perplexus.info :: Numbers : Four + Five = Nine

Four + Five = Nine (Posted on 2009-07-26)

Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this alphametic equation. None of the numbers can contain any leading zero.

FOUR + FIVE = NINE, where FOUR is divisible by 4, FIVE is divisible by 5 and, NINE is divisible by 3.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

Answer

| Comment 1 of 2

FOUR=1960.

FIVE=1275.

NINE=3235.

By inspection, R=0, so E=5 and O=9.

Then either F=1, N=3, U+V=13, or F=2, N=5, U+V=15, or F=3, N=7, U+V=17, or F=4, N=9, U+V=19.

The 2nd and 4th cases are impossible since 5 and 9 are already assigned. The 3rd case is eliminated since one of U,V must be 9, which is already assigned.

Since U is even, (U,V) must be either (8,5), (6,7), (4,9), of which only the 2nd case is possible.

NINE divisible by 3 requires I=2.

Posted by xdog on 2009-07-26 16:19:23

Please log in:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info