Puzzles
A bit of Spanish
Each of the letters P, O, C, M, U and H represent a different digit from 0 to 9.
Which digit does each letter represent?
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POCO is 4595 and MUCHO is 68925.
Extension
The question could be written as \(POCO\times 15=MUCHO\).
For which values of \(n\) are the letters uniquely defined by \(POCO\times n = MUCHO\)?
Odd squares
Prove that 1 and 9 are the only square numbers where all the digits are odd.
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If \(n^2\) has all odd digits then the units digit of \(n\) must be odd. It can be checked that \(n\) cannot be a one digit number (except 1 or 3 as given in the question) as the tens digit will be even.
Therefore \(n\) can be written as \(10A+B\) where \(A\) is a positive integer and \(B\) is an odd positive integer.
$$n^2=(10A+B)^2\\=100A+20AB+B^2$$
Now consider the tens digit of this.
\(100A\) has no effect on this digit. The tens digit of \(20AB\) will be the units digit of \(2AB\) which will be even. The tens digit of \(B^2\) is even (as checked above). Therefore the tens digit of \(n^2\) is even.
Hence 1 and 9 are the only square numbers where all the digits are odd.
Extension
For which bases is this not true?
37
Take a three digit number where all the digits are the same (eg. 888).
Divide this number by the sum of its digits (eg. 888÷24).
Your answer is 37.
Prove that the answer will always be 37.
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Let the chosen digit be \(n\). The three digit number will be \(100n+10n+1n\) or \(111n\). The sum of the digits will be \(n+n+n\) or \(3n\). Therefore the division is:
$$\frac{111n}{3n}=\frac{111}{3}\\=37$$
Extension
If the numbers are written is a base other than 10, will this trick work? How could it be adapted to work in any base?
Four integers
\(a\), \(b\), \(c\) and \(d\) are four positive (and non-zero) integers.
$$abcd+abc+bcd+cda+dab+ab+bc+cd+da+ac+bd\\+a+b+c+d=2009$$
What is the value of \(a+b+c+d\)?
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$$abcd+abc+bcd+cda+dab+ab+bc+cd+da+ac+bd\\+a+b+c+d=(a+1)(b+1)(c+1)(d+1)-1$$
So:
$$(a+1)(b+1)(c+1)(d+1)=2010\\=2\times 3\times 5\times 67$$
Therefore \(a+b+c+d=1+2+4+66=73\).
Extension
Which numbers could 2009 be replaced with so that the problem still has a unique solution?
Half digits
Can you use each of the digits 1 to 9 to make a fraction which is equal to a half?
Frogs
Two frogs and two toads are standing on five lily pads.
The frogs and toads need to pass each other. They can only move by jumping one or two lily pads forward. In jumping two pads forwards they can jump over other frogs or toads.
How many jumps need to be made to get the frogs and toads past each other?
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Representing the frogs as \(F\), the toads as \(T\) and the spaces as \(\), the solution is as follows:
$$
F\ F\ \_\ T\ T\\
F\ \_\ F\ T\ T\\
F\ T\ F\ \_\ T\\
F\ T\ F\ T\ \_\\
F\ T\ \_\ T\ F\\
\_\ T\ F\ T\ F\\
T\ \_\ F\ T\ F\\
T\ T\ F\ \_\ F\\
T\ T\ \_\ F\ F\\
$$
Eight moves are required.
Extension
If there are three frogs on each side, how many moves are needed?
If there are three frogs on one side and two on the other, how many moves are needed?
If there are \(n\) frogs on one side and \(m\) on the other, how many moves are needed?
1089
Take a three digit number. Reverse the digits then take the smaller number from the larger number.
Next add the answer to its reverse.
For example, if 175 is chosen:
$$571-175=396$$
$$396+693=1089$$
What numbers is it possible to obtain as an answer, and when will each be obtained?
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Call the digits of the starting number \(A\), \(B\) and \(C\).
If the number is a palindrome (if \(A=C\)) then the answer to the subtraction will be 0, so the answer is 0.
If the first and last digits differ one, we can consider the case \(A=C+1\). The same analysis will apply to \(C=A+1\).
$$100C+10B+A-(100A+10B+C)=99C-99A$$
$$=99A+99-99A$$
$$=99$$
Then, \(99+99=198\) so the answer is 198
If the first and last digits differ by at least two, we can consider the case \(A=C+n\), where \(2\leq n\leq 9\). The same analysis will apply to \(C=A+n\).
$$100C+10B+A-(100A+10B+C)=99C-99A$$
$$=99(A+n)-99A$$
$$=99n$$
This number will be 198, 297, 396, 495, 594, 693, 792, 891 or 990. In each of these numbers, the middle digit is 9 and the two other digits add up to 9, so they can be written \(100n+90+(9-n)\). If we add this to its reverse:
$$100n+90+(9-n)+100(9-n)+90+n$$
$$=100n+99-n+900-100n+90+n$$
$$=900+99+90+100n-100n+n-n$$
$$=1089$$
So the three possible answers are 0, 99 and 1089 which will occur when the first and list digits of the number differ by 0, 1 or more respectively.
Extension
Which answers are possible if the numbers are written in different bases?
Which answers are possible if four digit numbers are taken? Or five digit numbers?
Dirty work
Timothy, Urban, and Vincent are digging identical holes in a field.
When Timothy and Urban work together, they dig 1 hole in 4 days.
When Timothy and Vincent work together, they dig 1 hole in 3 days.
When Urban and Vincent work together, they dig 1 hole in 2 days.
Working alone, how long does it take Timothy to dig one hole?
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Let \(t\), \(u\) and \(v\) be the prortion of a hole dug in one day by Timothy, Urban and Vincent respectively. Then the following equations hold:
$$4t+4u=1$$$$3t+3v=1$$$$2u+2v=1$$
These can be solved to find that \(t=\frac{1}{24}\) so it will take Timothy 24 days to dig a hole.
Extension
Working alone, how long does it take Urban to dig one hole?
Working alone, how long does it take Vincent to dig one hole?
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