Progressive Intersections

Number Theory Level 4

${AP}_{ 1 }=\quad 10,\quad 17,\quad 24,\quad 31,\quad \ldots\\ { AP }_{ 2 }=\quad 9,\quad 14,\quad 19,\quad 24,\quad \ldots\\ { AP }_{ 3 }= \quad 5,\quad 9,\quad 13,\quad 17,\quad\ldots\\ { AP }_{ 4 }=\quad 5,\quad 8,\quad 11,\quad 14,\quad \ldots$

Consider the four arithmetic progressions above. What is the smallest 4-digit number common to all these progressions?

This problem is a part of my set The Best of Me .

The answer is 1109.

1 solution

Siddharth G
Jun 9, 2014

The following series can be written in the form-
$AP_{ 1 }\rightarrow \quad x\quad \equiv \quad 3\quad (mod\quad 7)\\ \ { AP }_{ 2 }\rightarrow \quad x\quad \equiv \quad 4\quad (mod\quad 5)\\ \ { AP }_{ 3 }\rightarrow \quad x\quad \equiv \quad 1\quad (mod\quad 4)\\ \ { AP }_{ 4 }\rightarrow \quad x\quad \equiv \quad 2\quad (mod\quad 3)$

With the help of Chinese Remainder Theorem (CRT), we get -

$x\quad \equiv \quad 2369\quad (mod\quad 420)\\ \ x\quad \equiv \quad 269\quad \equiv \quad 1109\quad (mod\quad 420)$

Hence the answer is $\boxed {1109}$
P.S Thanks to Shubham Dahiya for correcting me! Please post alternate solutions!

My solution was same but like I didn't know CRT I did it the long way.

A Former Brilliant Member - 6 years, 2 months ago

There's no harm in posting one! :)

Brilliant Mathematics Staff - 6 years, 1 month ago

Progressive Intersections

This problem is a part of my set The Best of Me .

The answer is 1109.

1 solution

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