Math Puzzle

[greatwhiteshark]

Find two-ten digit numbers which become nine times as large if the order of the digits is reversed?

 

[tkhunny]

Not a hard problem. Find a representation of the numbers, a representation that will allow you to separate the digits and switch them around. That's ALL the tricky part there is.

 

[greatwhiteshark]

ok

I will try what you said.

 

[Gene]

I had tried TKH's method but the ten unknowns were too much for me. As a logic problem...
The first small number S₀ has to be 1 which makes S₉ = 9. That's the only multiplier that gives x1.
S₁ could be 0 or 1 but 1 seemed unlikely so I tried 0. S₈ = 9x+8 (carried) = ?0 so x=8 and 8 is carried again.
So far S = 1,0??,???,?89. Continuing that idea (with multiple choices for next numbers where necessary) I came up with
1,089,001,089
1,098,910,989
1,099,999,989
I would be interested in seeing TKH's algebraic solution. I'm not proud of any (semi) trial and error solution. I may look at it again now that I have it down to six unknowns. I had a few thoughts that I abandoned.

 

[tkhunny]

Backed myself into a corner on that one, didn't I? I entirely missed the word "ten", simply thinking "two" was repeated. That would be easy, since there aren't any such pairs.

OK, it must be time to slow down and read more carefully.

 

[Gene]

Pheew! I was afraid I was missing some clever manipulation that made it as easy as two. I may be back.
-------------------------
Gene

 

[Denis]

So far S = 1,0??,???,?89. Continuing that idea (with multiple choices for next numbers where necessary) I came up with
1,089,001,089
1,098,910,989
1,099,999,989

Just a play around idea:
If x = your ?????? above, and y = x in reverse, then we get:
9(10^9 + 100x + 89) = 98(10^8) + 100y + 1
simplifies to: 9x - y = 7999992
So both x anf y are multiples of 9 (can be seen in your 3 numbers above).

 

[soroban]

Hello, Janet!

Find two-ten digit numbers which become nine times as large if the order of the digits is reversed

I wouldn't try to solve this with "conventional" algebra.
As you said, ten variables is a bit too much.

This seems to be more of a "cryptarithm" puzzle, where you replace letters with digits.

I'll demonstrate part of it with a <u>five</u>-digit number.


We have this arithmetic problem. (I numbered the columns for reference.)

. . . 1 .2 .3 .4 .5
. . . A .B .C .D .E
. . . x . . . . . . . 9
. . .-----------------
. . . E .D .C .B .A

We see that A = 1.
. . Otherwise, in column 1: .9 x A would be a two-digit number.
. . So we have: . A = 1, .E = 9
. . And there is no "carry" from column 2. .Hence: .B = 0


So we have:

. . . 1 .2 .3 .4 .5
. . . 1 .0 .C .D .9
. . . x . . . . . . . 9
. . .-----------------
. . . 9 .D .C .0 .1

In column 5: . 9 x 9 .= .81 . (carry 8)

Then in column 4: .9 x D + 8 .ends in 0.
. . Hence, D = 8

And we have:

. . . 1 .2 .3 .4 .5
. . . 1 .0 .C .8 .9
. . . x . . . . . . . 9
. . .----------------
. . . 9 .8 .C .0 .1


As it turns out, this particular example has <u>no</u> solution,
. . but I hope you get the idea of how to work out the digits one-at-a-time.

 

[greatwhiteshark]

Thank you

I will keep your notes. Soroban, thank you for your steps.