Prove: Is ${2017 \choose 652}$ Divisible by 343?

[lfdahl]

Is the binomial coefficient: ${2017 \choose 652 }$ divisible by $343$?

Please prove your statement.

 

[kaliprasad]

Is the binomial coefficient: ${2017 \choose 652 }$ divisible by $343$?

Please prove your statement.

Spoiler

Ley us see how many powers of 7 are there in 2017!

2017 = 288 * 7 +1
288 = 41 * 7 + 1
41 = 5 * 7 + 6
so 288 numbers are divisible by 7. 41 by 49 and 5 by 343 so 288 + 41 + 5 or 334

so 2017! is divisible by $7^{334}$

denominator

652 = 93 * 7 + 1
93 = 13 * 7 + 2
13 = 7* 1 + 6
so 652! is divisible by $7^{107}$

2017 - 652 = 1365

1365 = 195 * 7
195 = 27 * 7 + 6
27 = 3 * 7 + 6
so 1365! is divsible by $7^{225}$ and hence highest power of 7 that devides denominator of ${2017 \choose 652 }$ is 332 and highest power deviding numerator = 334 so $7^2$ devides the numbeer and not $7^3$ and the ans is no

 

[lfdahl]

Spoiler

Ley us see how many powers of 7 are there in 2017!

2017 = 288 * 7 +1
288 = 41 * 7 + 1
41 = 5 * 7 + 6
so 288 numbers are divisible by 7. 41 by 49 and 5 by 343 so 288 + 41 + 5 or 334

so 2017! is divisible by $7^{334}$

denominator

652 = 93 * 7 + 1
93 = 13 * 7 + 2
13 = 7* 1 + 6
so 652! is divisible by $7^{107}$

2017 - 652 = 1365

1365 = 195 * 7
195 = 27 * 7 + 6
27 = 3 * 7 + 6
so 1365! is divsible by $7^{225}$ and hence highest power of 7 that devides denominator of ${2017 \choose 652 }$ is 332 and highest power deviding numerator = 334 so $7^2$ devides the numbeer and not $7^3$ and the ans is no

Thankyou once again, kaliprasad for your clear cut solution, Nice!

Below is an alternative approach:

Spoiler

Solution:

No, the coefficient is not divisible by $343 = 7^3$.

This can be shown e.g. by means of Kummers theorem:

First, we express the entries in base $7$:

\[\binom{2017}{652}_{10} = \binom{1365+652}{652}=\binom{5611}{1621}_7=\binom{3660+1621}{1621}_7\]

Then, we add the two base $7$ numbers:\[\begin{matrix} \: \: \: \overset{1}{3} \overset{1}{6}60\\+1621\\ \: \: \: \: 5611 \end{matrix}\]The addition involves two carriers, hence by Kummers theorem
the highest power of $7$, that divides the binomial coefficient is $7^2$.