Find a six-digit even number containing no zeros and no repeating digits in which the first digit is four more than the second digit, the third digit is one less than the sixth digit, and the fourth and fifth digits when read as a single number equal the product of the first and sixth digits.
abcdef
a=b+4
c=f-1
de=af
Let's see what de can be, and what af could then be. Remember that a>=5, since b>=1
de: af
14: 72 b=3 c=1 no
15: 53 no, odd
16: 82 b=3 c=1 no
18: 92 b=5 c=1 no, or 63 b=2 c=1 no
21: 73 no, odd
24: 83 no, odd
27: 93 no, odd
28: 74 b=3 c=3 no
32: 84 b=4 no
36: 94 b=5 c=3 no
42: 67 no, odd; 76 b=3 c=5 YES
54: 69 no, odd; or 96 b=5 no
56: 78 b=3 c=7 no, or 87 b=4 c=6 no
63: 79 or 97 no, both odd
72: 89 no, odd; 98 b=5,c=7 no
So, we are left with 735426
To find a six-digit even number satisfying the given conditions, we can list out the possibilities step by step.
Step 1: The first digit is four more than the second digit.
Let's assume the second digit is x. Therefore, the first digit is x + 4.
Step 2: The third digit is one less than the sixth digit.
Let's assume the sixth digit is y. Therefore, the third digit is y - 1.
Step 3: The fourth and fifth digits, when read as a single number, equal the product of the first and sixth digits.
Let's assume the first digit is x + 4 and the sixth digit is y. Therefore, the fourth and fifth digits can be written as (x + 4) * y.
We have all the conditions now, and we need to find a six-digit even number. The even number property tells us that the last digit must be even. Therefore, y must be even.
Now, let's generate all possible values for x and y based on the given conditions:
1. If y = 2, then the third digit is 2 - 1 = 1.
The fourth and fifth digits will be (x + 4) * 2.
The only available digits left are 3, 5, 6, 7, 8, and 9 for x.
The only available even digit for y is 2.
2. If y = 4, then the third digit is 4 - 1 = 3.
The fourth and fifth digits will be (x + 4) * 4.
The only available digits left are 1, 2, 5, 6, 7, and 8 for x.
The available even digits for y are 2 and 4.
3. If y = 6, then the third digit is 6 - 1 = 5.
The fourth and fifth digits will be (x + 4) * 6.
Since we don't want any digit to be repeated, this case is not valid.
From the possibilities above, we need to find a combination that satisfies all the given conditions.
Let's consider the first case:
- For y = 2, the third digit is 1, and the fourth and fifth digits are (x + 4) * 2.
We cannot have any zero in the six-digit number, so neither x nor y can be 0. Since the second digit is x, we can eliminate 0 from the possible digits for x.
Now, let's find a combination of d igits for this first case. We can set up the equations using the given information.
x ≠ 0 (no zeros allowed)
1 ≠ x (no repeating digits allowed)
(x + 4) * 2 ≠ 0 (no zeros allowed)
(x + 4) * 2 ≠ x + 4 (no repeating digits allowed)
By solving these equations simultaneously, we can find the appropriate digits for x and y.