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Check Codes

Age 14 to 16
Challenge Level Yellow star
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Robert Goudie, age 17 from Madras College, Fife, Scotland and Andrei Lazanu, age 14 from School No. 205, Bucharest, Romania sent in correct solutions here.

US Postal Service Money Order
The sum of the first 10 digits of 123-123-123-45 is 22, and, divided by 9, I obtain the remainder 4, not 5. That is $$a_1 + a_2 + ... + a_{10} \equiv 4 \hbox{ mod 9}$$ so, this is not a correct US Money Order number.

Airline Tickets
The first 8 digits of the number are 12345678. This means

$$12345678 \equiv 2 \hbox{ mod 7}$$

and does not satisfy the condition to be congruent to 0 mod 7. This is not a correct airline ticket number.

Universal Product Code
In this case, I wrote the 12 numbers 654321123456, and verified if they match the condition.

$$3(6 + 4 + 2 + 1 + 3 + 5) + (5 + 3 + 1 +2 + 4 + 6) = 84 \equiv 4 \hbox{ mod 10}.$$

So, the code is incorrect.

Credit Card
From the number 1723-1001-2065-1098 I calculated $\alpha$, $\beta$ and $\gamma$.

$$\eqalign{ \alpha &= 2(1+2+1+0+2+6+1+9)=44\cr \beta &= 2 \ ({\rm 5\ and\ 9})\cr \gamma &= 7+3+0+1+0+5+0+8=24\cr \alpha+\beta+\gamma&=70 \equiv 0 {\rm \ mod 10}.}$$

Therefore 1723 1001 2065 1098 is a valid Credit Card number.

ISBN
For 0 5215 3612 4 the result of the summation is 148 and it isn't divisible by 11.

$$10 \times 0 + 9 \times 5 + 8\times 2 + 7\times 1 + 6\times 5 + 5\times 3 + 4\times 6 + 3\times 6 + 2\times 1 + 1\times 2 = 148$$

Therefore 0 5215 3612 4 is not a valid ISBN.

US Banks
The number 07100013 is not a valid US Bank number because it does not have 9 digits. If it had the digit 6 at the end then it would be valid because

$$7\times 0 + 3\times 7 + 9\times 1 + 7\times 0 + 3\times 0 + 9\times 0 + 7\times 1 + 3\times 3 = 46 \equiv 6 \hbox{ mod 10}.$$

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Mod 3

Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.

Novemberish

a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.

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