### Simple puzzle

Find a 6 digit number in which the first two digits are the product of the third and fourth digits, the first digit is double the second, the third digit is one less than the fourth, the last two digits are the sum of the third and fourth, the last digit is one less than the first and the fifth digit is two less than the last. The sum of all the digits is 23.

2 blogpoints to all correct answers prior to tomorrow evening.

## 2 comments:

Hey, an easy one for a change!

Thanks for leaving me time today to play some

Resistance!OK, here's my solution:

Let a be the 1st digit, b be the 2nd, and so on, thru f being the 6th digit.

Therefore we know the following:

10a + b = c * d

a = 2b

c = d - 1

10e + f = c + d

f = a - 1

e = f - 2

a + b + c + d + e + f = 23

Trying to isolate to one variable (b) gets us:

f = 2b - 1

a = 2b

e = (2b - 1) - 2 = 2b - 3

c + d = 10e + f = 10(2b - 3) + (2b - 1) = 22b - 31

Substituting into the final original equation:

a + b + (c + d) + e + f = 23

So (2b) + (b) + (22b - 31) + (2b - 3) + (2b - 1) = 23

So 29b = 58

So b = 2

This tells us:

a = 4

b = 2

e = 1

f = 3, and

c + d = 13, where c = d - 1

So c = 6

d = 7

Thus, the original six digit number is 426713.

Points off for not showing my work, but I came up with the same answer (426713) as Matt, although I reduced to 3b + 13f = 45 and solved for b and f via trial and error.

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