Puzzles
3 December
190 is the smallest multiple of 10 whose digits add up to 10.
What is the smallest multiple of 15 whose digits add up to 15?
2 December
What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 7, and 8?
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8×7×5×3=840. This is also a multiple of 4, 6, 2, and 1. If anything is removed from the product it will no longer be a multiple of all the numbers.
10 December
Today's number is the smallest multiple of 24 whose digits add up to 24.
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There are one three-digit even numbers whose digits add to 24: 798, 888 and 996. Of these, only 888 is a multiple of 24.
15 December
Today's number is smallest three digit palindrome whose digits are all non-zero, and that is not divisible by any of its digits.
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The number can contain no 1 (as all numbers are multiples of 1). It's first and last digits cannot be 2 (as it would then be even and so a multiple of 2).
The lowest three-digit palindrome containing no 0 or 1 and not ending in 2 is 323. This is not a multiple of 2 or 3 and so is the solution.
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?
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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?
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).
Combining multiples
In each of these questions, positive integers should be taken to include 0.
1. What is the largest number that cannot be written in the form \(3a+5b\), where \(a\) and \(b\) are positive integers?
2. What is the largest number that cannot be written in the form \(3a+7b\), where \(a\) and \(b\) are positive integers?
3. What is the largest number that cannot be written in the form \(10a+11b\), where \(a\) and \(b\) are positive integers?
4. Given \(n\) and \(m\), what is the largest number that cannot be written in the form \(na+mb\), where \(a\) and \(b\) are positive integers?
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1. 7
2. 11
3. 90
First, if \(n\) and \(m\) share a common factor other than 1, there will be no largest number: any number that is not a multiple of the common factor will be impossible to make.
If \(n\) and \(m\) are coprime, then considered the remainders when the multiples of \(n\) are divided by \(m\). For example, if \(n=3\) and \(m=7\):
| Multiple | Remainder |
| 0 | 0 |
| 3 | 3 |
| 6 | 6 |
| 9 | 2 |
| 12 | 5 |
| 15 | 1 |
| 18 | 4 |
| 21 | 0 |
Once a number with a given remainder is reached, then all other numbers with that remainder can be reached by repeatedly adding \(m\). Once \(mn\) is reached, the remainders column will repeat itself. Before \(mn\), all remainders will appear (this can be shown by showing that there are \(m\) rows which much all have different remainders). Hence above \(mn\) all numbers can be made.
In the 3,7 example, the last remainder to be hit is 4. The highest number that cannot be made will be the highest number with remainder 4 that is less than 18 (when remainder 4 is hit).
In general, the last remainder will be hit at \(mn-n\). The number before this with the same remainder will be \(mn-n-m\). This will be the highest number that cannot be made.
Extension
Given \(n\), \(m\) and \(k\), what is the largest number that cannot be written in the form \(na+mb+kc\), where \(a\), \(b\) and \(c\) are positive integers?
