A number theory problem by Piyush Kumar

Number Theory Level 4

Find the total number of positive integers $n$ less than 17 for which $n!+(n+1)!+(n+2)!$ is an integral multiple of 49.

The answer is 5.

3 solutions

Piyush Kumar
Sep 4, 2014

the given expression equals to n!{1+(n+1)(n+2)}=n!(n+2)^2 either 7 divides n+2 or 49 divides n! therefore, n=5,12,14,15 and 16 hence total no.of integers= 5

An exactly duplicate model of your solution is there on my copy:p..

Anik Mandal - 6 years, 9 months ago

whaddya mean?

Krishna Ar - 6 years, 9 months ago

Just meant that I did the same thin'.

Anik Mandal - 6 years, 9 months ago

n!(1 + n+1 + (n+1)(n+2) )

= n!(1 + n+1 + n^2 + 3n + 2)

= n!(n^2 + 4n + 3) = n!(n+1)(n+3)

Then, 11, 13, 14, 15, 16 will be integral multiples of 49.

Therefore, the answer is 5. You forgot the "(n+1)!"

Kartik Sharma - 6 years, 9 months ago

Check it... u missed to add 1... we get: n!(n+2)^{2}

Saket Sharma - 6 years, 8 months ago

Edwin Gray
Jul 11, 2018

Construct the following table: 1!.cong.1(mod(49),2!.cong.2(mod(49),3!.cong.6(mod 49), 4!.cong.-25(mod 49),5!.cong.22(mod(49),6!.cong.-15(mod 49),7!.cong. -7(mod 49), 8!.cong. -7(mod(49), 9!.cong. -14 (mod 49), 10!.cong. 7 (mod 49), 11!.cong.28 (mod(49), 12! .cong. -7 (mod 49), 13!.cong. 7 (mod 49), 14! .cong. 0 (mod 49), 15! .cong. 0 (mod 49), 16! .cong. 0 (mod(49). We look for the sum of three consecutive residues to equal 0. This happens for n =5,12,14,15, and 16. So answer is 5. Ed Gray

Vincent Miller Moral
Jul 2, 2015

Troll ``` def fact( n ): ans = 1 while n: ans = ans*n n = n - 1 return ans

ctr = 0
for i in range(1,17): foo = fact(i) + fact(i+1) + fact(i+2) if foo%49 == 0: ctr = ctr + 1

print ctr
```

A number theory problem by Piyush Kumar

The answer is 5.

3 solutions

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