is 8 then the number does not fit divisibility of 9. Hence only possibility is A=3 andB=8. Hence sufficient.

If teh 5 digit number above is divisible by 45, and A, B, and C are distinct digits, what is the value of the number?

I agree with Twixt. It looks as though 4 is a viable option for either A, B, or C. Therefore the answer is A because in statement (2) A can be either 8 or 4.

Quote:

I do not understand your point when you write divisible by 9 <=> sum of digits = 9

Twixt - This is a number properties rule, for every multiple of 9, the sum of the digits equals a multiple of 9. 54 -> 5+4=9, 108 ->1+0+8 = 9, 198 -> 1+9+8 = 18. I'm not quite sure of the mathematical proof as to why this is true, it just is. Same goes for 3...for every multiple of 3, the sum of the digits equals a multiple of 3.

I thought about this during my transportation today and I noticed It is true but I did not find any obvious reason !

So being a property of 9 this digits sum is a property of 3. I never noticed this one in any book...

If I try to summarize :

divisible by 2 : last digit = 0,2,4,6,8
3 : sum of digits is a multiple of 3
4 : last digit = 2 times 2 rule ?
5 : last digit = 0,5
6 : mix of 2 and 3
7 : do not know
8 : last digit = 3 times 2 rule ?
9 : sum of digits is a multiple of 9

I thought about this during my transportation today and I noticed It is true but I did not find any obvious reason !

So being a property of 9 this digits sum is a property of 3. I never noticed this one in any book...

If I try to summarize :

divisible by 2 : last digit = 0,2,4,6,8 3 : sum of digits is a multiple of 3 4 : last digit = 2 times 2 rule ? 5 : last digit = 0,5 6 : mix of 2 and 3 7 : do not know 8 : last digit = 3 times 2 rule ? 9 : sum of digits is a multiple of 9

Any volunteer to fill in the blanks ?

Here you go...

divisible by
2 : last digit = even
3 : sum of digits is a multiple of 3
4 : last two digits is a multiple of 4
5 : last digit = 0,5
6 : mix of 2 and 3 (last digit is even and sum is divisible by 3)
9 : sum of digits is a multiple of 9
10: last digit is 0
12: mix of 3 and 4 (last two digits are divisible by 4 and sum is divisible by 3)

If teh 5 digit number above is divisible by 45, and A, B, and C are distinct digits, what is the value of the number?

I agree with Twixt. It looks as though 4 is a viable option for either A, B, or C. Therefore the answer is A because in statement (2) A can be either 8 or 4.

divisible by 2 : last digit = 0,2,4,6,8 3 : sum of digits is a multiple of 3 4 : last digit = 2 times 2 rule ? 5 : last digit = 0,5 6 : mix of 2 and 3 7 : do not know 8 : last digit = 3 times 2 rule ? 9 : sum of digits is a multiple of 9

I'm not familiar with the last digit = 2 times 2 rule or 3 times 2 rule. For 4s and 8s i see if the entire # is divisble by 2 twice or 3 times respectively. Going by the last digit doesn't work for 4s and 8s. Also if the last 2 digits are divisible by 4, the number is divisible by 4 (i.e. 1028...28 is divisible by 4 so 1028 is too).

Also for 100-999...if the first and last digit add up to the middle digit, the number is divisible by 11.