A,B,C,D and E are digits and we know that ABC/5 = DE and ABC/6 = ED. Find the value of A,B,C,D and E.?

Question

Math homework.. appreciate any help.
Thanx

Answer 1

Since ABC divided by 5 and by 6 are both integers and that 5 and 6 are relatively prime, it follows that ABC is divisible by 30, so C is 0.

By the division, DE and ED are integers with DE the larger and divisible by 6 and ED the smaller and divisible by 5. Also we can deduce that DE = (6/5)*ED.

Since ED is divisible by 5, then D is either 0 or 5, but if it were 0, then DE would be a one digit number and less than ED. So D must be 5.

So we are looking for a single digit E so that 5E = (6/5)*E5, and that 5E > E5. Since we only have 1, 2, 3, and 4 as our candidates (0 is right out by a similar one digit number argument as above). We can check each case to find that E=4.

So ABC = 6*ED = 6*45 = 270. Thus the numbers are

A = 2
B = 7
C = 0
D = 5
E = 4.

Cheers!

Answer 2

This is easier than it looks. Since A,B,C,D & E are digits, we know that the value of "DE" is really 10D+E. Similarly, the value of ED is really 10E+D. We can rewrite our equations as follows:

ABC = 5(DE)
ABC = 6(ED)

So we know that 5(10D+E) = 6(10E+D)
50D+5E = 60E+6D
44D = 55E
4D=5E

If D and E are digits, the only way this can be true is if they are both zero, which leads to the solution ABCDE all being zero, OR if D=5 and E=4.

So ABC = 5(54) = 270 = 6(45)

A=2, B=7, C=0, D=5, E=4

Answer 3

A=2
B=7
C=0
D=5
E=4

That was fun!