Do you really like 11 so much?

Number Theory Level 4

A number is formed using the digits $2,4,7$ exactly once, and using as many zeroes as required for adjustment.

Then $S$ be the set of remainders which can NOT be obtained on dividing the number by $11$ . Find the sum of all elements of $S$

Details and assumptions :-

$\bullet\quad$ A set contains each element exactly once, no repetitions.

$\bullet \quad S$ only contains some integers $k_i$ in the range $0\leq k_i \leq 10$ , as remainder is , by definition, $0\leq remainder < divisor$ .

$\bullet \quad$ You are expected to find the sum of all elements in $S$ , that is the remainders which can't be obtained by dividing the number by $11$ .

$\bullet\quad$ The number $2470$ is a a valid number for consideration, and so is $204007$ and all others which contain only the digits $2,4,7,0$ and $2,4,7$ are exactly once. (Thus, $2247$ is not a valid number)

This is a part of set 11≡ awesome (mod remainders)

The answer is 22.

1 solution

Aditya Raut
Sep 17, 2014

Remainder modulo $11$ is obtained by the following,

$\text{Sum of digits having odd power 10 - Sum of digits having even power of 10}$

(This is because $10^{2k}\equiv 1 \pmod{11}$ , and $10^{2k+1}\equiv -1 \pmod{11}$

There are $3$ ways of splitting the digits in sets of $2$ and $1$ , which are $(i) \quad (2,4),7 \implies \text{remainders (1,10) possible} \\ (ii) \quad (4,7),2 \implies \text{remainders (2,9) possible} \\ (iii) \quad (7,2),4 \implies \text{remainders (5,6) possible}$

Putting all of them at same places, like in $20407$ , remainder is $2$ which has been counted.

(In these, the negative of a remainder can be obtained ust by putting a 0 at end of number, for example $247$ has remainder $5$ and $2470$ has remainder $-5\equiv 6 \pmod{11}$

Thus, we see that $(3,8 4,7)$ can NOT be obtained. Their sum is $\boxed{22}$

Yeah, like pairing up the residues :) Nice hint you gave there!!

Krishna Ar - 6 years, 8 months ago

Used the condition of divisibility by 11 that if the difference between the sum of values at odd places and even places is divisible by 11 ( in this case possible difference is 0 and 11 only) , then the number is also divisible by 11. :)

Sandeep Bhardwaj - 6 years, 8 months ago

nice solution

math man - 6 years, 8 months ago

same way-nice problem though

Ashu Dablo - 6 years, 8 months ago

Do you really like 11 so much?

The answer is 22.

1 solution

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