## 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.

5 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).

fun! says

June 5, 2020 @ 19:39

Complicated but fun.

Greg says

July 27, 2020 @ 22:53

I agree that this statement is not supported by the facts given:

“The only numbers that work for C and E are 2 and 6 or 4 and 8”

There is no correlation between C and E from the start that expressly states that C must be 4 less than E, (or E, 4 more than C).

E does have to be even, and F has to be evenly divisible by 4 (being 4 times the value of another integer), while 4*C cannot be >9 (the same is true for 4*A, as shown below), and C can’t be 0, so C can only be a 1 or a 2, making F a 4 or an 8.

C cannot equal 1, because if C=1, then F=4, which would mean that the viable pairings that add up to 4 are, 1&3 only, and 1 is used by C which isn’t used in the (A+D), (C+E) = F parings. Therefor, F must equal 8 and C must equal 2, which makes E = 6 (for 2+6=8), and leaves the following possible pairings for A+D (1, 7 or 3, 5).

(4*A) must also be less than 10, otherwise D would be 2 digits in : AB = DE / 4, which makes A a 1 or a 2 (and C is already a 2), which means A=1, and D=7.

At this point, we have: A=1, B=?, C=2, D=7, E=6, F=8

Going back to: AB = DE / 4 (1B = 76 / 4 = 19), Thus B=9

192768

Leave a comment