Puzzles
6 December
Write down the numbers from 12 to 22 (including 12 and 22). Under each number, write down its largest odd factor*.
Today's number is the sum of all these odd factors.
* If a number is odd, then its largest odd factor is the number itself.
Digitless factor
Ted thinks of a three-digit number. He removes one of its digits to make a two-digit number.
Ted notices that his three-digit number is exactly 37 times his two-digit number. What was Ted's three-digit number?
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Ted's number was 925: \(925\div25=37\).
If Ted had removed the final digit of his number, then he would be looking for a solution of \(ABC = 37\times AB\). But \(ABC\)
is between 10 and 11 times \(AB\) (it is \(10\times AB + C\)) and so cannot be 37 times \(AB\). So Ted cannot have removed the final digit.
Therefore, Ted must have removed one of the first two digits: so two- and three- digit numbers have the same final digit (\(C\)).
The final digit of the three-digit number (\(C\)) will be the final digit of \(7\times C\) (7 times the final digit of the two digit number).
This is only possible if the final digit is \(C\) is 0 or 5.
This only leaves four possible solutions—10, 15, 20 and 25—as \(30\times37>1000\). Of these only \(925=37\times25\) works.
Extension
How many three-digit numbers are there that are a multiple of one of the two-digit numbers you can make by removing a digit?
24 December
Today's number is the smallest number with exactly 28 factors (including 1 and the number itself as factors).
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If \(p\) and \(q\) are prime numbers, then the number \(p^a\times q^b\) will have \((a+1)(b+1)\) factors. This is because all its factors are
of the form \(p^\alpha\times q^\beta\), with \(\alpha=0,1,...,a\) and \(\beta=0,1,...,b\). The same idea can be used on numbers with three or more prime factors; in general the number \(p_1^{a_1}\times...\times p_n^{a_n}\) has \((a_1+1)\times...\times(a_n+1)\) factors.
28 can be written as: 28, 14×2, 7×4, or 7×2×2. Therefore the following numbers have 28 factors:
$$2^{27},\quad
2^{13}\times3^1,\quad
2^{6}\times3^3,\quad
2^{6}\times3^1\times5^1
$$
and any other number with 28 factors will have larger prime factors, so will be larger.
These numbers are 134217728, 24576, 1728 and 960. Therefore the smallest number with 28 factors is 960.
21 December
The factors of 6 (excluding 6 itself) are 1, 2 and 3. \(1+2+3=6\), so 6 is a perfect number.
Today's number is the only three digit perfect number.
20 December
What is the largest number that cannot be written in the form \(10a+27b\), where \(a\) and \(b\) are nonnegative integers (ie \(a\) and \(b\) can be 0, 1, 2, 3, ...)?
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Any number can be written as \(10a+27b\) with integer \(a\) and \(b\), since \(1=3\times27-8\times10\).
So the problem may be thought of as asking when one of \(a\) and \(b\) must be negative.
Given one way of writing a number, you can get the others by shifting by 14*29. For example,
$$219 = 10\times30 - 27\times3$$ $$= 10 + 10\times29 - 27\times3 $$ $$= 1\times10 + 27\times11$$
So the question now becomes: When does this adjustment fail to eliminate negative numbers?
This is when you are at what Pedro calls "limit coefficients":
$$10\times(-1) + 27\times13 = 10\times28 + 27\times(-1) = 233$$
So the answer is 233.
Extension
Let \(n\) and \(m\) be integers. What is the largest number that cannot be written in the form \(na+mb\), where \(a\) and \(b\) are nonnegative integers?
Elastic numbers
Throughout this puzzle, expressions like \(AB\) will represent the digits of a number, not \(A\) multiplied by \(B\).
A two-digit number \(AB\) is called elastic if:
- \(A\) and \(B\) are both non-zero.
- The numbers \(A0B\), \(A00B\), \(A000B\), ... are all divisible by \(AB\).
There are three elastic numbers. Can you find them?
Show answer & extension
Hide answer & extension
15, 18 and 45 are elastic.
15's factors are 5 and 3. 105, 1005, 10005, etc will all be multiples of 5 (because they end in 5) and multiples of 3 (as their digits add to 6). Hence they are all multiples of 15.
Similarly, 108, 1008, 10008, etc are all multiples of 9 (adding digits) and 2 (they are even), so they are multiples of 18; and 405, 4005, 40005, etc are all multiples of 9 (adding digits) and 5 (last digits are 5), so they are multiples of 45.
Extension
How many elastic numbers are there in other bases?
16 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10.
Today's number is the largest number than can be made from the digits in red boxes.
| × | | × | | = 6 |
| × | | × | | × | |
| × | | × | | = 180 |
| × | | × | | × | |
| × | | × | | = 336 |
=
32 | | =
70 | | =
162 | |
14 December
Today's number is the largest number that cannot be written in the form \(27a+17b\), where \(a\) and \(b\) are positive integers (or 0).
