There is a set of digits from 1 to 9 in front of me, with no repeats. I have selected 5 of these digits and I’ve formed a 5-digit number that fulfils all of the following criteria:
1. The second digit is the fourth digit minus the third digit.
2. The fourth digit is double the sum of the third and fifth digits.
3. The sum of the first and fifth digits is equal to the sum of the third and fourth digits.
4. The sum of the four unused digits is equal to an even number.
5. The first digit is the ten-thousands digit and the fifth digit is the units (or ones) digit.
What number did I form?
Scroll down for a clue and further down for the answer.
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Clue: Label the digits as ‘a’ through to ‘e’. Now, place the first three sentences into three equations. This should reduce the answer to four options for the values of b, c, d and e.
Scroll down for the answer.
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Answer: 67,183.
If we label the 1st digit= a, the second digit= b etc. Then…
From statement 1. b= d-c
From statement 2. d= 2c + 2e
From statement 3. a + e = c + d
From statement 2. we can deduce that the 4th digit must be a 6 or an 8.
Using statement 1. to get b, we can narrow down 4 possibilities:
b=5, c=1, d=6, e=2
or
b=4, c=2, d=6, e=1
or
b=5, c=3, d=8, e=1
or
b=7, c=1, d=8, e=3
If we now use statement 3. to determine values for a:
a=5, b=5, c=1, d=6, e=2
or
a=7, b=4, c=2, d=6, e=1
or
a=10, b=5, c=3, d=8, e=1
or
a=6, b=7, c=1, d=8, e=3
We can eliminate the first of these possibilities as two digits are the same number.
We can eliminate the third of these possibilities because a=10 which is two digits.
This leaves us with:
a=7, b=4, c=2, d=6, e=1
and
a=6, b=7, c=1, d=8, e=3
The first of these options has the remaining digits: 3, 5, 8 and 9 which have an odd sum (25).
The answer is therefore a=6, b=7, c=1, d=8, e=3 which forms the number: 67183.
Puzzle a Day 