Puzzles
4 December
If \(n\) is 1, 2, 4, or 6 then \((n!-3)/(n-3)\) is an integer. The largest of these numbers is 6.
What is the largest possible value of \(n\) for which \((n!-123)/(n-123)\) is an integer?
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We can split this fraction into:
$$\frac{n!}{n-123}-\frac{123}{n-123}$$
As long as \(n\) is larger than 123, \(n-123\) is positive and less than \(n\). This means that \(n!/(n-123)\) is an integer, as \(n-123\) will be one of the numbers multiplied together in \(n!\).
(\((124!-123)/(124-123)\) is an integer, so the answer is at least 124 and \(n!/(2-123)\) is an integer when \(n\) is the answer.)
We now need to answer the simpler question: What is the largest value of \(n\) for which \(123/(n-123)\) is an integer? This will be when \(n-123\) is equal to 123, and so \(n\) is 246.
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
Holly adds up the first six even numbers, then adds on half of the next even number. Her total is 49.
Next, Holly adds up the first \(n\) even numbers then adds on half of the next even number. This time, her total is 465124. What is \(n\)?
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If we add up the first \(n\) even numbers then add on half of the next even number, we get \((n+1)^2\). This means that Holly added up the first \(\sqrt{465124}-1\) or 681 even numbers.
Extension
Can you show why adding up the first \(n\) even numbers and half of the next even number gives \((n+1)^2\)?
23 December
How many numbers are there between 100 and 1000 that contain no 0, 1, 2, 3, or 4?
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These numbers all have 3 digits are there are 5 choices for each digit (5, 6, 7, 8, or 9). The number of numbers is therefore 5×5×5=125.
22 December
Ivy makes a sequence by starting with the number 35, then repeatedly making the next term by reversing the digits of the current number and adding 6.
The first few terms of this sequence are:
$$35$$
$$53+6 = 59$$
$$95+6 = 101$$
What is the first number in Ivy's sequence that is smaller than the previous term?
21 December
In the annual tournament of Christmas puzzles, each player must play one puzzle match
against each other player. Last year there were four entrants into the tournament (A, B, C, and D),
and so 6 matches were played: A vs B, C vs D, A vs D, A vs C, D vs B, and finally B vs C.
This year, the tournament has grown in popularity and 22 players have entered. How many
matches will be played this year?
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Each player must play 21 other players: this makes 22×21=462 matches. But this includes (eg) A vs B and B vs A as two different matches: every match is
included twice, so the number of matches if 462/2=231.
19 December
120 is the smallest number with exactly 16 factors (including 1 and 120 itself).
What is the second smallest number with exactly 16 factors (including 1 and the number itself)?
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If \(p_1^{a_1}\times p_2^{a_2}\times\dots\times p_n^{a_n}\) is the prime factorisation of a number, then the number has \((a_1+1)(a_2+1)\dots(a_n+1)\) factors.
The prime factorisation of 120 is \(2^3\times3\times5\). The next smallest number with 16 factors must be one of:
- \(2\times3^3\times5=270\)
- \(2^3\times3\times7=168\)
- \(2^3\times3^3=216\)
The smallest of these is 168.
18 December
Noel writes the integers from 1 to 1000 in a large triangle like this:
| | | | 1 | | | |
| | | 2 | 3 | 4 | | |
| | 5 | 6 | 7 | 8 | 9 | |
| 10 | 11 | 12 | 13 | ... | | |
The number 12 is directly below the number 6. Which number is directly below the number 133?
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The rightmost entry in the \(n\)th row is \(n^2\). A number below a number in the \(n\)th row is \(2n\) larger than the number in the \(n\)th row.
The smallest square number larger than 133 is 144, so 133 is in the 12th row. This means that the number under it will be 24 greater than it: 133+24=157.
