Exporting Grain

Algebra Level 5

In the seventh month, the flow of rice entering the Imperial Capital (Kaifeng) via the Grand Canal arrives at a set pattern for fifteen days.

On the first day, one small ship carrying 24 sacks of rice reaches Kaifeng.

On the second day, five small ships carry 120 sacks of rice to Kaifeng.

On the third day, six medium-sized ships carry 360 sacks of rice to Kaifeng.

On the fourth day, seven large ships carry 840 sacks of rice to Kaifeng.

If this pattern continues, how many sacks were imported from day 1 to day 15?

Note: Kaifeng was the imperial capital of the Northern Song Dynasty


The answer is 279072.

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2 solutions

Tijmen Veltman
Aug 21, 2014

The most logical pattern to spot is:

24 × 5 1 = 120 120 × 6 2 = 360 360 × 7 3 = 840 840 × 8 4 = 1680 24\times\frac51 = 120\\ 120\times\frac62=360\\ 360\times\frac73=840\\ 840\times\frac84=1680\\

and so on. The total number of sacks is therefore:

24 n = 0 14 k = 1 n k + 4 k = 24 n = 0 14 ( n + 4 n ) = 24 11628 = 279072 24\sum_{n=0}^{14}\prod_{k=1}^n \frac{k+4}k \\ = 24\sum_{n=0}^{14}{{n+4}\choose n} \\ = 24 * 11628 = \boxed{279072}

this solution assumes that the size of the ship has no bearing. It looks purely like the # of ships by the day count multiple by the # of sacks in that day.... This is not the only answer if we are just looking for a number pattern... for instance you can also convert all the ships to the small ship equivalent and then use the pattern of (#of ships X 2)+5 as the number of small ships arriving pattern. Then multiple that by 24. The total sacks then becomes 3930264. The series then looks like this: 1, 5, 15, 35, 75, 155, 315... 24, 120, 360, 840, 1800, 3720, 7560...

Randy Yap - 6 years, 8 months ago
Jaiveer Shekhawat
Oct 24, 2014

Just take a look on the data given about the sacks in the question..

24, 120, 360, 840, ???????????

Thus,

24=(1)(2)(3)(4)

120=(2)(3)(4)(5)

360=(3)(4)(5)(6)

840=(4)(5)(6)(7)

Thus, it is nothing other than:

k = 1 15 k ( k + 1 ) ( k + 2 ) ( k + 3 ) \displaystyle \sum_{k=1}^{15}k(k+1)(k+2)(k+3)

= k = 1 15 k 4 \displaystyle \sum_{k=1}^{15}k^{4} + k = 1 15 6 k 3 \displaystyle \sum_{k=1}^{15}6k^{3} + k = 1 15 11 k 2 \displaystyle \sum_{k=1}^{15}11k^{2} + k = 1 15 6 k \displaystyle \sum_{k=1}^{15}6k

= n ( n + 1 ) 30 \frac{n(n+1)}{30} ( 6 n 3 6n^{3} + 9 n 2 9n^{2} +n-1) + 6 n ( n + 1 ) 2 \frac{n(n+1)}{2} X n ( n + 1 ) 2 \frac{n(n+1)}{2} + 11 n ( n + 1 ) ( 2 n + 1 ) 6 \frac{n(n+1)(2n+1)}{6} + 6 n ( n + 1 ) 2 \frac{n(n+1)}{2}

= 15 ( 16 ) 30 \frac{15(16)}{30} (6( 1 5 3 15^{3} )+9( 1 5 2 15^{2} )+15-1) + 6 15 ( 16 ) 2 \frac{15(16)}{2} X 15 ( 16 ) 2 \frac{15(16)}{2} + 11 15 ( 16 ) ( 31 ) 6 \frac{15(16)(31)}{6} + 6 15 ( 16 ) 2 \frac{15(16)}{2}

= 178312+86400+13640+720

= 279072 \huge{279072}

It was my easiest Level 5 problem...

F \color{#D61F06}{F} E \color{#3D99F6}{E} E \color{#20A900}{E} L \color{#333333}{L} I \color{#EC7300}{I} N \color{#69047E}{N} G \color{#624F41}{G}

H \color{#69047E}{H} A \color{#E81990}{A} P \color{#D61F06}{P} P \color{#3D99F6}{P} Y \color{#333333}{Y}

jaiveer shekhawat - 6 years, 7 months ago

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I like the colours.

Steven Zheng - 6 years, 7 months ago

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what about my solution???

jaiveer shekhawat - 6 years, 7 months ago

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