lfdahl said:Is the binomial coefficient: ${2017 \choose 652 }$ divisible by $343$?
Please prove your statement.
Thankyou once again, kaliprasad for your clear cut solution, Nice!kaliprasad said: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
To prove if ${2017 \choose 652}$ is divisible by 343, we need to use the binomial theorem and the properties of divisibility. We can expand the binomial coefficient and then use modular arithmetic to determine if the result is divisible by 343.
The value of ${2017 \choose 652}$ is a large number, approximately 2.31 x 10^468. This can be calculated using the formula ${n \choose k} = \frac{n!}{k!(n-k)!}$ where n is the total number of items and k is the number of items chosen.
Yes, we can use the property that ${n \choose k} = {n \choose n-k}$ to simplify ${2017 \choose 652}$ to ${2017 \choose 1365}$, which can make it easier to determine divisibility by 343.
343 is a prime number and also a perfect power of 7. This means that it cannot be broken down into smaller factors, making it a useful number for determining divisibility.
Yes, we can use the divisibility rule for 7 which states that if the alternate sum of the digits is divisible by 7, then the number is divisible by 7. Since 343 has 3 digits, we can group the digits of ${2017 \choose 652}$ into groups of 3 and apply this rule to determine if it is divisible by 343.