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We have $X$ cubes with $8000\le X\le10000$. We have built columns with $2×2$ bases, leaving 2 cubes. We have also built columns with $3×3$ and $5×5$ bases, leaving 4 cubes in these cases.

How can we calculate the number of cubes?

I have created the equations $$n\equiv2\bmod4$$ $$n\equiv4\bmod9$$ $$n\equiv4\bmod25$$ but I am not sure how to proceed in calculating the right number. What is the best way to calculate it? Thanks for help.

  • 0
    Have you tried [Chinese remainder theorem](https://en.wikipedia.org/wiki/Chinese_remainder_theorem)? – 2017-01-13
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    @JaroslawMatlak The solution isn't unique, actually. 900 (the product of the coprime moduli) is less than half of the given range of 2000. – 2017-01-13
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    @ParclyTaxel - so it seems, that there are at least two different solutions :) – 2017-01-13

3 Answers 3

1

Since 4, 9 and 25 are relatively prime, the Chinese remainder theorem guarantees that the number of blocks $X$ is unique modulo $4\cdot9\cdot25=900$.

$X\equiv4\bmod9$ and $X\equiv4\bmod25$ imply $X\equiv4\bmod225$ (the product of 9 and 25). Add 225 repeatedly to 4 until arriving at a number that is $2\bmod 4$: $$4+225+225=454\equiv2\bmod4$$ This is $X\bmod900$; now add 900 repeatedly to 454 until arriving at a number between 8000 and 10000. The possible values for $X$ are thus $$454+900\cdot9=8554$$ $$454+900\cdot10=9454$$

0

The Chinese Remainder is used for calculating a number $n$ in this case that when divided by $4$ has remainder $2$, by $9$ remainder $4$ and by $25$ remainder $4$.

Then $lcm(4,9,25) = 900$ and we need $a+b+c \equiv r \pmod{900}$

Let's start calculating $a=2\cdot9\cdot25\cdot t$ where $t_1 \equiv (9\cdot25)^{-1} \pmod 4$.

Then $t_1 \equiv 225^1 \equiv 1 \pmod 4$ and $a=450$

For $b=4\cdot4\cdot25\cdot t_2$ where $t_2 \equiv (4\cdot25)^{-1} \pmod 9$.

Then $t_2 \equiv 1 \equiv 100^5 \pmod 9$ and $b=400$

For $c=4\cdot4\cdot9\cdot t_3$ where $t_3 \equiv (4\cdot9)^{-1} \pmod 9$

Then $t_3 \equiv 16 \equiv 36^{19} \pmod{25}$ and $c=2304$

Thus $454 \equiv a+b+c \equiv 3154 \pmod{900}$ so $n=454$ but we aren't done yet.

Since you want that $n$ satisfies $8000

0

$n \equiv 4 \pmod 9\\n \equiv 4 \pmod {25}$

So
$n \equiv 4 \pmod {225}$

Possible numbers are $8104, 8329, 8554, 8779, 9004, 9229, 9454, 9679, 9904$

Last two digits of above numbers are
$04, 29, 54, 79, 04, 29, 54, 79, 04 $

Last two digits added by $2$ of above numbers are
$06,31,56,81,06,31,56,81,06$

Only $56$ is divisible by $4$

Corresponding numbers are $8554$ and $9454$
these are the required numbers