Sum of Digit Permutations

Question

The question simply states "Let a secret three digit number be $cba$. If the sum of $cab + bac + bca + abc + acb = 2536$, what is $cba$?" I have no idea how to approach this problem. Any hints or help would be greatly appreciated.

Answer 1

The sum of all six numbers is 222c + 222b + 222a If we omit cba = (100c + 10b + a), what is left is N = 122c + 212b + 221a which is what we are given.

Observing 221 = 13*17, we have N = 5c + 4b mod 13 and N = 3c + 8b mod 17 (We could have used factors of 122 or 212, but 2*61 and 2*2*53 are not nearly so convenient as 13 and 17.)

For N = 2536, we have N mod 13 = 1 and N mod 17 = 3 So we have 1 = 5c + 4b mod 13 and 3 = 3c + 8b mod 17

Since b and c are digits in range 0-9, we can try each value of b for 0 through 9 and complete the table below as follows:

• Compute 4 x b and 4 x b mod 13.
• Subtract that from 1 to (5c mod 13).
• Then find a suitable value for c which satisfies that.
• For example 5c = 1 mod 13 -> 5c = 40, c = 8.
• Then compute 3c+8b (mod 17) and find the one which equals 3.

b 4b 4b mod 13 5c mod 13 5c c (3c + 8b) mod 17 0 0 0 1 40 8 7 1 4 4 10 10 2 14 2 8 8 6 45 9 9 3 12 12 2 15 3 16 4 16 3 11 50 10 11 5 20 7 7 20 4 1 6 24 11 3 55 11 13 7 28 2 12 25 5 3 8 32 6 8 60 12 15 9 36 10 4 30 6 5

Examining the table, we see it is the row where b = 7 and c = 5. So we know that the original number was 57a. To determine a, we take 2536 - 122*5 - 212*7 = 2536 - 610 - 1484 = 442 Dividing by 221, we get a = 2, so the original number was 572. We can verify that 527+725+752+257+275 = 2536.

Answer 2

HINT:

$$222(a+b+c)-(cba)=2536$$

Try with multiples of $222$ nearing $2536$

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OR

$$\iff220a+202b+22c=2536\iff202b+22c=2536-220a$$

Answer 3

Since $cab + bac + bca + abc + acb = 2536$ and we know that the sum of digits are

Ones digit: $2c + 2b + a$

Tens digit: $2c + b + 2a$

Hundreds digit: $c + 2b + 2a$

So now we have

$\quad \quad 100(c + 2b + 2a) + 10(2c + b + 2a) + (2c + 2b + a) = 2536$

$\implies 122c + 212b + 221a = 2536$

Answer 4

$cab+bac+bca+abc+acb$ $+cba$ $=2536$+$cba$

$\displaystyle\Rightarrow \frac{2536+(100c+10b+a)}{111}=(2a+2b+2c)$

$\displaystyle 14 \le (a+b+c)\le 20$

and $\quad cba=222(a+b+c)-2536$

Checking with $\quad (a+b+c)=14\quad $ found $\quad cba=572$