Earlier today I set you two puzzles that the late John Horton Conway suggested for this column:
1) The Miracle Builders
I had a window in the north wall of my house. It was a perfect square, 1 metre wide and 1 metre high. But this window never let in enough light. So I hired this firm, the Miracle Builders, who performed the impossible. They remodeled the window so it let in more light. When when they’d finished the window was a perfect square, 1 metre high and 1 metre wide.
How did they do it?
Solution.
The original position of the square was such that the top corner was directly above the opposite corner, and the left corner on the same horizontal level as the right corner. The square was a perfect square, but it was not in the orientation most people assume. In other words, the height of the square, 1m, was the diagonal of the square.
The Miracle Builders built a 1m square that was positioned with two sides horizontal and two sides vertical, the diagonal would be √2m. This square was bigger than the original and let in more light.
2) The Ten Divisibilities
I have a ten digit number, abcdefghij. Each of the digits is different, and
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a is divisible by 1
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ab is divisible by 2
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abc is divisible by 3
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abcd is divisible by 4
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abcde is divisible by 5
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abcdef is divisible by 6
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abcdefg is divisible by 7
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abcdefgh is divisible by 8
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abcdefghi is divisible by 9
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abcdefghij is divisible by 10
What’s my number?
[To clarify: a, b, c, d, e, f, g, h, i, and j are all single digits. Each digit from 0 to 9 is represented by exactly one letter. The number abcdefghij is a ten-digit number whose first digit is a, second digit is b, and so on. It does not mean that you multiply a x b x c x…]
Solution 3816547290.
(The proof uses the following divisibility rules, which I gave as a hint below the line when I set the problem: If you add all the digits in a number, and the total is divisible by 3, then that number is also divisible by 3. If the last two digits of a number are divisible by 4, then that number is divisible by 4. If the last three digits of a number are divisible by 8, then that number is divisible by 8.)
This proof is quite long, and you need to use a calculator in the later stages, so congratulate yourself if you only got a part of the way.
Step 1 Any number divisible by 10 must end in a zero, so j = 0. Any number divisible by 5 must end either in a 0 or a 5. So e = 5.
Step 2 If a number is divisible by an even number, then that number must itself be even, which means that b, d, f and h must all be even. So b, d, f and h are some combination of 2, 4, 6 and 8. The remaining unknowns, a, c, g and i, are therefore some combination of the remaining (odd) numbers 1, 3, 7 and 9.
We know that abcd is divisible by 4. So cd is also divisible by 4. The only combinations in which c is odd, d is even, and cd divides by 4 are 12, 16, 32, 36, 72, 76, 92 and 96. This reveals that d is either 2 or 6.
Step 3 The test for divisibility by 3 states that if the sum of a number’s digits is divisible by 3, then that number is also divisible by 3. It also works the other way around: if a number is divisible by 3, then so is the sum of its digits. So, a + b + c is divisible by 3.
Anything divisible by 6 is also divisible by 3, so a + b + c + d + e + f is also divisible by 3.
If two numbers are divisible by 3, then if you subtract the smaller from the larger, the result must also be divisible by 3. So, the number a + b + c + d + e + f – (a + b + c) = d + e + f is divisible by 3.
We know that d is either 2 or 6; we know that e is 5; and we know that f is either 2, 4, 6 or 8.
If d is 2, then 2 + 5 + f must be divisible by 3. So f must be 8. (It cannot be 2, because d is 2 and each digit appears only once. It cannot be 4 or 6, since 11 and 13 are not divisible by 3.)
If d is 6, then 6 + 5 + f must be divisible by 3, and by the same reasoning above f must be 4.
So we have two options for the middle three digits. Either def is 258 or it is 654. Let’s now try each one.
Step 4 Assume def is 258.
From the divisibility by 8 rule, if an eight-digit number abcdefgh is divisible by 8, then the three-digit number fgh is also divisible by 8. So, 8gh is divisible by 8. g is 1, 3, 7 or 9 and h is one of the remaining even numbers, 4 or 6. The number 8g4 is not divisible by 8 for any value of g, so h must be 6. We have accounted for 2, 6 and 8, which means that b, the final even number, must be 4.
Anything divisible by 9 must be divisible by 3. So a + 4 + c + 2 + 5 + 8 + g + 6 + i is divisible by 3. Since anything divisible by 6 is also divisible by 3, we know that a + 4 + c + 2 + 5 + 8. As we saw above, if two numbers are divisible by 3, then if you subtract the smaller from the larger, the result must also be divisible by 3. So:
g + 6 + i must be divisible by 3.
So, g + i must be divisible by 3. We have to select g and i from 1, 3, 7 and 9, which means that g and i are each either 3 or 9. So a and c are each either 1 or 7. We therefore have four possibilities for the final number, when a, c, g and iare:
1, 7, 3, 9 (so the number is 1472583690)
7, 1, 3, 9 (so the number is 7412583690)
1, 7, 9, 3 (so the number is 1472589630)
7, 1, 9, 3 (so the number is 7412589630)
Reach for your calculator, and test these numbers to see if they conform to the rules stated in the question. You will find that none of them do.
1472583690 fails because 14725836 is not divisible by 8.
7412583690 fails because 7412583 is not divisible by 7.
1472589630 fails because 1472589 is not divisible by 7.
7412589630 fails because 7412589 is not divisible by 7.
Dead end. This doesn’t work. We can conclude that def is not 258.
Step 5. We now know that def is 654.
From the divisibility by 8 rule, if an eight-digit number abcdefgh is divisible by 8, then the three-digit number fgh is also divisible by 8. So, 4gh is divisible by 8. Since 4gh = 400 + gh and 400 is divisible by 8, we know that gh is divisible by 8.
g is 1, 3, 7 or 9 and h is one of the remaining even numbers, 2 or 8. The number gh is not divisible by 8 when h is 8, so h must be 2. Which means that b must be 8.
So, now our number looks like this: a8c654g2i0
Anything divisible by 9 must be divisible by three. So a + 8 + c + 6 + 5 + 4 + g +2 + i is divisible by 3 and using the reasoning above we know that: g + 2 + i must be divisible by 3, where i and g are each one of 1, 3, 7 or 9.
The options for g and i must be one of these:
1 and 3, 3 and 1, 1 and 9, 9 and 1, 3 and 7, 7 and 3, 7 and 9, 9 and 7.
If you go through these options and, using a calculator, test the numbers to see if they conform to the rules stated in the question, the only pair that works is 7 and 9
Thus the answer is 3816547290.
We got there! Thanks for persevering.
If you knew John Conway and want to share a memory about him, please post a comment on the page on which I set these questions.
I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
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