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Alphanumerics (ABCD×4=DCBA)
By Joe Connelly on Saturday, July 21, 2001 - 01:49 pm:

Find replacements for each letter to make the problems true. For each problem the letter represents a different single digit and must be the same wherever it appears

A B C D × 4 = D C B A

and

R S S + S T = S R R

By The Editor

There are a couple of quite different methods suggested here, so if you don't like the first one, do read on to Arvan's message later.

By Jim Oldfield on Saturday, July 21, 2001 - 08:50 pm:

Start these problems by expressing them algebraically, eg for your second problem
100R+10S+S + 10S+T = 100S+10R+R
Þ 89R - 79S + T = 0
since R, S and T are each digits we know
0£R£9, 0£S£9, 0£T£9
So now try whole numbers in the equation above (try values of R and S and check that T is then between 0 and 9). When you have your values check them in the original equation.
You can use the same method for the first question (I think).

Jim

By Jim Oldfield on Saturday, July 21, 2001 - 09:09 pm:

For the first question I would leave the C's on the right hand side of the equation and other letters on the left. Then you can try values of A, B and D that will give a whole number for C (try dividing both sides of the equation by 60 before you put numbers in: this should make sense when you get there).
I should point out that A, D, R and S can't be 0 because they are the first digit in the numbers.
This method works, but not quite as well for the first question, anyone else have any ideas for solving this problem?

Jim

By Joe Connelly on Sunday, July 22, 2001 - 03:24 am:

Wow!
I think I can do it now.

Thank you Jim

By Jim Oldfield on Sunday, July 22, 2001 - 05:24 pm:

I think there is a solution to 1, here is what I got. There may be other solutions, I don't know.
If you don't want to see the answers, please look away now :)


1)
A=2, B=1, C=7, D=8
ABCD×4=DCBA
2178×4=8712

2)
R=7, S=8, T=9
RSS+ST=SRR
788+89=877

I got the following intermediate equations from the questions:
1) 3999A+390B-996D=60C (then ÷60)
2) 89R-79S+T=0 (I suppose 79S-89R=T is more useful).

By Arvan Pritchard on Monday, July 23, 2001 - 10:25 am:

I usually try to work out restrictions for each digit rather than turning the whole thing into an equation:

For 1:
ABCD×4=DCBA
So A (units of DCBA) is even and (thousands of ABCD) less than 3, so A=2.
Hence D=8, now we can put these in and simplify to
BC×4 + 3=CB
So B is odd (units of CB), less than 3 (tens of BC) so B is 1.
Hence C is 2 or 7 (units of 4C+3 are 1) and at least 4 (tens of CB)

So the solution is 2187×4 = 8712

By Jim Oldfield on Monday, July 23, 2001 - 11:31 pm:

That's a much better method Arvan, I like it.

Jim

By The Editor

This method is perhaps even more useful on the second question: write the sum in columns, and then compare the tens and the units columns. What does this tell you?

There are lots more alphanumeric problems on the NRICH site. Try these links:
Long multiplication
Skeleton (long division)
Kids (addition)
Tis Unique (addition)
Cayley (addition)
Aba (addition)
Two and two (addition)
Alphabet soup (inventing your own)
ABC (long multiplication)
Football sum (addition)
Star in the East (addition)
Many of these will respond better to Arvan's method than Jim's.