10
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  111
  333
  555
  777
+ 999
-----
1,111

The above addition is not correct. Can you fix it?

For which numbers $n$ can you change $n$ of the $15$ digits of the five three-digit numbers to $0$ so that the sum really is $1111$? You are allowed to change leading digits to $0$. Other than changing some of the digits to $0$, no other changes are allowed.

The following is a failed attempt for $n=3$:

  011
  300
  555
  777
+ 999
-----
1,111

This puzzle is a generalization of a puzzle by Dudeney.

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3 Answers 3

17
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First let's see what the surplus is:

$111+333+555+777+999-1111=1664$

Now let's see how we can distribute that surplus over the 3 columns:

Each column has a sum ranging from $0$ to $1+3+5+7+9=25$, so there can be a carry of $0$, $1$, or $2$. The carry from the hundreds to the thousands column must be $1$, so that leaves $9$ possible distributions:

1664  1664  1664  1664  1664  1664  1664  1664  1664
  04    14    24    04    14    24    04    14    24
 06    05    04    16    15    14    26    25    24
16    16    16    15    15    15    14    14    14

Those column sums can be made in several ways:

04 = 1+3
05 = 5
06 = 1+5
14 = 5+9
15 = 3+5+7    = 1+5+9
16 = 7+9      = 1+3+5+7
24 = 3+5+7+9
25 = 1+3+5+7+9
26 = impossible

Counting how many digits are used gives:

  1664  1664  1664  1664  1664  1664  1664  1664  1664
    04    14    24    04    14    24    04    14    24
   06    05    04    16    15    14    26    25    24
  16    16    16    15    15    15    14    14    14
n= 6/8   5/7   8/10  7/9   8     9     -     9     10

So $n$ can be any value in the range

$5$ to $10$.

Here is an explicit example of each:

5)    6)    7)    8)    9)    10)
  111   100   011   000   111   111
  333   330   033   030   030   300
  500   505   000   005   000   000
  077   077   077   077   070   700
  090   099   990   999   900   000
 ----+ ----+ ----+ ----+ ----+ ----+
 1111  1111  1111  1111  1111  1111

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5
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First note that $111+333+555+777+999 = 2775 = 1111+1664$

So we have 2 ways to find a solution

  • find $n$ numbers to make $1664$
  • find $m=15-n$ numbers to make $1111$

The "easiest" manual solution

$m=5 (n=10) : 1+10+100+300+700 = 1111$

If we follow the same manual approach to find 1664:

$1664=900+700+50+9+5 $ ($n=5$, equivalent to the solution of @Paul Palmpje)

Hmmm, there is no no_computer tag, I could not resist to launch an extensive search:

$n=5$
$1+10+100+3+30+300+500+7+70+90=1111$

$n=6$
$100+30+300+5+500+7+70+9+90=1111$

$n=7$
$10+30+300+5+50+7+700+9=1111$

$n = 8$
$1+3+30+300+7+70+700=1111$
$1+100+300+50+500+70+90=1111$
$1+10+3+300+7+700+90=1111$

$n = 9$
$1+100+3+300+7+700=1111$
$1+10+30+300+70+700=1111$
$300+5+7+700+9+90=1111$

$n = 10$
$1+10+100+300+700=1111$

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3
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Solution for n=5:

 111
 333
 500
 077
 090
====
1111

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