Question 171070: Find the value of the digit A if the 5-digit number 12A3B is divisible by both 4 and 9, and A does not equal B. I feel like I've tried every number and I can't figure it out. Thanks
Found 2 solutions by Earlsdon, Mathtut:
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Starting with 12A3B.
To be divisible by 4, the last two digits (3B) must be divisible by 4, so B must be either 2 (32 = 4X8) or 6 (36 = 9X4)
Let's assume that B is 2, so then we have:
12A32 Now for this to be divisible by 9, the sum of the digits must be divisible by 9 so that 1+2+A+3+2 is a multiple of 9.
Add the digits: 1+2+3+2+A = 8+A So A must be 1 because 8+1 = 9
Let's check:
 So it's divisible by 4. Now try 9
 and it's also divisible by 9.
A = 1 and B = 2
Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website! according to divisibility rules
for a number to be divisible by 4 the last 2 digits have to be divisible by 4
and for a number to be divisible by 9 all the digits added up have to be divisible by 9. so with that in mind.
:
lets check out 9's first
:
12A3B
:
so 1+2+A+3+B must be divisible by 9
:
1+2+3=6....meaning that A+B must equal either 3 or 12:
for A+B to equal 3..we have two possibilities, A=1 and B=2 or A=2 and B=1.
lets step in with the divisibility rule for 4 now . We need to test these two possibilities . We know we have a 3 in the tens digit position so our possibilities are 31,32 which need to be divisible by 4....31 is not but 32 is divisible by 4.
:
for A+B to equal 12....there would be 8 combinations with the following numbers
: 3 and 9....4 and 8....5 and 7.....6 and 6. but because A cannot equal B
the 6,6 combo is thrown out.
:
At this point lets bring in the divisibility rule for 4 for our 6 remaining combos. Again we know that the tens digit is 3....so the ones digit along with the tens digit must be divisible by
4. We have 6 possibilities from three remaining combinations 3 and 9, 4 and 8, 5 and 7. We must place each number in the ones position while 3 is in the 10's position......so we have 33,34,35,37,38,39.......none of these are divisible by 4......so we can throw them all out.
:
We have only one combo that survived that would be where B=2 and A=1.
:
remember A=6 and B=6 would have worked if they hadnt given us the statement that A unequal to B.... It is important in these problems to eliminate as well as find all the possibilities....which we have done.
:
so the number is 
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