All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Dark Divisibility (Posted on 2013-04-27)
The natural numbers a,b,c,d are such that their least common multiple equals a+b+c+d. Prove that abcd is divisible by 3 or by 5.

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 5.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Proof | Comment 14 of 18 |
Let n=a+b+c+d. Since n is the least common multiple of a, b, c, and d, n is divisible by a, b, c, and d. Therefore, a=n/w, b=n/x, c=n/y, and d=n/z for some positive integers w, x, y, and z. That means that n/w+n/x+n/y+n/z=n, so 1/w+1/x+1/y+1/z=1. Suppose n is not divisible by 3 or 5. Then, w, x, y, and z cannot be divisible by 3 or 5 since they divide n. Without loss of generality, let w<=x<=y<=z. Since 1/w+1/x+1/y+1/z, 1<w<=4. If w=4, then x=y=z=4, so a=b=c=d=n/4. However, if a=b=c=d, then lcm(a, b, c, d)=a, but a+b+c+d=4a. That is impossible, so w cannot be 4. Since w is not divisible by 3, w cannot be 3. Therefore, w=2, so 1/2+1/x+1/y+1/z=1. Then, 1/x+1/y+1/z=1/2, so 2<x<=6. Since x is not divisible by 3 or 5, x cannot be 3, 5, or 6. Therefore, x=4, so 1/4+1/y+1/z=1/2. Then, 1/y+1/z=1/4, so 4<y<=8. Since y is not divisible by 3 or 5, y cannot be 5 or 6. Therefore, y is either 7 or 8. 1/7+1/z=1/4 implies z=28/3, which is not an integer, so y cannot be 7. Therefore, y=8, so 1/8+1/z=1/4. Then, 1/z=1/8, so z=8. Since w=2, x=4, y=8, and z=8, a=n/2, b=n/4, c=n/8, and d=n/8. However, lcm(a, b, c, d)=n/2, not n. That is impossible, so n must be divisible by 3 or 5. Since n=lcm(a, b, c, d) is divisible by 3 or 5, either a, b, c, or d is divisible by 3 or 5. Therefore, abcd is divisible by 3 or 5.

 Posted by Math Man on 2013-05-09 13:22:15

 Search: Search body:
Forums (0)