## The Mystery Six-Digit Number

Find a six-digit number containing no zeros and no repeated digits that satisfies the following conditions:

1. The first and fourth digits sum to the last digit, as do the third and fifth digits.

2. The first and second digits when read as a two-digit number equal one quarter the fourth and fifth digits.

3. The last digit is four times the third digit.

192768.

If you call the number ABCDEF, then you get the following equations.

1. A + D = F and C + E = F

2. AB = DE / 4

3. F = 4 × C

The only numbers that work for C and E are 2 and 6 or 4 and 8, and in order to make F a single-digit number, we can deduce that C = 2, E = 6 and F = 8.

So far, our number is AB2D68.

We know A + D = 8 so A and D are both odd numbers. The only odd number less than 8 that we can use for D to make one-quarter of two-digit number D6 also be a two-digit number is 7, so D = 7 and A is 1. This makes the two-digit number AB 19.

3 Comments on "The Mystery Six-Digit Number"Cathy Lund says

February 28, 2017 @ 12:40

I enjoyed this. Thanks

Marrchelo says

July 2, 2019 @ 07:24

The answer also works for 121334

markinboone says

January 10, 2020 @ 12:13

I don’t know how “The only numbers that work for C and E are 2 and 6 or 4 and 8” is derived in the solution. C and E cannot be 4 and 8 because the sum is a 2-digit number.

You have to start with “F has to be a whole number divisible by 4.” So F can only be 8 or 4 to meet all the other requirements. 8 can be the sum of 3 pairs of unique numbers (4 can’t), so F must be 8 and C must be 2 (step 3).

For step 1 possible pairs are: 1+7, 2+6, 3+5.

So A must be 1, since C is 2 and it can’t be 3 (30 x 4 is a 3-digit number, step 2)

Now D must be 7 and E must be 6 (step 1).

ABCDEF is so far 1_2768, leaving 9 for B (76/4 step 2).

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