Puzzles
Square numbers
Towards the end of his life, Lewis Carroll recorded in his diary that he had discovered that double the sum of two square numbers could always be written as the sum of two square numbers. For example
$$2(3^2 +4^2 )=1^2 +7^2$$
$$2(5^2 +8^2 )=3^2 +13^2$$
Prove that this can be done for any two square numbers.
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Let \(a^2\) and \(b^2\) be the two square numbers.
$$2(a^2 +b^2 ) = 2a^2 +2b^2$$
$$= a^2 + 2ab + b^2 + a^2 - 2ab + b^2$$
$$= (a+b)^2 +(a-b)^2$$
Extension
Prove that 3 times the sum of 3 squares is also the sum of 4 squares.
N
Consider three-digit integers \(N\) such that:
(a) \(N\) is not exactly divisible by 2, 3 or 5.
(b) No digit of \(N\) is exactly divisible by 2, 3 or 5.
How many such integers \(N\) are there?
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(b) implies that the digits of \(N\) are all 1 or 7, so \(N\) can only be 111, 117, 171, 177, 711, 717, 771 or 777. These are all divisible by 3, so no such integers \(N\) exist.
Extension
Consider 21-digit integers \(N\) such that:
(a) \(N\) is not exactly divisible by 2, 3 or 5.
(b) No digit of \(N\) is exactly divisible by 2, 3 or 5.
How many such integers \(N\) are there?
Pocket money
When Dad gave out the pocket money, Amy received twice as much as her first brother, three times as much as the second, four times as much as the third and five times as much as the last brother. Peter complained that he had received 30p less than Tom.
Use this information to find all the possible amounts of money that Amy could have received.
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If Amy gets \(n\)p for pocket money then brother 1 gets \(\frac{n}{2}\)p, brother 2 gets \(\frac{n}{3}\)p, brother 3 gets \(\frac{n}{4}\)p and brother 4 gets \(\frac{n}{5}\)p.
If Tom is brother 1 and Peter is brother 2, then:
$$\frac{n}{2}-\frac{n}{3}=30$$
$$\frac{n}{6}=30$$
$$n=180$$
If Tom is brother 1 and Peter is brother 3, then:
$$\frac{n}{2}-\frac{n}{4}=30$$
$$\frac{n}{4}=30$$
$$n=120$$
If Tom is brother 1 and Peter is brother 4, then:
$$\frac{n}{2}-\frac{n}{5}=30$$
$$\frac{3n}{10}=30$$
$$n=100$$
If Tom is brother 2 and Peter is brother 3, then:
$$\frac{n}{3}-\frac{n}{4}=30$$
$$\frac{n}{12}=30$$
$$n=360$$
If Tom is brother 2 and Peter is brother 4, then:
$$\frac{n}{3}-\frac{n}{5}=30$$
$$\frac{2n}{15}=30$$
$$n=225$$
If Tom is brother 3 and Peter is brother 4, then:
$$\frac{n}{4}-\frac{n}{5}=30$$
$$\frac{n}{20}=30$$
$$n=600$$
So the possible amounts of money Amy could have received are £1.80, £1.20, £1, £3.60, £2.25 and £6.
Extension
Which values could the 30p be replaced with and still give a whole number of pence for all the possible answers?
Always a multiple?
Take a two digit number. Reverse the digits and add the result to your original number. Your answer is multiple of 11.
Prove that the answer will be a multiple of 11 for any starting number.
Will this work with three digit numbers? Four digit numbers? \(n\) digit numbers?
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Let the two digit number chosen by \(10a+b\), with \(a\) and \(b\) one digit integers. Reversed this will be \(10b+a\). The sum of these will be \(11a+11b\) which is divisible by 11.
This will work for any integer with an even number of digits. Let our number have \(2n\) digits. It can be written as:
$$\sum_{i=1}^{2n}10^{i-1} a_i $$
Adding it to its reverse, we get:
$$\sum_{i=1}^{2n}10^{i-1} a_i + \sum_{i=1}^{2n}10^{2n-i} a_i = \sum_{i=1}^{2n}(10^{i-1}+10^{2n-i}) a_i $$
\(10^{i-1}+10^{2n-i}\) is divisible by 11 (for \(n \in \mathbb{N}\), \(i \in \mathbb{N}\), \(i\leq 2n\)). This can be shown by induction on \(n\):
If \(n=1\): \(10^{i-1}+10^{2n-i} = 10^{i-1}+10^{2-i}=11\), which is clearly divisible by 11.
Suppose result is true for \(n-1\). Now consider \(10^{i-1}+10^{2n-i}\).
If \(i>1\), then \(10^{i-1}+10^{2n-i}= 10(10^{(i-1)-1}+10^{(2n-2)-(i-1)}\) which is divisible by 11 by the inductive hypothesis.
If \(i=1\), then:
$$10^{i-1}+10^{2n-i} = 1+10^{2n-1} = \sum_{j=1}^{j=2n-2}9\times 10^j +11$$
$$=\sum_{j=1}^{j=n-1}99\times 10^{2j} +11$$
$$=11\left(\sum_{j=1}^{j=n-1}9\times 10^{2j} +1\right)$$
Extension
Which numbers with an odd number of digits will be divisible by 11 when added to their reverse?
Cycling digits
I have in mind a number which when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth?
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A few examples are:
$$105263157894736842 \times 2 = 210526315789473684$$
$$210526315789473684 \times 2 = 421052631578947368$$
$$421052631578947368 \times 2 = 842105263157894736$$
Extension
I have in mind a number which when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 3. Am I telling the truth?
Mean, median, mode, range
A Find five one-digit positive integers which have a mean of 4, mode of 6, median of 4 and a range of 5.
B Find five one-digit positive integers which have a mean of 3, mode of 1, median of 1 and a range of 8.
C Find five one-digit positive integers which have a mean of 3, mode of 2, median of 2 and a range of 5.
Three digit numbers
Brigette wrote down a list of all 3-digit numbers. For each of the numbers on her list she found the product of the digits. She then added up all of these products. Which of the following is equal to her total?
A \(45\)
B \(45^2\)
C \(45^3\)
D \(2^{45}\)
E \(3^{45}\)
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The sum of the products is:
$$\sum_{i=1}^{9}\sum_{j=1}^{9}\sum_{k=1}^{9}ijk$$
$$=(\sum_{i=1}^{9}i)(\sum_{j=1}^{9}j)(\sum_{k=1}^{9}k)$$
$$=45\times45\times45$$
$$=45^3$$
Extension
Find the sum of the products of the digits of all n digit numbers.
Multiple sums
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
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The multiples of 3 less than 1000 are 3,6,9,...,999; the multiples of 5 are 5,10,15,...,995. Multiples of 15 (15,30,...,990) will appear in both lists so we are trying to find (3+6+9+...+999)+(5+10+15+...+995)-(15+30+...+990). This is:
$$\sum_{i=1}^{333}3i+\sum_{j=1}^{199}5j-\sum_{k=1}^{66}15k$$
$$=3\sum_{i=1}^{333}i+5\sum_{j=1}^{199}j-15\sum_{k=1}^{66}k$$
$$=3\times\frac{333\times334}{2}+5\times\frac{199\times200}{2}-15\times\frac{66\times67}{2}$$
$$=166833+99500-33165$$
$$=233168$$
Extension
Find the sum of all the multiples of 3 or 5 below \(n\).
