A number theory problem by Nikola Alfredi.

Number Theory Level pending

A function is defined as f ( x , y ) = 23 × 3 4 x + 5590 y + 588 \displaystyle f(x,y) = 23 \times 34^{x} + 5590 y + 588 .

Given that x , y Z + : f ( x , y ) 0 ( m o d 43 ) \displaystyle \exist {x,y} \in \mathbb{Z}^{ +} \ \ : f(x,y) \equiv 0 \pmod {43} .

If f ( x , y ) N ( m o d 559 ) \displaystyle f(x,y) \equiv N \pmod {559} . Then what is the value of N N . If you get multiple values of N N , then submit the answer as the sum of all the values N N .


The answer is 344.

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1 solution

5590 y + 559 0 ( m o d 43 ) 5590y+559\equiv 0\pmod {43} for all integer y y . So we have to find some positive integer x x for which 23 × 3 4 x + 29 0 ( m o d 43 ) 23\times 34^x+29\equiv 0\pmod {43} holds. By direct inspection we see that x = 2 x=2 is a candidate for this. For x = 2 , 23 × 3 4 x + 29 = 26617 344 ( m o d 559 ) x=2, 23\times 34^x+29=26617\equiv \boxed {344}\pmod {559} .

This is same, how I did.

Nikola Alfredi - 1 year, 3 months ago

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Did you check for the other possible values of x x ? Is it true that the result holds for all values of x x satisfying the congruence modulo 43 43 ?

A Former Brilliant Member - 1 year, 3 months ago

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I checked for quite few values ... then looked for its' multiples and came out with the result that x = 42 k + 2 k Z 0 x = 42k + 2 \ \ \ \forall k \in \mathbb{Z}^{\geq 0} .

Nikola Alfredi - 1 year, 3 months ago

And f ( x , y ) 0 , 344 ( m o d 559 ) x , y Z + \displaystyle f(x,y) \equiv 0, 344 \pmod {559} \ \ \ \forall {x,y} \in \mathbb{Z} ^{+} .

Nikola Alfredi - 1 year, 3 months ago

@Alak Bhattacharya [Check this out.](https://brilliant.org/problems/5-card-combintions-watch-out/?ref_id=1585091)

Nikola Alfredi - 1 year, 3 months ago

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What is the connection between this question and that I didn't get.

A Former Brilliant Member - 1 year, 3 months ago

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No connection, It is a fun question... It feels satisfying after finding the answer.

Nikola Alfredi - 1 year, 3 months ago

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