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## Math Reflections

Posted by JB on April 5, 2012 – 7:55 pm

What 5 digit number when multiplied by 54321 produces a 10 digit number end with 12345?

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This post is under “Tom” and has 4 respond so far.

### 4 Responds so far- Add one»

1. 1. Wizard of Oz Said：

99945

2. 2. Krazeedude Said：

99945

*was beaten by miles*

3. 3. Knightmare Said：

hi Krazeedude…you got to tell Miles not to treat you like that, no matter how much he wants to know the answer

4. 4. Chris Said：

I can’t find a nice (e.g. modulo arithmetic technique). So here’s a direct method.

Let the number be abcde i.e. abcde*54321 = y*10^5 + 12345
We don’t care what y is, so I’ll completely ignore it.

You might find it useful to practice long multiplications to follow this solution.
The method is to determine e, then use that to determine d, etc.

Do a long multiplication

```abcde
54321
-----
abcde i.e. 1*abcde
????0 i.e. 20*abcde
???00 i.e. 300*abcde
??000 i.e. 4000*abcde
?0000 i.e. 50000*abcde
-----
12345

So e+0+0+0+0 = 5 => e = 5
So use that and re-examine

abcd5
54321
-----
abcd5 1*abcd5
???00 20*5 = 100
???00
??000
?0000
-----
12345

So d+0+0+0+0 = 4 => d = 4.

Substitute and retry.

abc45
54321
-----
abc45 1*45 = 45
??900 20*45 = 900
??500 300*45 = 13500
??000
?0000
-----
12345

We need c + 9 + 5 = 3 (mod 10)
=> c = -11 = 9 (mod 10)
So substituting for c =>

ab945
54321
-----
ab945 1*945 = 945
?8900 20*945 = 18900
?3500 300*945 = 283500
?8000 4000*945 = 3780000
?0000
-----
12345

For the c column we have 9+9+5 = 23, so we carry 2 to the b column.

So we need 2 = 2 + b + 8 + 3 (mod 10) => b = 9

Rinse and repeat

a9945
54321
-----
a9945
98900 20*9945 = 198900. 20*a0000 contributes nothing
83500 300*9945 = 2983500
80000 4000*9945 = 39780000
50000 50000*9945 = 497250000
-----
12345

For the b column we have 9+8+3 = 20, so we carry 2 to the a column.

Need 1 = 2 + a + 9 + 8 + 8 + 5 (mod 10) => a = 9

Hence abcde = 99945
```

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